Salt Hydrolysis
Acid-base neutralization reactions are defined as an acid + base reacting to give water and a salt written generally as
\[ \mathrm{\textcolor{Red}{HA}}(aq) + \mathrm{\textcolor{green}{BOH}}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{\textcolor{Peach}{B}\textcolor{TealBlue}{A}}(aq)\]
where AB is the salt written in molecular form.
In the ionic form, the salt would be written as
\[\mathrm{\textcolor{Peach}{B}\textcolor{TealBlue}{A}} (aq) \longrightarrow \textcolor{Peach}{\mathrm{B^+}}(aq) + \textcolor{TealBlue}{\mathrm{A^-}}(aq)\]
The goal of this article is to address the question: “Will the components of a salt (the cation and/or the anion), react with water to change the pH of solution?”
Acidic Salt
Consider the reaction of the conjugate acid, B+, with water.
\[\textcolor{Peach}{\mathrm{B^+}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \textcolor{green}{\mathrm{B}}(aq)\] This reaction is equivalent to writing the following equation
\[\textcolor{Peach}{\mathrm{HB^+}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \textcolor{green}{\mathrm{B}}(aq)\]
The proton is now explicitly shown to illustrate the proton being donated to water. B+ and HB+ are the same (think of it as a NH4+ cation). The positive charge on B is due to the extra proton that is bound to it.
If the conjugate acid, HB+, is strong enough to react with water, it will produce H3O+ thereby lowering the pH of solution.
A salt that, when added to water, lowers the pH of the solution, is known as an acidic salt.
Basic salt
Consider the reaction of the conjugate base, A–, with water
\[ \textcolor{TealBlue}{\mathrm{A}^-}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^-} + \textcolor{Red}{\mathrm{HA}}\]
If the conjugate base, A–, is strong enough to react with water, it will produce OH–, thereby raising the pH of the solution.
A salt that, when added to water, raises the pH of the solution, is known as a basic salt.
Strong acid + strong base
A reaction between a strong acid and strong base gives a neutral solution.
\[\mathrm{H\textcolor{TealBlue}{Cl}}(aq) + \mathrm{\textcolor{Peach}{Na}OH}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{\textcolor{Peach}{Na}\textcolor{TealBlue}{Cl}}(aq)\]
Group 1A and 2A metals do not react with water to produce hydroxide. Na+, a group 1A metal, therefore will not undergo acid ionization and no OH– is produced and the pH does not change.
\[\mathrm{\textcolor{Peach}{Na^+}}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{no~reaction}\]
The anion, Cl–, does not undergo base ionization as it is too weak to react with water (its Kb << 1). The chloride anion is a conjugate base to a strong acid, and therefore, will never become HCl (strong acids dissociate completely). To illustrate the reaction that will not take place, we write the following
\[\mathrm{\textcolor{TealBlue}{Cl^-}}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{no~reaction}\]
Because neither component of the salt, NaCl, react with water, NaCl is a neutral salt.
Summary
- Group 1A and 2A metals do not react with water to change the pH (neutral cations)
- Conjugate bases of strong acids do not react with water to change the pH (neutral anions)
- Neutral salts do not affect the pH of a solution
Strong acid + weak base
A reaction between a strong acid and weak base gives a slightly acidic solution.
\[\mathrm{H\textcolor{TealBlue}{Cl}}(aq) + \mathrm{NH_3}(aq) \rightleftharpoons \mathrm{\textcolor{Peach}{NH_4}\textcolor{TealBlue}{Cl}}(aq)\]
The above reaction can be written out in two steps to illustrate the explicit production of both water and salt from this neutralization reaction.
\[\begin{align} \mathrm{NH_3}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{\textcolor{Peach}{NH_4}OH}(aq) \end{align}\] \[\begin{align} \mathrm{HCl}(aq) + \mathrm{\textcolor{Peach}{NH_4}OH}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{\textcolor{Peach}{NH_4}\textcolor{TealBlue}{Cl}}(aq) \end{align}\]
We have already discussed that Cl– will not react with water.
However, NH4+ will since it is a conjugate acid to a weak base.
The Ka of ammonium ion is less than 1. Since the following reaction will occur
\[\mathrm{\textcolor{Peach}{NH_4^+}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{\textcolor{green}{NH_3}}(aq)\]
NH4+ will cause the pH of the solution to decrease giving a slightly acidic solution.
Summary
- Conjugate acids to weak bases will react with water to lower the pH of solution
- Salts that lower the pH of solution are called acidic salts
- Cations that lower the pH of solution are conjugates to a weak base or small, high-charge metals (such as Al3+, Fe3+, Cr3+ and Sc3+) and are collectively called acidic cations.
