10.3: Hydrolysis of Salt Solutions
By the end of this section, you will be able to:
- Predict whether a salt solution will be acidic, basic, or neutral
- Calculate the concentrations of the various species in a salt solution
- Describe the acid ionization of hydrated metal ions
Salts with Acidic Ions
Salts are ionic compounds composed of cations and anions, either of which may be capable of undergoing an acid or base ionization reaction with water. Aqueous salt solutions, therefore, may be acidic, basic, or neutral, depending on the relative acid-base strengths of the salt's constituent ions. For example, dissolving ammonium chloride in water results in its dissociation, as described by the equation
The ammonium ion is the conjugate acid of the base ammonia, NH 3 ; its acid ionization (or acid hydrolysis) reaction is represented by
NH 4 + ( aq ) + H 2 O( l ) ⇌ H 3 O + ( aq ) + NH 3 ( aq )
Since ammonia is a weak base, K b is measurable and K a > 0 (ammonium ion is a weak acid).
The chloride ion is the conjugate base of hydrochloric acid, and so its base ionization (or base hydrolysis ) reaction is represented by
Since HCl is a strong acid, K a is immeasurably large and K b ≈ 0 (chloride ions don’t undergo appreciable hydrolysis).
Thus, dissolving ammonium chloride in water yields a solution of weak acid cations ( ) and inert anions (Cl − ), resulting in an acidic solution.
Example 14.15
Calculating the pH of an Acidic Salt Solution
Aniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of anilinium chloride
\[ \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3{ }^{+}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^+(a q)+\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2(a q) \]
Solution
The K a for anilinium ion is derived from the K b for its conjugate base, aniline (see Appendix H ):
\[ K_{\mathrm{a}}=\dfrac{K_{\mathrm{w}}}{K_{\mathrm{b}}}=\dfrac{1.0 \times 10^{-14}}{4.3 \times 10^{-10}}=2.3 \times 10^{-5} \]
Using the provided information, an Equilibrium table for this system is prepared:
| C 5 H 5 NH 3 + | H 2 O | C 6 H 5 NH 2 | H 3 O + | |
| Equlibrium Concentration (M) | 0.233 | X | X |
Substituting these equilibrium concentration terms into the K a expression gives
\[ \mathrm{K}_a=\dfrac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \right] \left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3^{+}\right]} \]
\[ \mathrm{2.3 \times 10^{-5}}=\dfrac{\mathrm{X}^2}{0.233} \]
X = 0.002
The ICE table defines x as the hydronium ion molarity, and so the pH is computed as
Check Your Learning
What is the hydronium ion concentration in a 0.100- M solution of ammonium nitrate, NH 4 NO 3 , a salt composed of the ions and Which is the stronger acid orAnswer:
[H 3 O + ] = 7.5 10 −6 M ; is the stronger acid.
Salts with Basic Ions
As another example, consider dissolving sodium acetate in water:
The sodium ion does not undergo appreciable acid or base ionization and has no effect on the solution pH. This may seem obvious from the ion's formula, which indicates no hydrogen or oxygen atoms, but some dissolved metal ions function as weak acids, as addressed later in this section.
The acetate ion, is the conjugate base of acetic acid, CH 3 CO 2 H, and so its base ionization (or base hydrolysis ) reaction is represented by
Because acetic acid is a weak acid, its K a is measurable and K b > 0 (acetate ion is a weak base).
Dissolving sodium acetate in water yields a solution of inert cations (Na + ) and weak base anions resulting in a basic solution.
Example 14.16
Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base
Determine the acetic acid concentration in a solution with and [OH − ] = 2.5 10 −6 M at equilibrium. The reaction is:
NaCH 3 CO 2 ( s ) ⇋ Na + ( aq ) + CH 3 CO 2 − ( aq )
Solution The provided equilibrium concentrations and a value for the equilibrium constant will permit calculation of the missing equilibrium concentration. The process in question is the base ionization of acetate ion, for which
\[ K_{\mathrm{b}}\left( \mathrm{CH}_3 \mathrm{CO}_2^{-}\right) = \dfrac{K_{\mathrm{w}}}{K_{\mathrm{a}}}=\frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}}=5.6 \times 10^{-10} \]
Substituting the available values into the K b expression gives
\[\begin{gathered}K_{\mathrm{b}}=\frac{\left[\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{CH}_3 \mathrm{CO}_2^{-}\right]}=5.6 \times 10^{-10} \\ =\frac{\left[\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}\right]\left(2.5 \times 10^{-6}\right)}{(0.050)}=5.6 \times 10^{-10}\end{gathered} \]
Solving the above equation for the acetic acid molarity yields [CH 3 CO 2 H] = 1.1 10 −5 M .
Check Your Learning
What is the pH of a 0.083- M solution of NaCN?Answer:
11.11
Salts with Acidic and Basic Ions
Some salts are composed of both acidic and basic ions, and so the pH of their solutions will depend on the relative strengths of these two species. Likewise, some salts contain a single ion that is amphiprotic, and so the relative strengths of this ion’s acid and base character will determine its effect on solution pH. For both types of salts, a comparison of the K a and K b values allows prediction of the solution’s acid-base status, as illustrated in the following example exercise.
Example 14.17
Determining the Acidic or Basic Nature of Salts
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:(a) KBr
(b) NaHCO 3
(c) Na 2 HPO 4
(d) NH 4 F
Solution
Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here:(a) The K + cation is inert and will not affect pH. The bromide ion is the conjugate base of a strong acid, and so it is of negligible base strength (no appreciable base ionization). The solution is neutral.
(b) The Na + cation is inert and will not affect the pH of the solution; while the anion is amphiprotic. The K a of is 4.7 10 −11 ,and its K b is
Since K b >> K a , the solution is basic.
(c) The Na + cation is inert and will not affect the pH of the solution, while the anion is amphiprotic. The K a of is 4.2 10 −13 ,
and its K b is Because K b >> K a , the solution is basic.
(d) The ion is acidic (see above discussion) and the F − ion is basic (conjugate base of the weak acid HF). Comparing the two ionization constants: K a of is 5.6 10 −10 and the K b of F − is 1.6 10 −11 , so the solution is acidic, since K a > K b .
Check Your Learning
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:(a) K 2 CO 3
(b) CaCl 2
(c) KH 2 PO 4
(d) (NH 4 ) 2 CO 3
Answer:
(a) basic; (b) neutral; (c) acidic; (d) basic