Acid/Base Relationships


The strength of an acid or base refers to its magnitude of dissociation in solution. For the rest of this article, we will simply consider aqueous solution (solvent is water). The stronger the acid or base, the more it dissociates in solution.

The relative strengths of some acids and their corresponding conjugate bases are given in the table below. All values are given at 25 °C.

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

We will discuss the relationships between these values.


Acid Ionization, Ka


Consider the following, generalized acid-ionization reaction.

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

We can write an equilibrium constant for this reaction as

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

The larger the acid-dissociation constant, Ka, the more the acid dissociates in water (the more A is produced). The stronger the acid, the larger the Ka value.

Ka values are fine but are generally inconvenient to use given their derivation from concentrations which are typically very small or very large numbers. Rather than fuss with scientific notation, we can convert the Ka scale to a logarithmic scale.

\[ \mathrm{p}K_{\mathrm{a}} = -\log K_{\mathrm{a}} \] The pKa of an acid can be converted to Ka by

\[K_{\mathrm{a}} = 10^{-\mathrm {p}K_{\mathrm{a}}}\]

Example of Ka vs. pKa

The Ka for acetic acid is 1.75 × 10–5.

The pKa for acetic acid is 4.76.

The pKa value is more convenient.


The important thing to note is that the pKa scale is opposite that of the Ka scale. The relationship is below:

  • Stronger acid = larger Ka = smaller pKa
  • Weaker acid = smaller Ka = larger pKa

Looking at the table at the top of this article, the strongest acid is at the top while the weakest acid is at the bottom.

When an acid, HA, dissociates, it loses a proton. The species that is left (that is, the deprotonated acid), is referred to as the conjugate base. Stronger acids give rise to weaker conjugate bases. Weaker acids give rise to stronger conjugate bases.


Base Ionization, Kb


We can consider a similar reaction for a base ionizing in water.

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

\[K_{\mathrm{b}}=\dfrac{\left[\textcolor{green}{\mathrm{HB}}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\textcolor{red}{\mathrm{B}}]}\] or, written in the alternative notation,

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

\[K_{\mathrm{b}}=\dfrac{\left[\textcolor{green}{\mathrm{HA}}\right]\left[\mathrm{OH}^{-}\right]}{[\textcolor{red}{\mathrm{A^-}}]}\]

The larger the base-dissociation constant, Kb, the more the base dissociates in water (the more HB+ or HA is produced). The stronger the base, the larger the Kb value.

Kb values are fine but are generally inconvenient to use given their derivation from concentrations which are typically very small or very large numbers. Rather than fuss with scientific notation, we can convert the Kb scale to a logarithmic scale.

\[ \mathrm{p}K_{\mathrm{b}} = -\log K_{\mathrm{b}} \] The pKb of an base can be converted to Kb by

\[K_{\mathrm{b}} = 10^{-\mathrm {p}K_{\mathrm{b}}}\]

Example of Kb vs. pKb

The Kb for the acetate ion is 5.71 × 10–10.

The pKb for the acetate ion is 9.24.

The pKb value is more convenient.


The important thing to note is that the pKb scale is opposite that of the Kb scale. The relationship is below:

  • Stronger base = larger Kb = smaller pKb
  • Weaker base = smaller Kb = larger pKb

Looking at the table at the top of this article, the weakest base is at the top and the strongest base is at the bottom.

When a base, B, ionizes, it gains a proton. The species that is created (that is, the protonated base), is referred to as the conjugate acid. Stronger bases give rise to weaker conjugate acids. Weaker bases give rise to stronger conjugate acids.


Relationship between Ka and Kb


Ka and Kb are related through Kw, the water-dissociation constant that is related to the following reaction (called the auto-ionization of water)

\[\mathrm{2H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH}^-(aq) \quad K_{\mathrm{w}} = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0\times 10^{-14}\]

The given Kw is at 25 °C. Water naturally undergoes this reaction. Note that the equilibrium constant is very small.

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

Water is amphiprotic meaning it acts as both an acid and a base. Notice how a water loses a proton (acid) while another water gains a proton (base).