Weak acid + strong base
A reaction between a weak and strong base gives a slightly basic solution.
\[\mathrm{H\textcolor{TealBlue}{F}}(aq) + \mathrm{\textcolor{Peach}{K}OH}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{\textcolor{Peach}{K}\textcolor{TealBlue}{F}}(aq)\]
We have already discussed that Group 1A and 2A metals do not react with water to change the pH of solution. Therefore, K+ does not affect the pH of solution. However, F– can react with water as it is a conjugate base to a weak acid, thereby raising the pH of the solution.
\[\mathrm{\textcolor{TealBlue}{F^-}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{\textcolor{red}{HF}} + \mathrm{OH^-}(aq)\]
Summary
- Conjugate bases to weak acids will react with water to raise the pH of solution
- Salts that raise the pH of solution are called basic salts
- Anions that are conjugates to weak acids (e.g. CN–, NO2–, CH3COO–) are called basic anions
Weak acid + weak base
In this scenario, one must examine the Ka and Kb values of the acid and base being mixed. Consider the following reaction
\[\mathrm{\textcolor{red}{HF}}(aq) + \mathrm{\textcolor{green}{NH_3}}(aq) \rightleftharpoons \mathrm{\textcolor{Peach}{NH_4}\textcolor{TealBlue}{F}}(aq)\]
Both components can react with water to change the pH of solution.
- F– is a conjugate base to the weak acid, HF. F– can react with water to produce OH– (raise pH). * NH4+ is a conjugate acid to a weak base, NH3. NH4+ can react with water to produce H3O+ (lower pH).
Since both components can raise/lower the pH of solution, which one will win? You have to compare the Ka of the acid to the Kb of the base.
- Ka(HF) = 3.5 × 10–4; therefore, Kb(F–) = 2.86 × 10–11
- Kb(NH3) = 1.76 × 10–5; therefore, Ka(NH4+) = 5.68 × 10–10
Since the Ka of NH4+ is greater than the Kb of F–, the acid (NH4+) will react slightly more with water to produce slightly more H3O+ giving a slightly acidic solution!
Summary
- If Ka of conjugate acid is greater than the Kb of the conjugate base, the solution will be slight acidic * If Ka of conjugate acid is less than the Kb of the conjugate base, the solution will be slight basic
Practice Problems
Question 1
Ammonium chloride, NH4Cl, is dissolved in pure water. Is the resulting solution acidic, basic, or neutral?
Solution
Acidic
NH4+ is an acidic cation that is a conjugate acid to a weak base, NH3. It is able to react with water to produce H3O+.
\[\mathrm{NH_4^+}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{NH_3}(aq)\]
Question 2
Sodium cyanide, NaCN, is dissolved in pure water. Is the resulting solution acidic, basic, or neutral?
Solution
Basic
CN– is a basic anion that is a conjugate base to a weak acid, HCN. It is able to react with water to produce OH–.
\[\mathrm{CN^-}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^-}(aq) + \mathrm{HCN}(aq)\]
Question 3
Potassium nitrate, KNO3, is dissolved in pure water. Is the resulting solution acidic, basic, or neutral?
Solution
Acidic
NO3– is an basic anion that is a conjugate base to a strong acid, HNO3 It will not react with water to produce OH–.
Question 4
What is the pH and pOH (to two decimal places) of a 50.0 mL 0.300 M KF aqueous solution (at 25 °C)? pKb(F–) = 10.83
Solution
Salt dissociates and delivers a weak base (F–).
\[\mathrm{NaF}(s) \longrightarrow \mathrm{Na^+}(aq) + \mathrm{F^-}(aq)\] This base is a conjugate to a weak acid (HF) and can therefore react with water to produce OH–.
\[\mathrm{F^-}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^-}(aq) + \mathrm{HF}(aq)\] Make ICE table. Proceeds right.
F–(aq) | + | H2O(l) | \(\rightleftharpoons\) | OH–(aq) | + | HF(aq) | |
---|---|---|---|---|---|---|---|
I | 0.300 | 0 | 0 | ||||
C | -x | +x | +x | ||||
E | 0.300 – x | x | x |
\[\begin{align*} K_{\mathrm{b}} &= 10^{-\mathrm{p}K_{\mathrm{b}}} \\ &= 10^{-10.83} \\ &= 1.48\times 10^{-11}\\[3ex] \dfrac{[\mathrm{OH^-}][\mathrm{HF}]}{[\mathrm{F^-}]} = K_{\mathrm{b}}\\[1.5ex] \dfrac{(x)(x)}{0.300-x} &= 1.48\times 10^{-11}\\[1.5ex] \dfrac{x^2}{0.300} &= 1.48\times 10^{-11}\\[1.5ex] x &= 2.11\times 10^{-6}\\[3ex] \% x &= \dfrac{2.11\times 10^{-6}}{0.300}\times 100\% = 7\times 10^{-4}\% \\[3ex] \mathrm{pOH} &= -\log(2.11\times 10^{-6}) \\ &= 5.68 \\[3ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH} \\ &= 14 - 5.68 \\ &= 8.32 \end{align*}\]
Timed Assessments
Acid/Base Equilibrium
Answers
- pH = 8.27; pOH = 5.73; basic
- pH = 8.39; pOH = 5.61; basic
- pH = 4.73; pOH = 9.07; acidic