Acid-Base Equilibrium

Weak or strong, its got a pH going on.

Key concepts:

• Acids react with water to produce hydronium and conjugate base
• Bases react with water to produce hydroxide and conjugate acid
• Acids and bases react with each other (neutralization reaction)

Note: Equilibrium values are given at 25 °C in water unless otherwise stated.

Terminology

We define the following notation for acids and bases

• Acid: HA, HB⁺
• Base: B, A^–

There are different definitions for acids and bases.

Arrhenius Acid/Base

An Arrhenius acid is is a substance that dissociates in water to produce H⁺ ions.

\[\color{green}{\mathrm{HCl}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \color{red}{\mathrm{Cl^-}}(aq)\]

or generally

\[\color{green}{\mathrm{HA}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \color{red}{\mathrm{A^-}}(aq)\]

An Arrhenius base is is a substance that dissociates in water to produce OH^– ions.

\[\color{red}{\mathrm{NH_3}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^-} + \color{green}{\mathrm{NH_4^+}}(aq)\]

or generally

\[\color{red}{\mathrm{B}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^-}(aq) + \color{green}{\mathrm{HB^+}}(aq)\]

Bronsted-Lowry Acid/Base

A Brønsted-Lowry acid is a substance that donates a proton (i.e. is a proton donor).

\[\color{green}{\mathrm{HCl}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \color{red}{\mathrm{Cl^-}}(aq)\]

Here, HCl, an acid, donates its proton to water. A general scheme can be written as

\[\color{green}{\mathrm{HA}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \color{red}{\mathrm{A^-}}(aq)\]

A Brønsted-Lowry base is a substance that accepts a proton (i.e. is a proton acceptor).

\[\color{red}{\mathrm{NH_3}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^-}(aq) + \color{green}{\mathrm{NH_4^+}}(aq)\]

Here, NH₃, an base, accepts a proton from water. A general scheme can be written as

\[\color{red}{\mathrm{B}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^-}(aq) + \color{green}{\mathrm{HB^+}}(aq)\] or

\[\color{red}{\mathrm{A^-}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^-}(aq) + \color{green}{\mathrm{HA}}(aq)\]

Lewis Acid/Base

A Lewis acid is a substance that accepts a pair of electrons (i.e. a lone-pair acceptor).

\[\color{green}{\mathrm{HCl}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \color{red}{\mathrm{Cl^-}}(aq)\]

Here, HCl dissociates into H⁺ and Cl^–. The oxygen on water donates a lone pair of electrons to the free proton and forms a bond to give the hydronium ion, H₃O⁺.

Lewis Acid Example

Water behaves as a Lewis base by donating an electron pair to HCl. HCl behaves as a Lewis acid by accepting the lone electron pair from water.

A Lewis base is a substance that donates a lone pair of electrons (i.e. lone-pair donor).

\[\color{red}{\mathrm{NH_3}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^–}(aq) + \color{green}{\mathrm{NH_4^+}}(aq)\]

Here, the lone pair on the nitrogen in ammonia, NH₃, is donated to a proton on water forming a bond to give NH₄⁺.

Lewis Base Example

Water behaves as a Lewis acid by accepting the lone electron pair from ammonia. Ammonia behaves as a Lewis base by donating an electron pair to water.

In this example, ammonia forms a Lewis adduct with boron triflouride. No proton is transferred in this addition reaction. The N–B covalent bond is a dative bond, one that forms from an electron pair from one species.

Practice

Determine the type of acid (or base) that HCN is behaving as. Which definition(s) does it comply with?

\[\mathrm{HCN}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{CN^-}(aq)\]

Answer

1. Arrhenius acid - HCN produces hydronium when dissolved in water
2. Brønsted-Lowry acid - HCN donates a proton to water
3. Lewis acid - HCN accepts a lone electron pair from water

Conjugate Acid/Base

The word conjugate means “joined together” (e.g. coupled) especially “in pairs.” In acid/base chemistry, a conjugate is a pairing of reactant and product. For example, a deprotonated acid, A^–), is the conjugate base of the starting acid, HA.

\[\begin{align*} \color{green}{\mathrm{HA}} ~~~ &\rightleftharpoons ~~~~~ \mathrm{H^+} &&+~~~~~~~~ \color{red}{\mathrm{A^-}}\\[1.5ex] \mathrm{acid} ~~~ &\rightleftharpoons ~ \mathrm{proton} &&+~ \mathrm{conjugate~base} \end{align*}\]

An acid can also be generalized as HB⁺.

\[\begin{align*} \color{green}{\mathrm{HB^+}} ~~~ &\rightleftharpoons ~~~~~ \mathrm{H^+} &&+~~~~~~~~ \color{red}{\mathrm{B}}\\[1.5ex] \mathrm{acid} ~~~~~ &\rightleftharpoons ~ \mathrm{proton} &&+~ \mathrm{conjugate~base} \end{align*}\]

To determine the conjugate base of any acid, simply deprotonate the acid (remove an H⁺). The resulting species is the conjugate base.

A conjugate acid is the opposite of a conjugate base. A conjugate acid can be generally written as HB⁺ or HA.

\[\begin{align*} \color{red}{\mathrm{B}} &~~+~~~ \mathrm{H^+} &&\rightleftharpoons ~~~~~~~~~ \color{green}{\mathrm{HB^+}} \\[1.5ex] \mathrm{base} &~~+~ \mathrm{proton} &&\rightleftharpoons ~~~\mathrm{conjugate~acid} \end{align*}\]

\[\begin{align*} \color{red}{\mathrm{A^-}} &~~+~~~~~ \mathrm{H^+} &&\rightleftharpoons ~~~~~~~~~ \color{green}{\mathrm{HA}} \\[1.5ex] \mathrm{base} &~~+~ \mathrm{proton} &&\rightleftharpoons ~~~\mathrm{conjugate~acid} \end{align*}\]

To determine the conjugate acid of any base, simply protonate the base (add an H⁺). The resulting species is the conjugate acid.

Practice

Identify the conjugate (and type) to each of the following:

1. HNO₃ (acid)
2. NH₄⁺ (acid)
3. OH^– (base)

Answer

1. NO₃^–   ;   conjugate base
2. NH₃   ;   conjugate base
3. H₂O   ;   conjugate acid

Amphoterism

An amphoteric substance is one that can behave both as an acid and a base depending on the environment it is in. An amphiprotic substance is a type of amphoteric substance and is one that can accept or donate a proton, H⁺.

One fundamental example is water, an amphiprotic substance. When an acid is added to water, water accepts a proton (and behaves as a base). When a base is added to water, water donates a proton (and behaves as an acid).

Aluminum hydroxide is an example of an amphoteric substance. When reacting with an acid, such as HCl, Al(OH)₃ acts as a Lewis base.

\[\mathrm{Al(OH)_3}(aq) + \mathrm{3HCl}(aq) \rightleftharpoons \mathrm{AlCl_3}(aq) + \mathrm{3H_2O}(l)\]

Al(OH)₃ behaves as a Lewis acid when reacting with NaOH (molecular equation given below).

\[\mathrm{Al(OH)_3}(aq) + \mathrm{NaOH}(aq) \rightleftharpoons \mathrm{Na[Al(OH)_4]}(aq)\]

We could write the above reaction in its net ionic form as

\[\mathrm{Al(OH)_3}(aq) + \mathrm{OH^-}(aq) \rightleftharpoons \mathrm{Al(OH)_4^-}(aq)\]

Acid-Base Ionization Reactions

The acid-ionziation reaction is defined as an acid reacting with a solvent such as water.

\[\color{green}{\mathrm{HA}}(aq) + \mathrm{H_2O}(aq) \rightleftharpoons \mathrm{H_3O^+} + \color{red}{\mathrm{A^-}}\]

and has the equilibrium expression

\[K_{\mathrm{a}}=\dfrac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\color{red}{\mathrm{A}^{-}}\right]}{[\color{green}{\mathrm{HA}}]}\]

A strong acid has a K_a that is large (generally 1 or greater) whereas a weak acid has a K_a that is less than 1.

The base-ionization reaction is defined as a base reacting with a solvent such as water.

\[\color{red}{\mathrm{B}}(aq) + \mathrm{H_2O}(aq) \rightleftharpoons \mathrm{OH^–} + \color{green}{\mathrm{HB^+}}\]

and has the equilibrium expression

\[K_{\mathrm{b}}=\dfrac{\left[\mathrm{OH}^{-}\right]\left[\color{green}{\mathrm{HB}}^{+}\right]}{[\color{red}{\mathrm{B}}]}\]

Alternatively, we can express the reaction as

\[\color{red}{\mathrm{A^-}}(aq) + \mathrm{H_2O}(aq) \rightleftharpoons \mathrm{OH^–} + \color{green}{\mathrm{HA}}\]

\[K_{\mathrm{b}}=\dfrac{\left[\mathrm{OH}^{-}\right]\left[\color{green}{\mathrm{HA}}\right]}{[\color{red}{\mathrm{A^-}}]}\] A strong base has a K_b that is large (generally 1 or greater) whereas a weak base has a K_b that is less than 1.

Acid-Base Relationships

Below is a table of some acids and their conjugate bases.

Acid		Ka	pKa	C. Base	Kb	pKb
HCl	hydrochloric acid	2 × 106	–6.3	Cl–	5 × 10–21	20.30
H2SO4	sulfuric acid	1 × 103	–3.00	HSO4–	1 × 10–17	17.00
HNO3	nitric acid	2 × 101	–1.30	NO3–	5 × 10–16	15.30
H3O+	hydronium ion	1	0	H2O	1 × 10–14	14.00
HF	hydrofluoric acid	6.76 × 10–4	3.17	F–	1.48 × 10–11	10.83
HNO2	nitrous acid	5.13 × 10–4	3.29	NO2–	1.95 × 10–11	10.71
CH3COOH	acetic acid	1.75 × 10–5	4.76	CH3COO–	5.71 × 10–10	9.24
H2CO3	carbonic acid	4.26 × 10–7	6.37	HCO3–	2.35 × 10–8	7.63
NH4+	ammonium	5.62 × 10–10	9.25	NH3	1.78 × 10–5	4.75
H2O	water	1 × 10–14	14	OH–	1	0
NH3	ammonia	1 × 10–37	37	NH2–	1 × 1023	–23.00

Equations

K_a and pK_a

\[\begin{align*} \mathrm{p}K_{\mathrm{a}} &= -\log K_{\mathrm{a}} \\[1.5ex] K_{\mathrm{a}} &= 10^{-\mathrm{p}K_{\mathrm{a}}} \end{align*}\]

K_b and pK_b

\[\begin{align*} \mathrm{p}K_{\mathrm{b}} &= -\log K_{\mathrm{b}} \\[1.5ex] K_{\mathrm{b}} &= 10^{-\mathrm{p}K_{\mathrm{b}}} \end{align*}\]

pK_a, pK_b, and pK_w (K_a, K_b, and K_w)

\[\begin{align*} \mathrm{p}K_{\mathrm{a}} + \mathrm{p}K_{\mathrm{b}} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] K_{\mathrm{a}} \times K_{\mathrm{b}} &= K_{\mathrm{w}} \end{align*}\]

pH, pOH, and pK_w

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \end{align*}\]

Practice

Formic acid has a K_a = 1.9 × 10^–4. What is the pK_a of the acid?

Answer

\[\begin{align*} \mathrm{p}K_{\mathrm{a}} &= -\log K_{\mathrm{a}} \\[1.5ex] &= -\log (1.9\times 10^{-4}) \\[1.5ex] &= 3.72 \end{align*}\]

Practice

Hydrosulfuric acid (H₂S) has a pK_a of 7.0. What is the K_a of the acid?

Answer

\[\begin{align*} K_{\mathrm{a}} &= 10^{-\mathrm{p}K_{\mathrm{a}}} \\[1.5ex] &= 10^{-7} \\[1.5ex] &= 1\times 10^{-7} \end{align*}\]

Practice

Rank the acids from strongest to weakest.