Looking at the table above, take any Ka and corresponding Kb value and multiply them. You should get the value for Kw. Therefore, if you know the Ka of an acid, you can easily find the Kb of the corresponding conjugate base. Similarly, if you know the Kb value of a base, you can easily get the Ka value of the corresponding conjugate acid.

Kw can also be expressed in log form to give

\[\mathrm{p}K_{\mathrm{w}} = -\log K_{\mathrm{w}}\]

where pKw = 14 at 25 °C.

The pKw of water can be converted to Kw by

\[K_{\mathrm{w}} = 10^{-\mathrm {p}K_{\mathrm{w}}}\]

Remember that all of these equilibrium constants, K, are temperature dependent. Below is a table of Kw and pKw values for pure water at various temperatures.

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 1.01 ×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

Note that even though pH decreases with increasing temperature, the water is not getting more acidic. In all cases above, [H3O+] = [OH], giving a neutral solution.

Below is a plot of pKw vs. temperature.


Relationship between pKa and pKb, pKw


The following equation describes the relationship between pKa and pKb.

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

You should notice in the table at the top of this article that adding any pKa to its corresponding pKb results in 14.


Example at standard temperature

The Ka for acetic acid is 1.75 × 10–5. What is the pKa of acetic acid at 25 °C?

\[\begin{align*} \mathrm{p}K_{\mathrm{a}} &= -\log K_{\mathrm{a}}\\ &= -\log 1.75\times 10^{-5} \\ &= 4.76 \end{align*}\]

What is the pKb of the conjugate base, CH3COO, of acetic acid at 25 °C?

\[\begin{align*} \mathrm{p}K_{\mathrm{a}} + \mathrm{p}K_{\mathrm{b}} &= \mathrm{p}K_{\mathrm{w}}\\ \mathrm{p}K_{\mathrm{b}} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{p}K_{\mathrm{a}}\\ &= 14 - 4.76 \\ &= 9.24 \end{align*}\]


Example at non-standard temperature

The pKa of acetic acid was found to be 1.633 × 10–5 at 50 °C. What is the pKb of the conjugate base, CH3COO, of acetic acid at 50 °C?

\[\begin{align*} \mathrm{p}K_{\mathrm{a}} &= -\log K_{\mathrm{a}}\\ &= -\log 1.633\times 10^{-5} \\ &= 4.79\\[2ex] \mathrm{p}K_{\mathrm{a}} + \mathrm{p}K_{\mathrm{b}} &= \mathrm{p}K_{\mathrm{w}}\\ \mathrm{p}K_{\mathrm{b}} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{p}K_{\mathrm{a}}\\ &= 13.26 - 4.79 \\ &= 8.47 \end{align*}\]


Percent ionization


A measure of the amount of ionization of an acid can be expressed as

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


Strong acids and bases


Below are some common strong acids and strong bases.

Strong Acid Strong Base
HClO4 perchloric acid LiOH lithium hydroxide
HCl hydrochloric acid NaOH sodium hydroxide
HBr hydrobromic acid KOH potassium hydroxide
HI hydroiodic acid Ca(OH)2 calcium hydroxide
HNO3 nitric acid Sr(OH)2 strontium hydroxide
H2SO4 sulfuric acid Ba(OH)2 barium hydroxide


pH and pOH


The pH and pOH scales are a measure of the hydronium ion concentration (in M) or hydroxide ion concentration (in M) in solution. Note that this has nothing to do with the strength of an acid or base. It simply is a measure of the aforementioned concentrations in solution.

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

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

At 25 °C,

\[ \mathrm{pH} + \mathrm{pOH} = 14\]

The lower the pH of a solution, the more hydronium ions are present (the more acidic the solution). The lower the pOH of a solution, the more hydroxide ions are present (the more basic the solution).


Acidic, Basic or Neutral?


Solutions can be described as being acidic, basic or neutral.

Solution pHa pOHa [H3O+]a [OH]a
Neutral [H3O+] = [OH] 7 7 =10–7 =10–7
Acidic [H3O+] > [OH] <7 >7 >10–7 <10–7
Basic [H3O+] < [OH] >7 <7 <10–7 >10–7

a At 25 °C.

