A salt is made of ions, and some ions are weak acids or weak bases. Using Ka × Kb = Kw and the weak-base method, you can predict and compute the pH of a salt solution.
Key idea
Ions from strong acids and strong bases ($\mathrm{Cl^-}$, $\mathrm{NO_3^-}$, $\mathrm{Na^+}$, $\mathrm{K^+}$) do not change the pH. The conjugate base of a weak acid ($\mathrm{F^-}$, $\mathrm{CH_3COO^-}$, $\mathrm{CN^-}$) makes the solution basic. The conjugate acid of a weak base ($\mathrm{NH_4^+}$) makes it acidic, and so do small, highly charged metal ions such as $\mathrm{Al^{3+}}$ and $\mathrm{Fe^{3+}}$.
Why it works
$\mathrm{Cl^-}$ is the conjugate base of HCl, a strong acid, so its $K_b$ is essentially zero: it does not take protons from water. $\mathrm{CH_3COO^-}$ is the conjugate base of a weak acid, so its $K_b = K_w/K_a$ is small but real, and it makes a little $\mathrm{OH^-}$. A metal ion such as $\mathrm{Al^{3+}}$ is surrounded by water molecules; its strong positive charge pulls on their O–H bonds, and one of those waters can release a proton: $$\mathrm{[Al(H_2O)_6]^{3+}} + \mathrm{H_2O} \rightleftharpoons \mathrm{[Al(H_2O)_5(OH)]^{2+}} + \mathrm{H_3O^+}$$
Worked examples
Worked example: A basic salt
Find the pH of $0.10\ \text{M}$ $\mathrm{NaCH_3COO}$ ($K_a$ of acetic acid $= 1.8 \times 10^{-5}$).
- $\mathrm{Na^+}$ is a spectator. Acetate is a weak base with $K_b = \dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$.
- $$[\mathrm{OH^-}] = \sqrt{(5.6 \times 10^{-10})(0.10)} = 7.5 \times 10^{-6}\ \text{M}$$; $\text{pOH} = 5.13$.
- $\text{pH} = 14.00 - 5.13 = 8.87$.
Answer: $\text{pH} = 8.87$.
Worked example: An acidic salt
Find the pH of $0.20\ \text{M}$ $\mathrm{NH_4Cl}$ ($K_b$ of $\mathrm{NH_3} = 1.8 \times 10^{-5}$).
- $\mathrm{Cl^-}$ is a spectator. $\mathrm{NH_4^+}$ is a weak acid: $K_a = \dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$.
- $$[\mathrm{H_3O^+}] = \sqrt{(5.6 \times 10^{-10})(0.20)} = 1.05 \times 10^{-5}\ \text{M}$$
- $\text{pH} = 4.98$.
Answer: $\text{pH} = 4.98$.
| Salt | Cation | Anion | Solution |
|---|---|---|---|
| $\mathrm{NaCl}$ | neutral | neutral | neutral |
| $\mathrm{KF}$ | neutral | weak base | basic |
| $\mathrm{NH_4NO_3}$ | weak acid | neutral | acidic |
| $\mathrm{AlCl_3}$ | acidic (hydrated ion) | neutral | acidic |
| $\mathrm{NH_4CN}$ | weak acid, $K_a = 5.6 \times 10^{-10}$ | weak base, $K_b = 2.0 \times 10^{-5}$ | basic ($K_b > K_a$) |
For a salt whose cation and anion both react, compare the two constants: the ion with the larger $K$ wins. In $\mathrm{NH_4CN}$, cyanide's $K_b$ is far larger than ammonium's $K_a$, so the solution is basic.
Common mistake
Thinking every salt solution is neutral because salts come from neutralization. Neutralizing a weak acid with a strong base leaves the weak acid's conjugate base, which reacts with water. Check: ask for each ion, "is it the partner of a weak acid or base?"