Acid	Ka	pKa
ammonium ion	5.65 × 10-10
formic acid		3.72
acetic acid	1.76 × 10-5
benzoic acid		4.19

Answer

Convert everything to K_a or pK_a. Higher K_a values correspond to lower pK_a values and correspond to stronger acids.

Acid	Ka	pKa
formic acid	1.9 × 10-4	3.72
benzoic acid	6.46 × 10-5	4.19
acetic acid	1.76 × 10-5	4.75
ammonium ion	5.65 × 10-10	9.25

Strong Acids and Bases

A list of strong acids and bases are given below. Strong acids generally have a pK_a of around –1 or lower. Strong bases generally have a pK_b of –1 or lower.

Strong Acid	Formula	Strong Base	Formula
Hydrochloric acid	HCl	Lithium hydroxide	LiOH
Hydrobromic acid	HBr	Sodium hydroxide	NaOH
Hydroiodic acid	HI	Potassium hydroxide	KOH
Perchloric acid	HClO4	Rubidium hydroxide	RbOH
Chloric	HClO3	Cesium hydroxide	CsOH
Sulfuric	H2SO4	Magnesium hydroxide	Mg(OH)2
Nitric	HNO3	Calcium hydroxide	Ca(OH)2
		Strontium hydroxide	Sr(OH)2
		Barium hydroxide	Ba(OH)2

Auto-ionization of Water

Water naturally auto-ionizes to produce equal concentrations of hydronium and hydroxide. The extent of this reaction is very small.

\[\mathrm{2H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH}^-(aq)\]

\[K_{\mathrm{w}} = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0\times 10^{-14} ~~~(\mathrm{at}~25~^{\circ}\mathrm{C})\]

Note that water is amphiprotic in this reaction. One water donates a proton while another water accepts the proton.

The relationship of the strength of an acid to its conjugate base is related through K_w.

\[K_{\mathrm{a}} \times K_{\mathrm{b}} = K_{\mathrm{w}}\]

or in log form

\[\mathrm{p}K_{\mathrm{a}} + \mathrm{p}K_{\mathrm{b}} = \mathrm{p}K_{\mathrm{w}}\]

We can see this relationship clearly by writing out the acid- and base-ionization reactions and adding them together.

\[\begin{align*} \color{green}{\mathrm{HA}}(aq) + \mathrm{H_2O}(aq) &\rightleftharpoons \mathrm{H_3O^+} + \color{red}{\mathrm{A^-}} \\ \color{red}{\mathrm{A^-}}(aq) + \mathrm{H_2O}(aq) &\rightleftharpoons \mathrm{OH^–} + \color{green}{\mathrm{HA}}\\[1.5ex] \mathrm{2H_2O}(l) &\rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH}^-(aq) \end{align*}\]

Recall that adding two equations together has an equilibrium constant that is the result of multiplying the original equilibrium constants together.

\[\begin{align*} K_{\mathrm{w}} &= K_{\mathrm{a}} \times K_{\mathrm{b}} \\[1.5ex] &=\dfrac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\color{red}{\mathrm{A}^{-}}\right]}{[\color{green}{\mathrm{HA}}]} \times \dfrac{\left[\color{green}{\mathrm{HA}}\right]\left[\mathrm{OH}^{-}\right]}{[\color{red}{\mathrm{A^-}}]}\\[2.5ex] &= [\mathrm{H_3O^+}][\mathrm{OH^-}] \end{align*}\]

Increasing the temperature of water slightly shifts K_w to the right.

T (°C)	Kw	pKw	pH	pOH
100	5.13 ×10–13	12.29	6.14	6.14
50	5.48 ×10–14	13.26	6.63	6.63
40	2.92 ×10–14	13.54	6.77	6.77
30	1.47 ×10–14	13.83	6.92	6.92
25	1.00 ×10–14	14.00	7.00	7.00
20	6.81 ×10–15	14.17	7.08	7.08
10	2.93 ×10–15	14.53	7.27	7.26
0	1.14 ×10–15	14.94	7.47	7.47

Notice that pK_w is equal to 14 at only one temperature… 25 °C. For all temperatures, pH is equal to pOH for pure water. This means that for any temperature, pure water is a neutral solution.

Endo or Exothermic?

Is the auto-ionization of water an endo or exothermic process? If water is heated, which way will the reaction go according to Le Chatelier’s Principle?

\[\mathrm{2H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH}^-(aq) \quad K_{\mathrm{w}} = [\mathrm{H_3O^+}][\mathrm{OH^-}]\]

According to the data, pK_w decreases as temperature increases. Recall that K_w and pK_w have an inverse relationship. If pK_w decreases, K_w is increasing (with increasing temperature). Therefore, we can conclude that this process is endothermic.

Practice

What is the K_b for the conjugate base of acetic acid at 50 °C?

K_a(CH₃COOH) = 1.633 × 10^–5 and K_w(H₂O) = 5.48 × 10^–14 (at 50 °C).

Answer

K_b(CH₃COO^– = 3.36 × 10^–9 at 50 °C.

\[\begin{align*} K_{\mathrm{a}} \times K_{\mathrm{b}} &= K_{\mathrm{w}} \\[1.5ex] K_{\mathrm{b}} &= \dfrac{K_{\mathrm{w}}}{K_{\mathrm{a}}} \\[1.5ex] &= \dfrac{5.48\times 10^{-14}}{1.633\times 10^{-5}} \\[1.5ex] &= 3.36\times 10{-9} \end{align*}\]

pH and pOH

The pH of an aqueous solution is a measure of the hydronium ion concentration.

\[\mathrm{pH} = -\log [\mathrm{H_3O^+}]\] and

\[[\mathrm{H_3O}^+] = 10^{-\mathrm{pH}}\]

The pOH is a measure of the hydroxide ion concentration

\[\mathrm{pOH} = -\log [\mathrm{OH^-}]\] and

\[[\mathrm{OH}^-] = 10^{-\mathrm{pOH}}\]

The pH and pOH of solution is related through pK_w such that

\[\mathrm{p}K_{\mathrm{w}} = \mathrm{pH} + \mathrm{pOH}\]

Acidic, basic, and neutral solutions are defined by the relative amounts of H₃O⁺ and OH^– in the solution.

Solution		at 25 °C
Acidic	[H3O+] > [OH–]	pH < 7, pOH > 7
Basic	[H3O+] < [OH–]	pH > 7, pOH < 7
Neutral	[H3O+] = [OH–]	pH = 7, pOH = 7

The image below illustrates a wide variety of substances across the pH scale of 0 to 14.

Given that pH is a log scale of hydronium ion concentration, it is important to understand that a small change in pH results in a large change in acidity.

Practice

The pH of an aqueous acidic solution is 3.45 at 25 °C. What is the pOH?

Answer

The pK_w of water is 14 at 25 °C. Therefore,

\[\begin{align*} \mathrm{p}K_{\mathrm{w}} &= \mathrm{pH} + \mathrm{pOH} \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14.00 - 3.45 \\[1.5ex] &= 10.55 \end{align*}\]

Practice

The pH of an aqueous acidic solution is 3.45 at 50 °C. What is the pOH?

Answer

The pK_w of water is 13.26 at 50 °C. Therefore,

\[\begin{align*} \mathrm{p}K_{\mathrm{w}} &= \mathrm{pH} + \mathrm{pOH} \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 13.26 - 3.45 \\[1.5ex] &= 9.81 \end{align*}\]

Practice

The pH of pure water at 50 °C is 6.63. Is the water acidic, basic, or neutral?

Answer

Neutral. pH is equal to pOH.

Percent Ionization

The percent ionization of an acid can be determined using

\[\%~\mathrm{ionization} = \dfrac{[\mathrm{H_3O^+}]_{\mathrm {eq}}}{[\mathrm{HA}]_0} \times 100\%\]

This is a ratio of the hydronium ion concentration at equilibrium (can be directly determined from a pH measurement) and the original acid concentration before dissociation.

Clearly, strong acids and bases will have (essentially) 100% ionization whereas weak acids and bases will have a small percent ionization.

Practice

A 0.1 M acetic acid solution (CH₃COOH; K_a = 1.75 × 10^–5) has a pH of 2.88. What is the percent ionization of the acid?

Answer

Determine the hydronium ion concentration at equilibrium.

\[\begin{align*} [\mathrm{H_3O^+}] &= 10^{-\mathrm{pH}} \\[1.5ex] &= 10^{-2.88} \\[1.5ex] &= 1.31\times 10^{-3}~M \end{align*}\]

Determine the percent ionization of the acid.

\[\begin{align*} \%~\mathrm{ionization} &= \dfrac{[\mathrm{H_3O^+}]_{\mathrm {eq}}}{[\mathrm{HA}]_0} \times 100\% \\[1.5ex] &= \dfrac{1.31\times 10^{-3}~M}{0.1} \times 100\% \\[1.5ex] &= 1.31\% \end{align*}\]

pH or pOH of Acid/Base Aqueous Solutions

The pH of an acid or base solution can be determined by using the K_a, K_b, pK_a, or pK_b of the acid or base. Solve for the hydronium or hydroxide ion concentration via a typical equilibrium problem (and ICE table) and convert to pH or pOH.

Strong Acid

Example

What is the pH and pOH of a 0.1 M HCl aqueous solution (at 25 °C)?
K_a(HCl) = 2 × 10⁶

Full Version

Strong acids (essentially) dissociate completely to give

\[\mathrm{HA}(aq) \longrightarrow \mathrm{H^+}(aq) + \mathrm{A^-}(aq)\] or with water explicitly in the chemical equation

\[\mathrm{HA}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{H_3O^+}(aq) + \mathrm{A^-}(aq)\]

Therefore, the concentration of the strong acid, [HA], will equal the product concentrations at equilibrium.

\[[\mathrm{HA}]_{\mathrm{i}} = [\mathrm{H^+}]_{\mathrm{eq}}/[\mathrm{H_3O^+}]_{\mathrm{eq}} = [\mathrm{A^-}]_{\mathrm{eq}}\]

To demonstrate, I present the work for this problem below. It is a typical equilibrium problem. Note that water already contains some small amount of H₃O⁺ and OH^–. At 25 °C, these concentrations are equal to 1 × 10^–7 M.

	HCl(aq)	+	H2O(l)	⇌	H3O+(aq)	+	Cl–(aq)
I	0.1				1.00E-07		0
C	-x				+x		+x
E	0.1-x				1.00E-07 + x		x

Solve for x in the equilibrium expression.

\[\begin{align*} \dfrac{[\mathrm{H_3O^+}][\mathrm{Cl^-}]}{[\mathrm{HCl}]} &= K \\[1.5ex] \dfrac{(1.00\times 10^{-7} + x)(x)}{0.1-x} &= 2\times 10^{6} \\[1.5ex] 1.00\times 10^{-7}x + x^2 &= 2\times 10^{6} (0.1-x) \\[1.5ex] x^2 + 1.00\times 10^{-7}x &= 2\times 10^{5} - 2\times 10^{6}x\\[1.5ex] x^2 + (2\times 10^{6}x + 1\times 10^{-7}x) - 2\times 10^{5} &= 0 \\[1.5ex] x^2 + 2\times 10^{6}x - 2\times 10^{5} &= 0 \\[1.5ex] x &= 0.1 \end{align*}\]

See quadratic solution at WolframAlpha.

\[\begin{align*} [\mathrm{HCl}]_{\mathrm{eq}} &= 0~M \\ [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 0.1~M\\ [\mathrm{Cl^-}]_{\mathrm{eq}} &= 0.1~M \end{align*}\]

Notice that all the strong acid converted into products.

Solve for the pH and pOH. Note that pK_w = 14 at 25 °C.