It is important to note that none of these solutions are defined by their pH value as this is temperature dependent. An acidic, basic, or neutral solution is defined by the relative amounts of hydronium and hydroxide ions it contains, regardless of pH! A neutral solution has equal concentrations of hydronium and hydroxide ions. It just so happens that at 25 °C the pH is 7.


Recap


You should recognize by now that if you have just one of the values given in a row from the table at the top of this article, you can mathematically determine any other value in that row using the equations discussed above.


Practice Problems


Question 1

What are the hydronium and hydroxide ion concentrations (to two decimal places) in a sample of pure water at 25 °C?

Solution
2H2O(l) \(\rightleftharpoons\) H3O+(aq) + OH(aq)
I 0 0
C +x +x
E x x

 

\[\begin{align*} [\mathrm{H_3O^+}][\mathrm{OH^-}] &= K_{\mathrm{w}} \\ (x)(x) &= 1.00\times 10^{-14} \\ x^2 &= 1.00\times 10^{-14} \\ x &= 1.00\times 10^{-7} \end{align*}\]


Question 2

A sample of pure water is spiked with a strong base resulting in a 1.00 × 10–4 M concentration at 25 °C. What is the concentration (to two decimal places) of H3O+ ? What is the pH (to two decimal places) of solution?

Solution

H2O(l) \(\rightleftharpoons\) H3O+(aq) + OH(aq)       Kw = 1.00 × 10–14

2H2O(l) \(\rightleftharpoons\) H3O+(aq) + OH(aq)
I 0 0
C +x +x
E x 1.00 × 10–4

 

\[\begin{align*} [\mathrm{H_3O^+}][\mathrm{OH^-}] &= K_{\mathrm{w}} \\ (x)(1.00\times 10^{-4}) &= 1.00\times 10^{-14} \\ x &= 1.00\times 10^{-10} \\[3ex] \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\ &= -\log (1.00\times 10^{-10})\\ &= 10.00 \end{align*}\]


Question 3

What is the pH (to two decimal places) of pure water at 37 °C? Is the water acidic, basic, or neutral? Kw = 2.32 × 10–14

Solution

2H2O(l) \(\rightleftharpoons\) H3O+(aq) + OH(aq)       Kw = 2.23 × 10–14

2H2O(l) \(\rightleftharpoons\) H3O+(aq) + OH(aq)
I 0 0
C +x +x
E x x

 

\[\begin{align*} [\mathrm{H_3O^+}][\mathrm{OH^-}] &= K_{\mathrm{w}} \\ (x)(x) &= 2.32\times 10^{-14} \\ x^2 &= 2.32\times 10^{-14} \\ x &= 1.52\times 10^{-7} \\[3ex] \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\ &= -\log (1.52\times 10^{-7}) \\ &= 6.82 \end{align*}\]

The water is neutral.


Question 4

The pH of an aqueous solution is 7.82 at 25 °C. What is the pOH (to two decimal places)?

Solution

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\ \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} -\mathrm{pH} \\ &= 14 - 7.82 \\ &= 6.18 \end{align*}\]


Question 5

The Kb for NO2 is 2.17 × 10–11. What is the Ka and pKa (to two decimal places) for its conjugate acid at 25 °C?

Solution

\[\begin{align*} K_{\mathrm{a}} \times K_{\mathrm{b}} &= K_{\mathrm{w}} \\[3ex] K_{\mathrm{a}} &= \dfrac{K_{\mathrm{w}}}{K_{\mathrm{b}}} \\ &= \dfrac{1.0\times 10^{-14}}{2.17\times 10^{-11}} \\ &= 4.61\times 10^{-4} \\[3ex] \mathrm{p}K_{\mathrm{a}} &= -\log K_{\mathrm{a}} \\ &= -\log (4.61\times 10^{-4})\\ &= 3.34 \end{align*}\]


Question 6

What is the percent ionization (to one decimal place) of a 0.125 M solution of nitrous acid (HNO2) with a pH of 2.09? Is the acid strong or weak?