\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\[1.5ex] &= -\log (0.1~M) \\[1.5ex] &= 1 \\[1.5ex] \end{align*}\]

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 -1 \\[1.5ex] &= 13 \end{align*}\]

Recognizing that a strong acid completely converts into hydronium

\[[\mathrm{HA}]_{\mathrm{i}} = [\mathrm{H_3O^+}]_{\mathrm{eq}}\] allows one to skip all the work and simply solve for the pH and pOH.

Short Version

Given that the initial H₃O⁺ concentration is very small compared to to the starting acid, we could approximate this concentration to be zero. This cleans up the math/work nicely. I will solve using this approximation to demonstrate.

	HCl(aq)	+	H2O(l)	⇌	H3O+(aq)	+	Cl–(aq)
I	0.1				~0		0
C	-x				+x		+x
E	0.1-x				x		x

Solve for x in the equilibrium expression.

\[\begin{align*} \dfrac{[\mathrm{H_3O^+}][\mathrm{Cl^-}]}{[\mathrm{HCl}]} &= K_{\mathrm{a}} \\[1.5ex] \dfrac{x^2}{0.1-x} &= 2\times 10^{6} \\[1.5ex] x^2 &= 2\times 10^{6} (0.1-x) \\[1.5ex] &= 2\times 10^{5} - 2\times 10^{6}x\\[1.5ex] x^2 + 2\times 10^{6}x - 2\times 10^{5} &= 0 \\[1.5ex] x &= 0.1 \end{align*}\]

See quadratic solution at WolframAlpha.

\[\begin{align*} [\mathrm{HCl}]_{\mathrm{eq}} &= 0~M \\ [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 0.1~M\\ [\mathrm{Cl^-}]_{\mathrm{eq}} &= 0.1~M \end{align*}\]

Notice that all the strong acid converted into products just as we solved before. This is a fantastic approximation to make.

Solve for the pH and pOH. Note that pK_w = 14 at 25 °C.

\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\[1.5ex] &= -\log (0.1~M) \\[1.5ex] &= 1 \\[1.5ex] \end{align*}\]

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 -1 \\[1.5ex] &= 13 \end{align*}\]

Recognizing that a strong acid completely converts into hydronium

\[[\mathrm{HA}]_{\mathrm{i}} = [\mathrm{H_3O^+}]_{\mathrm{eq}}\] allows one to skip all the work and simply solve for the pH and pOH.

Short Short Version

Recognize that all strong acid converts to hydronium and conjugate base:
0.1 M HCl → 0 M HCl, 0.1 M H₃O⁺, and 0.1 M Cl^–

\[\begin{align*} [\mathrm{HCl}]_{\mathrm{eq}} &= 0~M \\ [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 0.1~M\\ [\mathrm{Cl^-}]_{\mathrm{eq}} &= 0.1~M \end{align*}\]

Solve for the pH and pOH. Note that pK_w = 14 at 25 °C.

\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\[1.5ex] &= -\log (0.1~M) \\[1.5ex] &= 1 \\[1.5ex] \end{align*}\]

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 - 1 \\[1.5ex] &= 13 \end{align*}\]

Weak Acid

Example

What is the pH and pOH of a 0.1 M CH₃COOH aqueous solution (at 25 °C)?
K_a(CH₃COOH) = 1.75 × 10^–5

Full Version

Weak acids do not dissociate completely. Determine the hydronium ion concentration (via ICE table in an equilibrium problem) and find the pH and pOH.

	CH3COOH(aq)	+	H2O(l)	⇌	H3O+(aq)	+	CH3COO–(aq)
I	0.1				1.00E-07		0
C	-x				+x		+x
E	0.1-x				1.00E-07+x		x

\[\begin{align*} \dfrac{[\mathrm{H_3O^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} &= K_{\mathrm{a}} \\[1.5ex] \dfrac{(1\times 10^{-7} + x)(x)}{0.1-x} &= 1.75\times 10^{-5} \\[1.5ex] x^2 + 1\times 10^{-7}x &= 1.75\times 10^{-5}(0.1-x) \\[1.5ex] x^2 + 1\times 10^{-7}x &= 1.75\times 10^{-6} - 1.75\times 10^{-5}x \\[1.5ex] x^2 + (1\times 10^{-7}x + 1.75\times 10^{-5}x) - 1.75\times 10^{-6} &= 0 \\[1.5ex] x^2 + 1.76\times 10^{-5}x - 1.75\times 10^{-6} &= 0 \\[1.5ex] x &= 1.314\times 10^{-3} \end{align*}\]

See quadratic solution at WolframAlpha.

\[\begin{align*} [\mathrm{CH_3COOH}]_{\mathrm{eq}} &= 9.87\times 10^{-2}~M \\ [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 1.31\times 10^{-3}~M\\ [\mathrm{CH_3COO^-}]_{\mathrm{eq}} &= 1.31\times 10^{-3}~M \end{align*}\]

Solve for the pH and pOH. Note that pK_w = 14 at 25 °C.

\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\[1.5ex] &= -\log (1.31\times 10^{-3}~M) \\[1.5ex] &= 2.88 \\[1.5ex] \end{align*}\]

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 - 2.88 \\[1.5ex] &= 11.12 \end{align*}\]

Short Version

Weak acids do not dissociate completely. Determine the hydronium ion concentration (via ICE table in an equilibrium problem) and find the pH and pOH.

	CH3COOH(aq)	+	H2O(l)	⇌	H3O+(aq)	+	CH3COO–(aq)
I	0.1				~0		0
C	-x				+x		+x
E	0.1-x				x		x

\[\begin{align*} \dfrac{[\mathrm{H_3O^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} &= K_{\mathrm{a}} \\[1.5ex] \dfrac{x^2}{0.1-x} &= 1.75\times 10^{-5} \\[1.5ex] \dfrac{x^2}{0.1} &= 1.75\times 10^{-5} \\[1.5ex] x^2 &= 1.75\times 10^{-6} \\[1.5ex] x &= 1.32\times 10^{-3} \quad (4.18\%) \end{align*}\]

\[\begin{align*} [\mathrm{CH_3COOH}]_{\mathrm{eq}} &= 9.87\times 10^{-2}~M \\ [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 1.32\times 10^{-3}~M\\ [\mathrm{CH_3COO^-}]_{\mathrm{eq}} &= 1.32\times 10^{-3}~M \end{align*}\]

Solve for the pH and pOH. Note that pK_w = 14 at 25 °C.

\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\[1.5ex] &= -\log (1.32\times 10^{-3}~M) \\[1.5ex] &= 2.88 \\[1.5ex] \end{align*}\]

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 - 2.88 \\[1.5ex] &= 11.12 \end{align*}\]

Strong Base

Example

What is the pH and pOH of a 0.1 M NaOH aqueous solution (at 25 °C)?

Short Version

Note that NaOH is a soluble, basic salt. It should (essentially), dissociate completely in water.

\[\mathrm{NaOH}(s) \overset{\mathrm{H_2O}}\longrightarrow \mathrm{Na^+}(aq) + \mathrm{OH^-}(aq)\]

Therefore, the concentration of NaOH will equal the OH^– concentration.

\[[\mathrm{NaOH}]_{\mathrm{i}} = [\mathrm{OH^-}]_{\mathrm{eq}} = 0.1~M\]

Since the problem states that the concentration of NaOH is 0.1 M, that means that the hydroxide ion concentration will be 0.1 M (at equilibrium)! Of course, water contains a small amount of OH^– (1 × 10^–7 at 25 °C). However, 0.1 + 1 × 10^–7 ≈ 0.1. Therefore, we will ignore the small contribution of OH^– from water.

Note that our solution now contains H₂O(l), H₃O⁺(aq), OH^–(aq), and Na⁺(aq). However, Na⁺ will not be involved in the acid-base reaction between hydronium and hydroxide, so we will ignore it. The proper reaction is given in the ICE table below.

	H2O(l)	⇌	H3O+(aq)	+	OH–(aq)
I			~0		0.1
C			+x		+x
E			+x		0.1+x

Use K_w for pure water to determine the hydronium and hydroxide concentrations.

\[\begin{align*} [\mathrm{H_3O^+}][\mathrm{OH^-}] &= K_{\mathrm{w}} \\[1.5ex] x (0.1+x) &= 1\times 10^{-14} \\[1.5ex] x (0.1) &= 1\times 10^{-14} \\[1.5ex] x &= 1\times 10^{-14} (0.1) \\[1.5ex] x &= 1\times 10^{-15} \quad (1\times 10^{-12}\%) \end{align*}\]

\[\begin{align*} [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 1\times 10^{-15}~M \\ [\mathrm{OH^-}]_{\mathrm{eq}} &= 0.1 + 1\times 10^{-15}\\ &=\approx 0.1~M \end{align*}\]

Since we know that, at equilibrium, the hydroxide ion concentration is equal to the initial concentration of the strong base, we would skip the ICE table stuff. Therefore, we simply solve for the pOH of the solution and convert this to a pH.

Solve for the pH and pOH. Note that pK_w = 14 at 25 °C.

\[\begin{align*} \mathrm{pOH} &= -\log [\mathrm{OH^-}] \\[1.5ex] &= -\log (0.1~M) \\[1.5ex] &= 1 \\[1.5ex] \end{align*}\]

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH} \\[1.5ex] &= 14 - 1 \\[1.5ex] &= 13 \end{align*}\]

Weak Base

Example

What is the pH of a 0.1 M NH₃ aqueous solution (at 25 °C)?
K_b(NH₃) = 1.77 × 10^–5

Full Version

Weak bases do not dissociate completely. Determine the hydroxide ion concentration (via ICE table in an equilibrium problem) and find the pOH. We will account for the small amount of hydronium and hydroxide from water for this example.

	NH3(aq)	+	H2O(l)	⇌	OH–(aq)	+	NH4+(aq)
I	0.1				1.00E-07		0
C	-x				+x		+x
E	0.1-x				1.00E-07 + x		x

\[\begin{align*} \dfrac{[\mathrm{OH^-}][\mathrm{NH_4^+}]}{[\mathrm{NH_3}]} &= K_{\mathrm{b}} \\[1.5ex] \dfrac{(1\times 10^{-7} + x)(x)}{0.1-x} &= 1.77\times 10^{-5} \\[1.5ex] 1\times 10^{-7}x + x^2 &= 1.77\times 10^{-5}(0.1-x) \\[1.5ex] x^2 + 1\times 10^{-7}x &= 1.77\times 10^{-6} - 1.77\times 10^{-5}x \\[1.5ex] x^2 + (1\times 10^{-7}x + 1.77\times 10^{-5}x) - 1.77\times 10^{-6} &= 0 \\[1.5ex] x^2 + 1.78\times 10^{-5}x - 1.77\times 10^{-6} &= 0 \\[1.5ex] x &= 1.32\times 10^{-3} \end{align*}\]

See quadratic solution at WolframAlpha.

\[\begin{align*} [\mathrm{NH_3}]_{\mathrm{eq}} &= 9.87\times 10^{-2}~M \\ [\mathrm{OH^-}]_{\mathrm{eq}} &= 1.32\times 10^{-3}~M \\ [\mathrm{NH_4^+}]_{\mathrm{eq}} &= 1.32\times 10^{-3}~M \end{align*}\]

Solve for the pH and pOH. Note that pK_w = 14 at 25 °C.

\[\begin{align*} \mathrm{pOH} &= -\log [\mathrm{OH^-}] \\[1.5ex] &= -\log (1.32\times 10^{-3}~M) \\[1.5ex] &= 2.88 \\[1.5ex] \end{align*}\]

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH} \\[1.5ex] &= 14 - 2.88 \\[1.5ex] &= 11.12 \end{align*}\]

Short Version

In this version, we will approximate the hydroxide ion concentration to be zero.

	NH3(aq)	+	H2O(l)	⇌	OH–(aq)	+	NH4+(aq)
I	0.1				~0		0
C	-x				+x		+x
E	0.1-x				x		x

\[\begin{align*} \dfrac{[\mathrm{OH^-}][\mathrm{NH_4^+}]}{[\mathrm{NH_3}]} &= K \\[1.5ex] \dfrac{x^2}{0.1-x} &= 1.77\times 10^{-5} \\[1.5ex] \dfrac{x^2}{0.1} &= 1.77\times 10^{-5} \\[1.5ex] x^2 &= 1.77\times 10^{-6} \\[1.5ex] x &= 1.33\times 10^{-3} \quad (1.33\%) \end{align*}\]

\[\begin{align*} [\mathrm{NH_3}]_{\mathrm{eq}} &= 9.87\times 10^{-2}~M \\ [\mathrm{OH^-}]_{\mathrm{eq}} &= 1.33\times 10^{-3}~M\\ [\mathrm{NH_4^+}]_{\mathrm{eq}} &= 1.33\times 10^{-3}~M \end{align*}\]

Notice that the concentrations here are essentially the same as they were in the Full Version of the solution. The approximation works well.

Solve for the pH and pOH. Note that pK_w = 14 at 25 °C.

\[\begin{align*} \mathrm{pOH} &= -\log [\mathrm{OH^-}] \\[1.5ex] &= -\log (1.33\times 10^{-3}~M) \\[1.5ex] &= 2.88 \\[1.5ex] \end{align*}\]

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH} \\[1.5ex] &= 14 - 2.88 \\[1.5ex] &= 11.12 \end{align*}\]

Note how our pH and pOH matches exactly (to two decimal places) with the previous work done under the Full Version.

Oxyacids

Oxyacids (or oxoacids) are a class of inorganic acids containing oxygen, hydrogen, and another atom. There are many types of oxyacids.

Oxidation State

The strength of an oxyacid decreases with the oxidation state of the element. Below we see a series of oxyacids with the element chlorine. The acidic proton is given in red.

Name	Formula	Structure	Oxidation State	Ka	pKa
Perchloric acid	HClO4		7	1 × 108	–8
Chloric acid	HClO3		5	1 × 101	–1
Chlorous acid	HClO2		3	1 × 10-2	2.0
Hypochlorous acid	HClO		1	2.95 × 10-8	7.53

Notice that the oxidation state decreases for every oxygen that is removed. As the oxidation state of chlorine decreases, the pK_a increases (i.e. the strength of the acid decreases). To put another way, as each oxygen is removed, the ability of the acidic hydrogen to dissociate from the complex decreases.

Electronegativity of Central Atom

The electronegativity of the central atom also plays a role in the strength of the oxyacid. As the electronegativity increases, the acid strength increases. This is due to the ability for the central atom to pull surrounding electron density toward itself, thereby increasing the polarity of the O–H bond and making the acidic hydrogen more easily ionizable.

Below is a series of oxyacids with a central halogen atom. The oxidation state for the central atom is 1 for all of these. The electronegativity of the central atom is given.

Name	Formula	Structure	Electronegativity	Ka	pKa
Hypochlorous acid	HClO		3.16	4.0 × 10-8	7.40
Hypobromous acid	HBrO		2.96	2.8 × 10-9	8.55
Hypoiodous acid	HIO		2.66	3.2 × 10-11	10.5

Polyprotic oxyacids

Polyprotic oxyacids contain more than one acidic proton. The strength of the acid decreases with the removal of each successive proton. Consider sulfuric acid, H₂SO₄. There are two acidic protons. Removal of these protons, in sequence, gives

\[\begin{align*} \mathrm{H_2SO_4}(aq) + \mathrm{H_2O}(l) &\rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{HSO_4^-} &&\quad K_{\mathrm{a}} = 1.0\times 10^{3}\\ \mathrm{HSO_4^-}(aq) + \mathrm{H_2O}(l) &\rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{SO_4^{2-}} &&\quad K_{\mathrm{a}} = 1.2\times 10^{-2}\\ \end{align*}\]

Notice the dramatic decrease in the equilibrium constant as each proton is removed. H₂SO₄ is a much stronger acid than HSO₄^–. The data is tabulated below.

Name	Formula	Structure	Ka	pKa
Sulfuric acid	H2SO4		1.0 × 103	3.00
Hydrogen sulfate	HSO4–		1.2 × 10-2	1.92

The table below outlines the acid strength of phosphoric acid, H₃PO₄, and its deprotonated analogues.

Name	Formula	Structure	Ka	pKa
Phosphoric acid	H3PO4		7.1 × 10-3	2.15
Dihydrogen phosphate	H2PO4–		6.3 × 10-8	7.20
Hydrogen phosphate	HPO4–		4.2 × 10-13	12.38

Salt Hydrolysis

Hydrolysis is the reaction of a compound with water where water breaks one or more chemical bonds in the reacting compound. Salt hydrolysis is, therefore, the reaction of salt with water, where one or more bonds is broken in the salt. A familiar example of this is the dissolution of sodium chloride in water where the ionic bond is broken.

\[\mathrm{NaCl}(s) \overset{\mathrm{H_2O}}\longrightarrow \mathrm{Na^+}(aq) + \mathrm{Cl^-}(aq)\]

Some salts, when dissolved in water, may affect the pH of the solution. Acidic salts lower the pH of solution, basic salts raise the pH of solution, and neutral salts have no effect on the pH of solution.

Acidic Salts

An acidic salt is any salt that lowers the pH of a solution. Consider ammonium chloride, NH₄Cl. The hydrolysis of this salt is given as

\[\mathrm{NH_4Cl}(s) \overset{\mathrm{H_2O}}\longrightarrow \mathrm{NH_4^+}(aq) + \mathrm{Cl^-}(aq)\] Let us consider the reaction of each product of this reaction (i.e. the cation and anion of the salt) with water.

Cation

The ammonium ion, NH₄⁺, is a conjugate acid to a weak base. It is the conjugate acid to ammonia, NH₃. We can easily recognize this fact by simply deprotonating ammonium (removing a proton) which leaves us with the weak base, NH₃! Next, recall that weak bases give rise to stronger conjugate acids. Since this is the case here, the acid (NH₄⁺) will behave as such and react with water to produce hydronium!

\[\mathrm{NH_4^+}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{NH_3}(aq)\]

This is further understood by applying the fact that weak bases can exist in water. Here, ammonia, NH₃ is being produced and rightfully so. When this reaction occurs, the pH of the solution is lowered due to the production of H₃O⁺.

Anion

Next, consider the anion of the salt. Will the chloride ion react with water to produce HCl?

\[\mathrm{Cl^-}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^-}(aq) + \mathrm{HCl}(aq)\] The answer is no. Given that Cl^– is a conjugate base to a strong acid (HCl), it is too weak of a base to react with water to produce hydroxide. Also, HCl is a strong acid and essentially does not exist in water. Therefore, we conclude that this reaction cannot occur as HCl would not be generated in an environment where it cannot exist! Given this, hydroxide would not be produced and the pH is not affected by the anion (i.e. a conjugate base to a strong acid).

Conclusion: Ammonium chloride is an acidic salt.

Basic Salts

A basic salt is any salt that raises the pH of a solution. Consider potassium fluoride, KF. The hydrolysis of this salt is given as

\[\mathrm{KF}(s) \overset{\mathrm{H_2O}} \longrightarrow \mathrm{K^+}(aq) + \mathrm{F^-}(aq)\]

Let us consider the reaction of each product of this reaction (i.e. the cation and anion of the salt) with water.

Cation

The potassium ion, K⁺, is a Group 1A ion. It is quite stable given that it contains 8 valence electrons (its octet is satisfied). It is also an ion with positive charge and it would not be sensible to assume that it will readily accept a proton. Therefore, the following reaction will not occur.

\[\mathrm{K^+}(aq) + \mathrm{H_2O}(l) \nrightarrow \mathrm{OH^-}(aq) + \mathrm{HK}(aq)\]

Anion

Next, consider the anion of the salt. Will the fluoride ion react with water to produce HF?

\[\mathrm{F^-}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^-}(aq) + \mathrm{HF}(aq)\]

The answer is yes. Given that F^– is a conjugate base to a weak acid (HF), it is a strong enough base to react with water to produce hydroxide. Also, HF is a weak acid and can exist in water. This reaction can occur and given the production of hydroxide, the pH of the solution increases.

Conclusion: Potassium fluoride is a basic salt.

Neutral Salts

Neutral salts are those that do not affect the pH of solution. This means that neither cation or anion of the salt will react with water. Sodium chloride is an excellent example.

\[\mathrm{NaCl}(s) \overset{\mathrm{H_2O}}\longrightarrow \mathrm{Na^+}(aq) + \mathrm{Cl^-}(aq)\]

Cation

The sodium cation is positively charged and stable due to its octet being satisfied. It does not react with water to produce NaH and OH^–.

Anion

The chloride anion, as already discussed, is a conjugate base to a strong acid and is therefore too weak to react with water.

Conclusion: Salts containing Group 1A or heavy 2A ions bonded to ions that are conjugates to a strong acid/base are neutral salts.

Acidic Metal Hydrides

Acidic metal hydrides are salts that contain a highly charged metal that, when dissolved, coordinates with water molecules and results in a lower the pH of solution. Highly charged means the metal contains a 3+ (and sometimes 2+) charge. Take for example the soluble aluminum chloride salt, AlCl₃.

\[\mathrm{AlCl_3}(s) \overset{\mathrm{H_2O}}\longrightarrow \mathrm{Al^{3+}}(aq) + \mathrm{3Cl^-}(aq)\] The small, highly charged metal coordinates strongly with the partially positive negative oxygen atom in water. There are six coordination sites and, therefore, six coordinating waters in solution. The hydration process is shown below.

\[\mathrm{Al^{3+}}(aq) + \mathrm{6H_2O}(l) \longrightarrow \mathrm{[Al(H_2O)_6]^{3+}}(aq)\]

The product of this reaction is aluminum hexahydrate. This hydrate is a weak acid and can react with water to produce hydronium.

\[\mathrm{[Al(H_2O)_6]^{3+}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{[Al(H_2O)_5OH]^{2+}} \quad K_{\mathrm{a}} = 1.4\times 10^{-5}\]

Conclusion: Salts containing small, high charge metals are acidic salts.

The table below gives the acidities of some metal hydrates.

Hydrate	Formula	Ka	pKa
Iron hexahydrate	[Fe(H2O)6]3+	1.84 × 10-3	2.74
Chromium hexahydrate	[Cr(H2O)6]3+	1.6 × 10-4	3.80
Dimercury dihydrate	[Hg2(H2O)2]2+	1 × 10-4	4
Aluminum hexahydrate	[Al(H2O)6]3+	1.4 × 10-5	4.85
Beryllium tetrahydrate	[Be(H2O)4]2+	1 × 10 -5	5

Why Do Metal Hydrates Form?

Consider the valence electron configuration for the chromium atom.

To create Cr³⁺, the three highest energy electrons are removed, one electron from the 4s orbital and 2 from the 3d orbitals.

This leaves eight unoccupied orbitals in the valence region for chromium. Water has two lone eletron pairs on oxygen. A water can donate an electron pair to an empty Cr³⁺ orbital, forming at dative bond.

Since six orbitals are unoccupited on Cr³⁺, six waters can coordinate to the transition metal ion.

Summary

• Acidic salts contain ions that are conjugates to weak bases
• Basic salts contain ions that are conjugates to weak acids
• Group 1A and heavy 2A metal ions do not react with water
• Salts containing high charge metals are acidic salts due to becoming weak acid hydrates

Ion	Type	Examples
Acidic cation	Conjugate to weak base	NH4+, CH3NH3+
	High charge metal (not 1A or heavy 2A)	Al3+, Fe3+, Cr3+, Sc3+
Basic anion	Conjugate to weak acid	CN–, NO2–, CH3COO–
Neutral cations	Group 1A or heavy Group 2A	Li+, Na+, K+, Ba2+
Neutral anions	Conjugate to strong acids	Cl–, NO3–, ClO4–

Buffer

A buffer is a type of solution that can resist a change in pH.