Solution

HNO2(aq) + H2O(l) \(\rightleftharpoons\) H3O+(aq) + NO2(aq)

\[\begin{align*} [\mathrm{H_3O^+}] &= 10^{-\mathrm{pH}} \\ &= 10^{-2.09}\\ &= 8.13\times 10^{-3} \\[3ex] \%~\mathrm{ionization} &= \dfrac{[\mathrm{H_3O^+}]_{\mathrm{eq}}}{[\mathrm{HA}]_0} \times 100 \\ &= \dfrac{8.13\times 10^{-3}}{0.125} \times 100 \\ &= 6.5\% ~\mathrm{<<~100\%~~~weak} \end{align*}\]


Question 7

What is the pH and pOH (to two decimal places) of a 0.100 M HCN aqueous solution? pKa = 9.21

Solution
HCN(aq) + H2O(l) \(\rightleftharpoons\) H3O+(aq) + CN(aq)
I 0.100 0 0
C -x +x +x
E 0.100 – x x x

 

\[\begin{align*} K_{\mathrm{a}} &= 10^{-\mathrm{p}K_{\mathrm{a}}} \\ &= 10^{-9.21} \\ &= 6.2\times 10^{-10}\\[3ex] \dfrac{[\mathrm{H_3O^+}][\mathrm{CN^-}]}{[\mathrm{HCN}]} &= K_{\mathrm{a}}\\[1.5ex] \dfrac{(x)(x)}{0.100-x} &= 6.2\times 10^{-10}\\[1.5ex] \dfrac{x^2}{0.100} &= 6.2\times 10^{-10}\\[1.5ex] x &= 7.87\times 10^{-6}\\[3ex] \% x &= \dfrac{7.87\times 10^{-6}}{0.100}\times 100\% = 7.87\times 10^{-3}\% \\[3ex] \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\ &= -\log(7.87\times 10^{-6}) \\ &= 5.10\\[3ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\ &= 14 - 5.10 \\ &= 8.90 \end{align*}\]


Question 8

What is the pH (to two decimal places) of a 25.0 mL 0.100 M HCl aqueous solution (at 25 °C)?

Solution

Strong acid; large Ka; assume complete dissociation of acid.

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

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


Question 9

What is the pH and pOH (to two decimal places) of a 50.0 mL 0.200 M NaOH aqueous solution (at 25 °C)?

Solution

Strong base; large Kb; assume complete dissociation of the salt. Therefore, the concentration of OH from NaOH is 0.200 M.

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

\[\begin{align*} \mathrm{pOH} &= -\log [\mathrm{OH^–}] \\ &= -\log (0.200) \\ &= 0.70 \\[3ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH} \\ &= 14 - 0.70 \\ &= 13.3 \end{align*}\]


Question 10

What is the pH (to two decimal places) of a 50.0 mL 1×10–6 M HCl aqueous solution (at 25 °C)?

Solution

Strong acid; large Ka; assume complete dissociation of acid.

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

\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}]\\ &= -\log (1\times 10^{-6}) \\ &= 6.00 \end{align*}\]


Question 11

What is the pH and pOH (to two decimal places) of a 25.0 mL 0.500 M HF aqueous solution (at 25 °C)? pKa = 3.17

Solution

Weak acid; small Ka; ICE table.

HF(aq) + H2O(l) \(\rightleftharpoons\) H3O+(aq) + F(aq)
I 0.500 0 0
C -x +x +x
E 0.500 – x x x

 

\[\begin{align*} K_{\mathrm{a}} &= 10^{-\mathrm{p}K_{\mathrm{a}}} \\ &= 10^{-3.17} \\ &= 6.76\times 10^{-4}\\[3ex] \dfrac{[\mathrm{H_3O^+}][\mathrm{F^-}]}{[\mathrm{HF}]} &= K_{\mathrm{a}}\\[1.5ex] \dfrac{(x)(x)}{0.500-x} &= 6.67\times 10^{-4}\\[1.5ex] \dfrac{x^2}{0.500} &= 6.67\times 10^{-4}\\[1.5ex] x &= 1.84\times 10^{-2}\\[3ex] \% x &= \dfrac{1.84\times 10^{-2}}{0.500}\times 100\% = 3.7\% \\[3ex] \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\ &= -\log(1.84\times 10^{-2}) \\ &= 1.74\\[3ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\ &= 14 - 1.74 \\ &= 12.26 \end{align*}\]


Question 12

What is the pH and pOH (to two decimal places) of a 50.0 mL 0.350 M HF aqueous solution (at 25 °C)? pKa = 3.17

Solution

Weak acid; small Ka; ICE table.