A solution designed to maintain a constant pH when small amounts of a strong acid or base are added. Buffers usually consist of a fairly weak acid and its salt with a strong base. Suitable concentrations are chosen so that the pH of the solution remains close to the pK_a of the weak acid. – CRC 2016

It is able to do this due to the presence of appreciable and similar amounts of acid and conjugate base (or base and conjugate acid) in the same solution. Clearly, for an acid or a base to exist in water, the acid or base must be weak. Similar amounts means the ratio of acid to conjugate base (or base to conjugate acid) must be a 10:1 or 1:10 ratio.

pH of a Buffer Solution

The pH or pOH of a buffer solution can be determined using the Henderson-Hasselbalch equation.

For a buffer made from acid

\[\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{\color{red}{A^-}}]}{\mathrm{[\color{green}{HA}]}}\]

and a buffer made from a base

\[\mathrm{pOH} = \mathrm{p}K_{\mathrm{b}} + \log\dfrac{[\mathrm{\color{green}{HB^+}}]}{\mathrm{[\color{red}{B}]}}\]

Creating a Buffer

A buffer solution can be made from

1. A weak acid and a salt containing its conjugate base
2. A weak acid and a strong base
3. A weak base and a salt containing its conjugate acid
4. A weak base and a strong acid

But what about a…

…buffer from a Strong Acid or Base?

Given that a buffer must contain both acid and its conjugate base, a buffer cannot be created from a strong acid (since a strong acid nearly completely converts into hydronium).

Consider a 0.1 M HCl aqueous solution (at 25 °C).
K_a(HCl) = 2 × 10⁶

	HCl(aq)	+	H2O(l)	⇌	H3O+(aq)	+	Cl–(aq)
I	0.1				~0		0
C	-x				+x		+x
E	0.1-x				x		x

Solve for x in the equilibrium expression.

\[\begin{align*} \dfrac{[\mathrm{H_3O^+}][\mathrm{Cl^-}]}{[\mathrm{HCl}]} &= K_{\mathrm{a}} \\[1.5ex] \dfrac{x^2}{0.1-x} &= 2\times 10^{6} \\[1.5ex] x^2 &= 2\times 10^{6} (0.1-x) \\[1.5ex] &= 2\times 10^{5} - 2\times 10^{6}x\\[1.5ex] x^2 + 2\times 10^{6}x - 2\times 10^{5} &= 0 \\[1.5ex] x &= 0.1 \end{align*}\]

See quadratic solution at WolframAlpha.

The equilibrium concentrations are

\[\begin{align*} [\mathrm{HCl}]_{\mathrm{eq}} &= 0~M \\ [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 0.1~M\\ [\mathrm{Cl^-}]_{\mathrm{eq}} &= 0.1~M \end{align*}\]

The amount of strong acid, HCl, at equilibrium is zero. Therefore, the acid-to-base ratio is zero and the solution is not a buffer!

Buffers cannot be created from strong acids or strong bases!

…buffer from only a Weak Acid?

Let us consider the weak acid, acetic acid (CH₃COOH), and create a 0.01 M aqueous solution. The K_a = 1.75 × 10^–5. Solve for the equilibrium concentrations.

	CH3COOH(aq)	+	H2O(l)	⇌	H3O+(aq)	+	CH3COO–(aq)
I	0.01				~0		0
C	-x				+x		+x
E	0.01 - x				x		x

\[\begin{align*} \dfrac{[\mathrm{H_3O^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} &= K \\[1.5ex] \dfrac{x^2}{0.01-x} &= 1.75\times 10^{-5} \\[1.5ex] \dfrac{x^2}{0.01} &= 1.75\times 10^{-5} \\[1.5ex] x^2 &= 1.75\times 10^{-7} \\[1.5ex] x &= 4.18\times 10^{-4} \quad (4.18\%) \end{align*}\]

\[\begin{align*} [\mathrm{CH_3COOH}]_{\mathrm{eq}} &= 9.58\times 10^{-3} \\ [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 4.18\times 10^{-4}\\ [\mathrm{CH_3COO^-}]_{\mathrm{eq}} &= 4.18\times 10^{-4} \end{align*}\]

Our weak acid solution contains both weak acid (CH₃COOH) and conjugate base (CH₃COO^–). Let us determine the ratio and ensure it fits our definition of a buffer.

\[\dfrac{[\mathrm{base}]}{[\mathrm{acid}]} = \dfrac{[\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} = \dfrac{4.18\times 10^{-4}}{9.58\times 10^{-3}} = 0.04\] or

\[\dfrac{[\mathrm{acid}]}{[\mathrm{base}]} = \dfrac{[\mathrm{CH_3COOH}]}{[\mathrm{CH_3COO^-}]} = \dfrac{9.58\times 10^{-5}}{4.18\times 10^{-4}} = 22.9\]

The ratio of acid to base lies outside the range of 0.1 to 10. A 0.01 M CH₃COOH aqueous solution is not a buffer.

…buffer from only a Weak Base?

Let us consider the weak base, ammonia (NH₃), and create a 0.01 M aqueous solution. The K_b(NH₃) = 1.77 × 10^–5. Solve for the equilibrium concentrations.

	NH3(aq)	+	H2O(l)	⇌	OH–(aq)	+	NH4+(aq)
I	0.01				~0		0
C	-x				+x		+x
E	0.01 - x				x		x

\[\begin{align*} \dfrac{[\mathrm{OH^-}][\mathrm{NH_4^+}]}{[\mathrm{NH_3}]} &= K \\[1.5ex] \dfrac{x^2}{0.01-x} &= 1.77\times 10^{-5} \\[1.5ex] \dfrac{x^2}{0.01} &= 1.77\times 10^{-5} \\[1.5ex] x^2 &= 1.77\times 10^{-7} \\[1.5ex] x &= 4.21\times 10^{-4} \quad (4.21\%) \end{align*}\]

\[\begin{align*} [\mathrm{NH_3}]_{\mathrm{eq}} &= 1.96\times 10^{-2}~M \\ [\mathrm{OH^-}]_{\mathrm{eq}} &= 4.21\times 10^{-4}~M\\ [\mathrm{NH_4^+}]_{\mathrm{eq}} &= 4.21\times 10^{-4}~M \end{align*}\]

\[\dfrac{[\mathrm{acid}]}{[\mathrm{base}]} = \dfrac{[\mathrm{NH_4^+}]}{[\mathrm{NH_3}]} = \dfrac{4.21\times 10^{-4}}{1.96\times 10^{-2}} = 0.02\]

or

\[\dfrac{[\mathrm{base}]}{[\mathrm{acid}]} = \dfrac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]} = \dfrac{1.96\times 10^{-2}}{4.21\times 10^{-4}} = 46.5\]

A 0.01 M NH₃ aqueous solution is not a buffer.

Buffer from a Weak Acid and Salt

Consider a 0.01 M acetic acid (CH₃COOH) aqueous solution containing 0.01 M sodium acetate (NaCH₃COO). K_a = 1.75 × 10^–5

This solution contains a weak acid and a salt containing the conjugate base to the weak acid. Determine the equilibrium concentrations of the weak acid and conjugate base.

Work

Full Version

	CH3COOH(aq)	+	H2O(l)	⇌	H3O+(aq)	+	CH3COO–(aq)
I	0.01				~0		0.01
C	-x				+x		+x
E	0.01 - x				x		0.01+x

\[\begin{align*} \dfrac{[\mathrm{H_3O^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} &= K \\[1.5ex] \dfrac{x(0.01+x)}{0.01-x} &= 1.75\times 10^{-5} \\[1.5ex] \dfrac{0.01x}{0.01} &= 1.75\times 10^{-5} \\[1.5ex] 0.01x &= 1.75\times 10^{-7} \\[1.5ex] x &= 1.75\times 10^{-5} \quad (0.18\%) \end{align*}\]

\[\begin{align*} [\mathrm{CH_3COOH}]_{\mathrm{eq}} &= 9.98\times 10^{-3} \\ [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 1.75\times 10^{-5}\\ [\mathrm{CH_3COO^-}]_{\mathrm{eq}} &= 1.00\times 10^{-2} \end{align*}\]

The ratio of acid to base is

\[\dfrac{[\mathrm{base}]}{[\mathrm{acid}]} = \dfrac{[\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} = \dfrac{1.00\times 10^{-2}}{9.98\times 10^{-3}} = 1.002\] or

\[\dfrac{[\mathrm{acid}]}{[\mathrm{base}]} = \dfrac{[\mathrm{CH_3COOH}]}{[\mathrm{CH_3COO^-}]} = \dfrac{9.98\times 10^{-3}}{1.00\times 10^{-2}} = 0.998\]

The ratio lies between the range of 0.1 to 10. Therefore, this solution is a buffer!

Use Henderson-Hasselbalch (for a buffer made from an acid) to determine the pH.

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{\color{red}{A^-}}]}{\mathrm{[\color{green}{HA}]}} \\[1.5ex] &= -\log (1.75\times 10^{-5}) + \log\dfrac{1.00\times 10^{-2}~M}{9.98\times 10^{-3}~M}\\[1.5ex] &= 4.76 \end{align*}\]

Short Version

Skip the ICE table and simply approximate the equilibrium concentrations of the weak acid, HA, and conjugate base, A^– to both be 0.01 M.

The ratio of acid to conjugate base lies between the range of 0.1 to 10 (it is exactly 1 in this example). Therefore, this solution is a buffer!

Use Henderson-Hasselbalch (for a buffer made from an acid) to determine the pH.

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{\color{red}{A^-}}]}{\mathrm{[\color{green}{HA}]}} \\[1.5ex] &= -\log (1.75\times 10^{-5}) + \log\dfrac{0.01~M}{0.01~M}\\[1.5ex] &= 4.76 \end{align*}\]

Buffer from a Weak Acid and Strong Base

Adding some strong base to a weak acid solution can create a buffer because an acid-base neutralization reaction will take place to consume weak acid and produce conjugate base.

\[\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{A^-}(aq)\]

Consider a 50.0 mL 0.01 M acetic acid (CH₃COOH; K_a = 1.75 × 10^–5) aqueous solution. Add 10.00 mL 0.01 M NaOH to the solution. The solution is a buffer!

Note: This type of problem introduces a new component to the procedure for solving. Analyze carefully.

Work

To solve this type of problem, we will first consider the acid-base neutralization reaction that is to take place. A strong base, OH^– (from NaOH), is added to a weak acid solution. The added base will react with the weak acid in solution such that

\[\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{A^-}(aq)\]

We treat this reaction as a limiting reactant type problem. Therefore, we must convert everything to moles and carry out the reaction.

\[\begin{align*} n_{\mathrm{HA}} &= (0.050~\mathrm{L})\left(0.01~\mathrm{mol~L}^{-1}\right)\\ &= 0.0005~\mathrm{mol~HA} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.01~\mathrm{L})\left(0.01~\mathrm{mol~L}^{-1}\right)\\ &= 0.0001~\mathrm{mol~OH}^{-} \end{align*}\]

Next, write out an IRF table (I = initial moles, R = reaction, F = final moles) and run the reaction to completion.

I	0.0005		0.0001				0
R	HA	+	OH–	→	H2O	+	A–
F	0.0004		0				0.0001

Here, the strong base is the limiting reactant.

Next, convert everything back to molarity. Note that the volume of the solution has increased since we added some strong base (10.0 mL) to the starting solution (50.0 mL)!

\[\begin{align*} V_{\mathrm{final}} &= 50.0~\mathrm{mL} + 10.0~\mathrm{mL} \\ &= 60.0~\mathrm{mL} \\ &= 0.0600~\mathrm{L} \\[2ex] [\mathrm{HA}] &= \dfrac{0.0004~\mathrm{mol}}{0.0600~\mathrm{L}}\\[1.5ex] &= 0.00667~M \\[1.5ex] [\mathrm{A^-}] &= \dfrac{0.0001~\mathrm{mol}}{0.0600~\mathrm{L}}\\[1.5ex] &= 0.00167~M \end{align*}\]

We can stop here and determine if we have a buffer. The ratio of acid to base is

\[\dfrac{[\mathrm{base}]}{[\mathrm{acid}]} = \dfrac{[\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} = \dfrac{1.67\times 10^{-3}}{6.67\times 10^{-3}} = 0.250\] or

\[\dfrac{[\mathrm{acid}]}{[\mathrm{base}]} = \dfrac{[\mathrm{CH_3COOH}]}{[\mathrm{CH_3COO^-}]} = \dfrac{6.67\times 10^{-3}}{1.67\times 10^{-3}} = 3.99\] Since the ratio of acid to base lies between 0.1 and 10, this solution is a buffer!

Use Henderson-Hasselbalch (for a buffer made from an acid) to determine the pH.

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{\color{red}{A^-}}]}{\mathrm{[\color{green}{HA}]}} \\[1.5ex] &= -\log (1.75\times 10^{-5}) + \log\dfrac{0.00167~M}{0.00667~M}\\[1.5ex] &= 4.16 \end{align*}\]

Buffer from a Weak Base and Salt

Consider a 0.01 M ammonia (NH₃; K_b(NH₃) = 1.77 × 10^–5) aqueous solution containing 0.01 M ammonium chloride (NH₄Cl).

This solution contains a weak base and a salt containing the conjugate acid to the weak base Determine the equilibrium concentrations of the weak base and conjugate acid

Work

Long Version

	NH3(aq)	+	H2O(l)	⇌	OH–(aq)	+	NH4+(aq)
I	0.01				~0		0.01
C	-x				+x		+x
E	0.01 - x				x		0.01+x

\[\begin{align*} \dfrac{[\mathrm{OH^-}][\mathrm{NH_4^+}]}{[\mathrm{NH_3}]} &= K \\[1.5ex] \dfrac{x(0.01+x)}{0.01-x} &= 1.77\times 10^{-5} \\[1.5ex] \dfrac{0.01x}{0.01} &= 1.77\times 10^{-5} \\[1.5ex] 0.01x &= 1.77\times 10^{-7} \\[1.5ex] x &= 1.77\times 10^{-5} \quad (0.18\%) \end{align*}\]

\[\begin{align*} [\mathrm{NH_3}]_{\mathrm{eq}} &= 9.98\times 10^{-3}~M \\ [\mathrm{OH^-}]_{\mathrm{eq}} &= 1.77\times 10^{-5}~M\\ [\mathrm{NH_4^+}]_{\mathrm{eq}} &= 1.00\times 10^{-2}~M \end{align*}\]

\[\dfrac{[\mathrm{acid}]}{[\mathrm{base}]} = \dfrac{[\mathrm{NH_4^+}]}{[\mathrm{NH_3}]} = \dfrac{1.00\times 10^{-2}}{9.98\times 10^{-3}} = 1.00\]

or

\[\dfrac{[\mathrm{base}]}{[\mathrm{acid}]} = \dfrac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]} = \dfrac{9.98\times 10^{-3}}{1.00\times 10^{-2}} = 0.998\]

This solution is a buffer!

Use Henderson-Hasselbalch (for a buffer made from an base) to determine the pOH.

\[\begin{align*} \mathrm{pOH} &= \mathrm{p}K_{\mathrm{b}} + \log\dfrac{[\mathrm{\color{red}{HB^+}}]}{\mathrm{[\color{green}{B}]}} \\[1.5ex] &= -\log (1.77\times 10^{-5}) + \log\dfrac{1.00\times 10^{-2}~M}{9.98\times 10^{-3}~M}\\[1.5ex] &= 4.75 \end{align*}\]

Find the pH (at 25 °C).

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH} \\[1.5ex] &= 14 - 4.75 \\[1.5ex] &= 9.25 \end{align*}\]

Short Version

Skip the ICE table and simply approximate the equilibrium concentrations of the weak acid, B, and conjugate base, HB⁺ to both be 0.01 M.

The ratio of base to conjugate acid lies between the range of 0.1 to 10 (it is exactly 1 in this example). Therefore, this solution is a buffer!

Use Henderson-Hasselbalch (for a buffer made from an base) to determine the pH.

\[\begin{align*} \mathrm{pOH} &= \mathrm{p}K_{\mathrm{b}} + \log\dfrac{[\mathrm{\color{red}{HB^+}}]}{\mathrm{[\color{green}{B}]}} \\[1.5ex] &= -\log (1.77\times 10^{-5}) + \log\dfrac{0.01~M}{0.01~M}\\[1.5ex] &= 4.75 \end{align*}\]

Find the pH (at 25 °C).

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH} \\[1.5ex] &= 14 - 4.76 \\[1.5ex] &= 9.25 \end{align*}\]

Buffer from a Weak Base and Strong Acid

Adding some strong acid to a weak base solution can create a buffer because an acid-base neutralization reaction will take place to consume weak base and produce conjugate acid.

\[\mathrm{B}(aq) + \mathrm{H_3O^+}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{HB^+}(aq)\]

Consider a 50.0 mL 0.01 M ammonia (NH₃; K_b = 1.77 × 10^–5) aqueous solution. Add 10.00 mL 0.01 M HCl to the solution. The solution is a buffer!

Note: This type of problem introduces a new component to the procedure for solving. Analyze carefully.

Work

To solve this type of problem, we will first consider the acid-base neutralization reaction that is to take place. A strong base, H₃O⁺ (from HCl), is added to a weak base solution. The added acid will react with the weak base in solution such that

\[\mathrm{B}(aq) + \mathrm{H_3O^+}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{HB^+}(aq)\]

We treat this reaction as a limiting reactant type problem. Therefore, we must convert everything to moles and carry out the reaction.

\[\begin{align*} n_{\mathrm{B}} &= (0.050~\mathrm{L})\left(0.01~\mathrm{mol~L}^{-1}\right)\\ &= 0.0005~\mathrm{mol~B} \\[2ex] n_{\mathrm{H_3O}^{+}} &= (0.01~\mathrm{L})\left(0.01~\mathrm{mol~L}^{-1}\right)\\ &= 0.0001~\mathrm{mol~H_3O}^{+} \end{align*}\]

Next, write out an IRF table (I = initial moles, R = reaction, F = final moles) and run the reaction to completion.

I	0.0005		0.0001				0
R	B	+	H3O+	→	H2O	+	HB+
F	0.0004		0				0.0001

Here, the strong acid is the limiting reactant.

Next, convert everything back to molarity. Note that the volume of the solution has increased since we added some strong base (10.0 mL) to the starting solution (50.0 mL)!

\[\begin{align*} V_{\mathrm{final}} &= 50.0~\mathrm{mL} + 10.0~\mathrm{mL} \\ &= 60.0~\mathrm{mL} \\ &= 0.0600~\mathrm{L} \\[2ex] [\mathrm{B}] &= \dfrac{0.0004~\mathrm{mol}}{0.0600~\mathrm{L}}\\[1.5ex] &= 0.00667~M \\[1.5ex] [\mathrm{HB^+}] &= \dfrac{0.0001~\mathrm{mol}}{0.0600~\mathrm{L}}\\[1.5ex] &= 0.00167~M \end{align*}\]

We can stop here and determine if we have a buffer. The ratio of acid to base is

\[\dfrac{[\mathrm{acid}]}{[\mathrm{base}]} = \dfrac{[\mathrm{NH_4^+}]}{[\mathrm{NH_3}]} = \dfrac{1.67\times 10^{-3}}{6.67\times 10^{-3}} = 0.250\]

or

\[\dfrac{[\mathrm{base}]}{[\mathrm{acid}]} = \dfrac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]} = \dfrac{6.67\times 10^{-3}}{1.67\times 10^{-3}} = 3.99\]

Since the ratio of acid to base lies between 0.1 and 10, this solution is a buffer!

Use Henderson-Hasselbalch (for a buffer made from an base) to determine the pH.

\[\begin{align*} \mathrm{pOH} &= \mathrm{p}K_{\mathrm{b}} + \log\dfrac{[\mathrm{\color{red}{HB^+}}]}{\mathrm{[\color{green}{B}]}} \\[1.5ex] &= -\log (1.77\times 10^{-5}) + \log\dfrac{1.67\times 10^{-3}~M}{6.67\times 10^{-3}~M}\\[1.5ex] &= 4.15 \end{align*}\]

Find the pH (at 25 °C).

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH} \\[1.5ex] &= 14 - 4.76 \\[1.5ex] &= 8.65 \end{align*}\]

Resisting a change in pH

A buffer contains both acid and base. When adding an acid to the buffer solution, the added acid can react with the base in the buffer. This effectively helps neutralize the added acid! Since the added acid is neutralized (to produce water and a salt), the change in pH is small.

Water has no buffering capability. Adding a strong acid to water dramatically changes its pH.

pH change of Strong Acid + Water

Add 0.01 M HCl to pure water (at 25 °C). Water initial has a pH of 7. Upon addition of the acid, the pH of the solution is 2. Water experiences a ΔpH = –5.

An acidic buffer has some buffering capability. Adding a strong acid to an acidic buffer will affect its pH slightly.

pH change of Strong Acid + Acidic Buffer

Add 10.0 mL 0.01 M HCl to a 50.0 mL buffer containing 0.01 M CH₃COOH and 0.01 M NaCH₃COO aqueous solution (at 25 °C). The solution initially has a pH of 4.76.

The added acid, H₃O⁺, will react with the base that is in solution, CH₃COO^–.

\[\mathrm{A^-}(aq) + \mathrm{H_3O^+}(aq) \longrightarrow \mathrm{H_2O}(l) + \mathrm{HA}(aq) \]

Convert everything to moles.

\[\begin{align*} n_{\mathrm{HA}} &= (0.050~\mathrm{L})\left(0.01~\mathrm{mol~L}^{-1}\right)\\ &= 0.0005~\mathrm{mol~HA} \\[2ex] n_{\mathrm{A^-}} &= (0.050~\mathrm{L})\left(0.01~\mathrm{mol~L}^{-1}\right)\\ &= 0.0005~\mathrm{mol~A^-} \\[2ex] n_{\mathrm{H_3O^+}^{-}} &= (0.01~\mathrm{L})\left(0.01~\mathrm{mol~L}^{-1}\right)\\ &= 0.0001~\mathrm{mol~H_3O^+} \end{align*}\]

I	0.0005		0.0001				0.0005
R	A–	+	H3O+	→	H2O	+	HA
F	0.0004		0				0.0006

Convert back to molarity.

\[\begin{align*} V_{\mathrm{final}} &= 50.0~\mathrm{mL} + 10.0~\mathrm{mL} \\ &= 60.0~\mathrm{mL} \\ &= 0.0600~\mathrm{L} \\[2ex] [\mathrm{HA}] &= \dfrac{0.0006~\mathrm{mol}}{0.0600~\mathrm{L}}\\[1.5ex] &= 0.01~M \\[1.5ex] [\mathrm{A^-}] &= \dfrac{0.0004~\mathrm{mol}}{0.0600~\mathrm{L}}\\[1.5ex] &= 0.00667~M \\[1.5ex] \end{align*}\]

We still have a buffer.

\[\dfrac{[\mathrm{base}]}{[\mathrm{acid}]} = \dfrac{[\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} = \dfrac{6.67\times 10^{-3}}{0.01} = 1.50\]

Solve for pH using the Henderson-Hasselbalch equation.

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{A^-}]}{\mathrm{[HA]}}\\[1.5ex] &= \mathrm{-\log(1.75\times 10^{-5})} + \log\dfrac{6.67\times 10^{-3}}{0.01} \\[1.5ex] &= 4.58 \end{align*}\]

The buffer solution experienced a pH change of only 0.18.

Similarly, adding a base to a buffer solution will cause an acid-base neutralization reaction. The added base will react with the acid in the buffer solution to produce water and a salt. The resulting pH change is small.

Water has no buffering capability. Adding a strong base to water dramatically changes its pH.

pH change of Strong Base + Water

Add 0.01 M NaOH to pure water (at 25 °C). Water initial has a pH of 7. Upon addition of the base, the pH of the solution is 12. Water experiences a ΔpH = 5.