HF(aq) + H2O(l) \(\rightleftharpoons\) H3O+(aq) + F(aq)
I 0.350 0 0
C -x +x +x
E 0.350 – x x x

 

\[\begin{align*} K_{\mathrm{a}} &= 10^{-\mathrm{p}K_{\mathrm{a}}}\\ &= 10^{-3.17} \\ &= 6.76\times 10^{-4}\\[3ex] \dfrac{[\mathrm{H_3O^+}][\mathrm{F^-}]}{[\mathrm{HF}]} &= K_{\mathrm{a}}\\[1.5ex] \dfrac{(x)(x)}{0.350-x} &= 6.67\times 10^{-4}\\[1.5ex] \dfrac{x^2}{0.350} &= 6.67\times 10^{-4}\\[1.5ex] x &= 1.54\times 10^{-2}\\[3ex] \% x &= \dfrac{1.54\times 10^{-2}}{0.350}\times 100\% = 4.4\% \\[3ex] \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\ &= -\log(1.54\times 10^{-2}) \\ &= 1.81 \\[3ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\ &= 14 - 1.81 \\ &= 12.19 \end{align*}\]


Question 13

What is the pH and pOH (to two decimal places) of a 50.0 mL 0.500 M HA aqueous solution (at 25 °C)?pKa = 4.80

Solution

Weak acid; small Ka; ICE table.

HA(aq) + H2O(l) \(\rightleftharpoons\) H3O+(aq) + A(aq)
I 0.500 0 0
C -x +x +x
E 0.500 – x x x

 

\[\begin{align*} K_{\mathrm{a}} &= 10^{-\mathrm{p}K_{\mathrm{a}}}\\ &= 10^{-4.80} \\ &= 1.58\times 10^{-5}\\[3ex] \dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]} &= K_{\mathrm{a}}\\[1.5ex] \dfrac{(x)(x)}{0.500-x} &= 1.58\times 10^{-5}\\[1.5ex] \dfrac{x^2}{0.500} &= 1.58\times 10^{-4}\\[1.5ex] x &= 8.89\times 10^{-3}\\[3ex] \% x &= \dfrac{8.89\times 10^{-3}}{0.500}\times 100\% = 1.78\% \\[3ex] \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\ &= -\log(8.89\times 10^{-3}) \\ &= 2.51 \\[3ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\ &= 14 - 2.51 \\ &= 11.95 \end{align*}\]


Question 14

What is the pH and pOH (to two decimal places) of a 50.0 mL 0.500 M HA aqueous solution (at 25 °C)? pKa = 6.5

Solution

Weak acid; small Ka; ICE table.

HA(aq) + H2O(l) \(\rightleftharpoons\) H3O+(aq) + A(aq)
I 0.500 0 0
C -x +x +x
E 0.500 – x x x

 

\[\begin{align*} K_{\mathrm{a}} &= 10^{-\mathrm{p}K_{\mathrm{a}}}\\ &= 10^{-6.5} \\ &= 3.16\times 10^{-7}\\[3ex] \dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]} &= K_{\mathrm{a}}\\[1.5ex] \dfrac{(x)(x)}{0.500-x} &= 3.16\times 10^{-7}\\[1.5ex] \dfrac{x^2}{0.500} &= 3.16\times 10^{-7}\\[1.5ex] x &= 3.98\times 10^{-4}\\[3ex] \% x &= \dfrac{3.98\times 10^{-4}}{0.500}\times 100\% = 0.08\% \\[3ex] \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\ &= -\log(3.98\times 10^{-4}) \\ &= 3.40 \\[3ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\ &= 14 - 3.40 \\ &= 10.60 \end{align*}\]


Timed Assessments


pH and pOH

VIDEO

Answers
  1. [H3O+] = 2.75 × 10–3
  2. [H3O+] = 8.91 × 10–3
  3. [OH] = 1.58 × 10–2

pH and pOH 2

VIDEO

Answers
  1. pH = 0.37; pOH = 13.63
  2. pH = 13.54; pOH = 0.46
  3. pH = 13.32; pOH = 0.68
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