A basic buffer has some buffering capability. Adding a strong base to an basic buffer will affect its pH slightly.

pH change of Strong Base + Acidic Buffer

Add 10.0 mL 0.01 M NaOH to a 50.0 mL buffer containing 0.01 M CH₃COOH and 0.01 M NaCH₃COO aqueous solution (at 25 °C). The solution initially has a pH of 4.76.

The added base, OH^–, will react with the acid that is in solution, CH₃COOH.

\[\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \longrightarrow \mathrm{H_2O}(l) + \mathrm{A^-}(aq) \]

Convert everything to moles.

\[\begin{align*} n_{\mathrm{HA}} &= (0.050~\mathrm{L})\left(0.01~\mathrm{mol~L}^{-1}\right)\\ &= 0.0005~\mathrm{mol~HA} \\[2ex] n_{\mathrm{A^-}} &= (0.050~\mathrm{L})\left(0.01~\mathrm{mol~L}^{-1}\right)\\ &= 0.0005~\mathrm{mol~A^-} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.01~\mathrm{L})\left(0.01~\mathrm{mol~L}^{-1}\right)\\ &= 0.0001~\mathrm{mol~OH}^{-} \end{align*}\]

I	0.0005		0.0001				0.0005
R	HA	+	OH–	→	H2O	+	A–
F	0.0004		0				0.0006

Convert back to molarity.

\[\begin{align*} V_{\mathrm{final}} &= 50.0~\mathrm{mL} + 10.0~\mathrm{mL} \\ &= 60.0~\mathrm{mL} \\ &= 0.0600~\mathrm{L} \\[2ex] [\mathrm{HA}] &= \dfrac{0.0004~\mathrm{mol}}{0.0600~\mathrm{L}}\\[1.5ex] &= 0.00667~M \\[1.5ex] [\mathrm{A^-}] &= \dfrac{0.0006~\mathrm{mol}}{0.0600~\mathrm{L}}\\[1.5ex] &= 0.01~M \end{align*}\]

We still have a buffer.

\[\dfrac{[\mathrm{base}]}{[\mathrm{acid}]} = \dfrac{[\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} = \dfrac{0.01}{6.67\times 10^{-3}} = 1.50\]

Solve for pH using the Henderson-Hasselbalch equation.

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{A^-}]}{\mathrm{[HA]}}\\[1.5ex] &= \mathrm{-\log(1.75\times 10^{-5})} + \log\dfrac{0.01}{6.67\times 10^{-3}} \\[1.5ex] &= 4.93 \end{align*}\]

The buffer solution experienced a pH change of only 0.17.

Buffer Capacity

Buffer capacity is the amount of acid or base a buffer is able to neutralize. Buffers containing higher concentrations of acid/c.base or base/c.acid have a larger buffering capacity.

Practice

Which buffer has a larger buffer capacity?

1. 0.001 M CH₃COOH + 0.001 M NaCH₃COO
2. 0.1 M CH₃COOH + 0.1 M NaCH₃COO

Answer

The second buffer

Titrations

Acid-base titrations is a quantitative analysis technique used to determine the concentration of an acid (or base) in an analyte (the unknown sample) by using a standard solution of base (or acid) to neutralize the unknown sample. The standard solution contains a known concentration of base (or acid). A pH meter is used throughout the titration to track the pH of the solution as base (or acid) is slowly added to the analyte. The resulting plot (pH vs. V_base added or V_acid added) results in the characteristic titration curve.

Titration Curves

Strong Acid + Strong Base

Suppose the analyte contained a strong acid. The pH of the solution is initially very low. The titration curve below indicates a strong acid solution initially (before any strong base is added).

Now, a strong base is slowly added to the analyte. The pH slowly rises as strong base is added. At a particular point, the pH jumps dramatically followed by a slow change in pH as base is added. The point at which the “jump” occurs is called the equivalence point and is the point at which all acid has been neutralized with added base. Therefore, at the equivalence point, the moles of acid equals the moles of base.

A strong acid/strong base titration will result in a pH of 7 at the equivalence point. The final pH of solution is quite high given that a strong base is added to the solution.

Weak Acid + Strong Base

Suppose the analyte contained a weak acid. The pH of the solution is initially somewhat low. The titration curve below indicates a weak acid solution initially (before any strong base is added).

A strong base is slowly added to the analyte. The pH slowly rises as strong base is added. The characteristic equivalence point appears but this time at a pH that is greater than 7. This is a feature of a weak acid/strong base titration.

Strong Base + Strong Acid

Suppose the analyte contained a strong base The pH of the solution is initially very high. The titration curve below indicates a strong acid solution initially (before any strong acid is added).

Now, a strong acid is slowly added to the analyte. The pH slowly decreases as strong acid is added. At a particular point, the pH drops dramatically followed by a slow change in pH as base is added. The point at which the “drop” occurs is called the equivalence point and is the point at which all base has been neutralized with added acid.

Weak Base + Strong Acid

Suppose the analyte contained a weak base The pH of the solution is initially somewhat large. The titration curve below indicates a weak base solution initially (before any strong base is added).

A strong acid is slowly added to the analyte. The pH slowly decreases as strong acid is added. The characteristic equivalence point appears but this time at a pH that is less than 7. This is a feature of a weak base/strong acid titration.

Diprotic Species

A titration involving a diprotic acid analyte will result in two equivalence points, the first for the first acidic proton and the second for the second acidic proton.

A titration involving a diprotic base analyte will result in two equivalence points.

Triprotic Species

A titration involving a triprotic acid analyte will result in three equivalence points.

A titration involving a triprotic base analyte will result in three equivalence points.

Calculating pH During Titration

When performing a titration, the pH of the solution is monitored via a pH probe, indicator, or some other form. Plotting the pH of solution vs. the amount of titrant added gives the characteristic titration curves (covered above).

We can also calculate the pH of the solution at any point of a titration and reproduce the titration curve plot without ever performing an actual titration!

These types of titration problems involving monoprotic acids can be broken down into four distinct pieces:

1. Initial pH
2. pH before equivalence point
3. pH at the equivalence point
4. pH beyond the equivalence point

Strong Acid/Strong Base Curve

Weak Acid/Strong Base Curve

The steps are outlined below.

Step 1: Determine acid/base reaction type

• Strong acid/strong base

  • \(\mathrm{H^+}(aq) + \mathrm{OH^-}(aq) \longrightarrow \mathrm{H_2O}(l)\)
  • \(\mathrm{H_3O^+}(aq) + \mathrm{OH^-}(aq) \longrightarrow \mathrm{2H_2O}(l)\)

• Weak acid/strong base

  • \(\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{A^-}(aq)\)

• Strong acid/weak base

  • \(\mathrm{H^+}(aq) + \mathrm{B}(aq) \rightleftharpoons \mathrm{HB^+}(aq)\)
  • \(\mathrm{H_3O^+}(aq) + \mathrm{B}(aq) \rightleftharpoons \mathrm{HB^+}(aq) + \mathrm{H_2O}(l)\)

Step 2: Determine molar changes (IRF table)

• React acid and base (in moles) if applicable

• Smaller amount is “wiped out”

Step 3: Determine final pH

• If “strong” (H₃O⁺ or OH^–) remains

  • Determine V_final

  • Calculate final [H₃O⁺] or [OH^–]

  • \(\mathrm{pH} = -\log{[\mathrm{H_3O^+}]}\) or \(\mathrm{pOH = -\log[\mathrm{OH^-}]}\)

• If “weak” pair present (HA/A^– or B/HB⁺), use Henderson-Hasselbalch

  • \(\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{A^-}]}{\mathrm{[HA]}}\)

  • \(\mathrm{pOH} = \mathrm{p}K_{\mathrm{b}} + \log\dfrac{[\mathrm{HB^+}]}{\mathrm{[B]}}\)

• If only one “weak” species present (HA, A^–, B, or HB⁺)

  • Determine V_final

  • Calculate concentration of “weak” species

  • Use ICE table to find pH

Strong Acid/Strong Base

Titrate 25.0 mL of 0.100 M HCl with 0.100 M NaOH. (at 25 °C).

Initial pH

What is the initial pH of solution?

Solution

Step 1: Recognize that this is a strong acid/strong base reaction.

Step 2: No molar changes occur since no strong base has been added yet.

Step 3: Determine final pH. Only a “strong” species is present (H₃O⁺).

Given that the initial solution contains a strong acid, recognize that all the strong acid reacts with water to produce hydronium and conjugate base.

\[\mathrm{HCl}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{H_3O^+}(aq) + \mathrm{Cl^-}(aq)\] Therefore, [H₃O⁺] = 0.100 M.

\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\ &= -\log (0.100) \\ &= 1.00 \end{align*}\]

pH before equivalence point

What is the pH after 15.0 mL of NaOH is added?

Solution

Step 2: Determine molar changes (IRF table)

To determine molar changes, convert concentration (M) to moles

\[\begin{align*} n_{\mathrm{H_3O}^{+}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.0150~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00150~\mathrm{mol} \end{align*}\]

React acid and base (in moles).

Initial moles	0.00250		0.00150
Reaction	H3O+	+	OH–	→	2H2O
Final moles	0.00100		0

Step 3: Determine final pH. Only a “strong” species is present (H₃O⁺).

A strong species (H₃O⁺) remains. Find the total volume of solution and convert moles back into molarity (M). Determine pH.

\[\begin{align*} V_{\mathrm{final}} &= 25.0~\mathrm{mL} + 15.0~\mathrm{mL} \\ &= 40.0~\mathrm{mL} \\ &= 0.04~\mathrm{L} \\[2ex] [\mathrm{H_3O^+}] &= \dfrac{0.00100~\mathrm{mol}}{\mathrm{0.040~L}}\\[1.5ex] &= 0.025~M \\[2ex] \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\ &= -\log (0.025) \\ &= 1.60 \end{align*}\]

pH at the equivalence point

What is the pH at the equivalence point?

Solution

Step 1: Determine acid/base reaction type

• This is a strong acid/strong base titration
• \(\mathrm{H_3O^+}(aq) + \mathrm{OH^-}(aq) \longrightarrow \mathrm{2H_2O}(l)\)

Step 2: Determine molar changes (IRF table)

To determine molar changes, convert concentration (M) to moles

\[\begin{align*} n_{\mathrm{H_3O}^{+}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol} \end{align*}\]

Note that a volume of added base was not given. However, we implement the definition of the equivalence point ([acid] = [base]) and recognize that there must be an equal number of moles of base as there is acid! Therefore,

\[\begin{align*} n_{\mathrm{OH^-}} &= 0.00250~\mathrm{mol} \\[2ex] \end{align*}\]

React acid and base (in moles).

Initial moles	0.00250		0.00250
Reaction	H3O+	+	OH–	→	2H2O
Final moles	0		0

Step 3: Determine final pH

Since no strong species (H₃O⁺ or OH^–) remains, the solution is neutral. All acid and base was converted to water. The only hydronium and hydroxide ions in solution come from pure water via the autoionization of water (at 25 °C)

\[2\mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH^-}(aq)\quad K_{\mathrm{w}} = 1.00\times 10^{-14}\]

\[\begin{align*} [\mathrm{H_3O^+}] &= [\mathrm{OH^-}] = 1\times 10^{-7}\\[2ex] \mathrm{pH} &= [\mathrm{H_3O^+}] \\ &= -\log (1\times 10^{-7}) \\ &= 7 \end{align*}\]

pH beyond the equivalence point

What is the pH after 30.0 mL of NaOH is added?

Solution

Step 1: Determine acid/base reaction type

• This is a strong acid/strong base titration
• \(\mathrm{H_3O^+}(aq) + \mathrm{OH^-}(aq) \longrightarrow \mathrm{2H_2O}(l)\)

Step 2: Determine molar changes (IRF table)

To determine molar changes, convert concentration (M) to moles

\[\begin{align*} n_{\mathrm{H_3O}^{+}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.0300~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00300~\mathrm{mol} \end{align*}\]

React acid and base (in moles).

Initial moles	0.00250		0.00300
Reaction	H3O+	+	OH–	→	2H2O
Final moles	0		0.0005

Step 3: Determine final pH

A strong species (OH^–) remains. Find the total volume of solution and convert moles back into molarity (M). Get pH.

\[\begin{align*} V_{\mathrm{final}} &= 25.0~\mathrm{mL} + 30.0~\mathrm{mL} \\ &= 55.0~\mathrm{mL}\\ &= 0.055~\mathrm{L} \\[2ex] [\mathrm{OH^-}] &= \dfrac{0.0005~\mathrm{mol}}{\mathrm{0.055~L}} \\[1.5ex] &= 0.0091~M \\[2ex] \mathrm{pOH} &= -\log [\mathrm{OH^-}] \\ &= -\log (0.0091) = 2.04 \\[2ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH}\\[1ex] &= 14 - 2.04 \\[1ex] &= 11.96 \end{align*}\]

Weak Acid/Strong Base

Titrate 25.0 mL of 0.100 M CH₃COOH with 0.100 M NaOH (at 25 °C).
K_a(CH₃OOH) = 1.80 × 10^–5

Initial pH

What is the initial pH of solution?

Solution

Step 1: Determine acid/base reaction type

• This is a weak acid/strong base titration

Step 2: No molar changes occur since no strong base has been added yet.

Step 3: Determine final pH

This is a typical equilibrium problem via an ICE table.

	CH3OOH	+	H2O	⇌	H3O+	+	CH3OO–
I	0.100				≈0		0
C	–x				+x		+x
E	0.100 - x				x		x

Solve the equilibrium expression.

\[\begin{align*} \dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{\mathrm{[HA]}} &= K_{\mathrm{a}} \\[1.5ex] \dfrac{x^2}{0.100-x} &= 1.80\times 10^{-5}\\[1.5ex] \dfrac{x^2}{0.100} &= 1.80\times 10^{-5}\\[1.5ex] x &= 1.34\times 10^{-3}\\[2ex] \dfrac{1.34\times 10^{-3}}{0.100} \times 100\% &= 1.34\% \end{align*}\]

Get pH. \[\begin{align*} \mathrm{pH} &= -\log[\mathrm{H_3O^+}] \\ &= -\log (1.3\times 10^{-3}) \\ &= 2.87 \end{align*}\]

pH before equivalence point

What is the pH after 15.0 mL of NaOH is added?

Solution

Step 1: Determine acid/base reaction type

• This is a weak acid/strong base titration
• \(\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{A^-}(aq)\)

Step 2: Determine molar changes (IRF table)

To determine molar changes, convert concentration (M) to moles

\[\begin{align*} n_{\mathrm{HA}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.0150~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00150~\mathrm{mol} \end{align*}\]

React acid and base (in moles).

Initial moles	0.00250		0.00150				0
Reaction	HA	+	OH–	→	H2O	+	A–
Final moles	0.00100		0				0.00150

Step 3: Determine final pH

If “weak” pair present (HA/A^– or B/HB⁺), determine of this is a buffer. If it is, use Henderson-Hasselbalch to get the pH.

\[\begin{align*} \dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]} = \dfrac{0.00150}{0.00100} = 1.5 \end{align*}\]

or

\[\begin{align*} \dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]} = \dfrac{0.00100}{0.00150} = 0.667 \end{align*}\]

Since the ratio of acid to conjugate base is between 0.1 and 10, we have a buffer and can use Henderson-Hasselbalch!

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{A^-}]}{\mathrm{[HA]}} \\[1ex] &= -\log(1.80\times 10^{-5}) + \log\dfrac{0.00150}{0.00100} \\[1ex] &= 4.92 \end{align*}\]

Why did you use moles??

I used mole quantities here in the Henderson-Hasselbalch equation because converting to concentration would net the same answer. Consider the two examples:

Example 1: Mole quantities

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{A^-}]}{\mathrm{[HA]}} \\[1.5ex] &= -\log(1.80\times 10^{-5}) + \log\dfrac{0.00150}{0.0010} \\[1.5ex] &= 4.92 \end{align*}\]

Example 2: Molar concentration quantities

\[V_{\mathrm{tot}} = 25.0~\mathrm{mL} + 15.0~\mathrm{mL} = 40.0~\mathrm{mL}\]

\[\begin{align*} [\mathrm{H_3O^+}] &= \dfrac{0.00100~\mathrm{mol~H_3O^+}}{0.040~\mathrm{L}}\\[1.5ex] &= 0.0250~M\\[2ex] [\mathrm{OH^-}] &= \dfrac{0.00150~\mathrm{mol~A^-}}{0.040~\mathrm{L}}\\[1.5ex] &= 0.0375~M \end{align*}\]

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{A^-}]}{\mathrm{[HA]}} \\ &= -\log(1.80\times 10^{-5}) + \log\dfrac{0.0375}{0.0250} \\ &= 4.92 \end{align*}\]

pH at equivalence point

What is the pH at the equivalence point?

Solution

Step 1: Determine acid/base reaction type

• This is a weak acid/strong base titration
• \(\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{A^-}(aq)\)

Step 2: Determine molar changes (IRF table)

To determine molar changes, convert concentration (M) to moles

\[\begin{align*} n_{\mathrm{HA}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol} \\[2ex] \end{align*}\]

Note that a volume of added base was not given. However, we implement the definition of the equivalence point ([acid] = [base]) and recognize that there must be an equal number of moles of base as there is acid! Therefore,

\[\begin{align*} n_{\mathrm{OH^-}} &= 0.00250~\mathrm{mol} \\[2ex] \end{align*}\]

React acid and base (in moles).

Initial moles	0.00250		0.00250				0
Reaction	HA	+	OH–	→	H2O	+	A–
Final moles	0		0				0.00250

Step 3: Determine final pH

Only one “weak” species (A^–) is present. Convert moles back into molarity (M).

First, determine the volume of base added.

\[\begin{align*} V_{\mathrm{NaOH}} &= \dfrac{0.00250~\mathrm{mol}}{0.100~\mathrm{mol~L^{-1}}} \\[1.5ex] &= 0.025~\mathrm{L} \\[1.5ex] &= 25.0~\mathrm{mL} \end{align*}\]

Next, find the total volume of solution and determine the concentration of A^–.

\[\begin{align*} V_{\mathrm{final}} &= 25.0~\mathrm{mL} + 25.0~\mathrm{mL} \\ &= 50.0~\mathrm{mL} \\ &= 0.0500~\mathrm{L} \\[2ex] [\mathrm{A^-}] &= \dfrac{0.00250~\mathrm{mol}}{0.0500~\mathrm{L}}\\[1.5ex] &= 0.0500~M \end{align*}\]

Set up an ICE table for A^–, a conjugate base, reacting with water.

	A–	+	H2O	⇌	OH–	+	HA
I	0.0500				≈0		0
C	–x				+x		+x
E	0.0500 - x				x		x

NOTE: Because we are analyzing a base-ionization reaction (since we have a base reacting with water), you must convert the acid-dissociation constant, K_a, to the base-ionization constant, K_b, of the corresponding conjugate base!

\[\begin{align*} K_{\mathrm{b}} &= \dfrac{K_{\mathrm{w}}}{K_{\mathrm{a}}} \\[1.5ex] &= \dfrac{1.00\times 10^{-14}}{1.80\times 10^{-5}} \\[1.5ex] &= 5.6\times 10^{-10} \end{align*}\]

Set up the equilibrium expression and solve.

\[\begin{align*} \dfrac{[\mathrm{OH^-}][\mathrm{HA}]}{[\mathrm{A^-}]} &= K_{\mathrm{b}}\\[1.5ex] \dfrac{x^2}{0.0500-x} &= 5.6\times 10^{-10}\\[1.5ex] \dfrac{x^2}{0.0500} &= 5.6\times 10^{-10} \\[1.5ex] x &= 5.29\times 10^{-6} (0.01\%) \end{align*}\]

Get pH.

\[\begin{align*} \mathrm{pOH} &= -\log[\mathrm{OH^-}] \\ &= -\log (5.29\times 10^{-6})\\ &= 5.28\\[2ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH}\\ &= 14 - 5.28\\ &= 8.72 \end{align*}\]

Beyond the equivalence point

What is the pH after 30.0 mL of NaOH is added?

Solution

Step 1: Determine acid/base reaction type

• This is a weak acid/strong base titration
• \(\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{A^-}(aq)\)

Step 2: Determine molar changes (IRF table)

To determine molar changes, convert concentration (M) to moles

\[\begin{align*} n_{\mathrm{HA}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol~HA} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.0300~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00300~\mathrm{mol~OH}^{-} \end{align*}\]

React acid and base (in moles).

Initial moles	0.00250		0.00300				0
Reaction	HA	+	OH–	→	H2O	+	A–
Final moles	0		0.0005				0.00250

Step 3: Determine final pH

A strong species (OH^–) remains. Find the total volume of solution and convert moles back into molarity (M). Get pH.

\[\begin{align*} V_{\mathrm{final}} &= 25.0~\mathrm{mL} + 30.0~\mathrm{mL}\\ &= 55.0~\mathrm{mL}\\ &= 0.055~\mathrm{L}\\[2ex] [\mathrm{OH^-}] &= \dfrac{0.0005~\mathrm{mol}}{\mathrm{0.055~L}}\\[1.5ex] &= 0.0091~M \\[2ex] \mathrm{pOH} &= -\log [\mathrm{OH^-}] \\ &= -\log (0.0091) = 2.04 \\[2ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH}\\ &= 14 - 2.04 \\ &= 11.96 \end{align*}\]

Practice

Below are tables for two different titrations (strong acid/strong base and weak acid/strong base). Try to replicate the pH in the last column by using only the data in column 1.

Strong acid/strong base titration

Titrate 25.0 mL of 0.100 M HCl with 0.100 M NaOH. (at 25 °C) What is the pH after X mL (from column 1) of NaOH is added?

VOH– (mL)	nOH– (mol)	nH+ remain (mol)	Vtot (mL)	[H+] (M)		pH
0	0	0.00250	25.0	0.100		1.00
5.0	0.00050	0.00200	30.0	0.0667		1.18
10.0	0.00100	0.00150	35.0	0.0429		1.36
15.0	0.00150	0.00100	40.0	0.0250		1.60
20.0	0.00200	0.00050	45.0	0.011		1.96
25.0	0.00250	0	50.0	1.0×10–7		7.00

VOH– (mL)	nOH– (mol)	Excess OH– (mol)	Vtot (mL)	[OH–] (M)	pOH	pH
30.0	0.00300	0.00050	55.0	0.0091	2.04	11.96
35.0	0.00350	0.00100	60.0	0.0167	1.778	12.22

Weak acid/strong base titration

Titrate 25.0 mL of 0.100 M CH₃COOH with 0.100 M NaOH (at 25 °C). What is the pH after X mL (from column 1) of NaOH is added? K_a(CH₃COOH) = 1.80 ×10^–5

VOH– (mL)	nOH– (mol)	CH3COOH remain (mol)	CH3COO– made (mol)	Vtot (mL)	pH
0	0	0.00250	0	25.0	2.87
5.0	0.00050	0.00200	0.00050	30.0	4.14
10.0	0.00100	0.00150	0.00100	35.0	4.56
15.0	0.00150	0.00100	0.00150	40.0	4.92
20.0	0.00200	0.00050	0.00200	45.0	5.34
25.0	0.00250	0	0.00250	50.0	8.72

VOH– (mL)	nOH– (mol)	Excess OH– (mol)	Vtot (mL)	[OH–] (M)	pOH	pH
30.0	0.00300	0.00050	55.0	0.0091	2.04	11.96
35.0	0.00350	0.00100	60.0	0.0167	1.778	12.22