Why is sodium acetate solution basic — explain salt hydrolysis

Question

Why is an aqueous solution of sodium acetate (CH₃COONa) basic? Explain using the concept of salt hydrolysis.

Solution — Step by Step

Sodium acetate (CH₃COONa) is a salt formed by neutralisation of:

Acetic acid (CH₃COOH) — a weak acid ( $K_a = 1.8 \times 10^{-5}$ )
Sodium hydroxide (NaOH) — a strong base

So sodium acetate is a salt of a weak acid and a strong base.

Sodium acetate is a strong electrolyte — it fully dissociates in water:

\text{CH}_3\text{COONa} \rightarrow \text{CH}_3\text{COO}^- + \text{Na}^+

The $\text{Na}^+$ ion comes from a strong base (NaOH) and has essentially no tendency to react with water. It is a spectator ion.

The $\text{CH}_3\text{COO}^-$ (acetate) ion comes from a weak acid. It has a significant tendency to react with water.

The acetate ion is the conjugate base of a weak acid. It acts as a base towards water:

\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-

This reaction is called salt hydrolysis (specifically, anionic hydrolysis — the anion hydrolyses).

The acetate ion accepts a proton from water, regenerating acetic acid and releasing hydroxide ion (OH⁻).

The accumulation of OH⁻ makes the solution basic (alkaline).

The degree of hydrolysis and the pH of the solution can be estimated. For the acetate ion:

K_h = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}

For a 0.1 M sodium acetate solution:

[\text{OH}^-] = \sqrt{K_h \times C} = \sqrt{5.56 \times 10^{-10} \times 0.1} = \sqrt{5.56 \times 10^{-11}} \approx 7.45 \times 10^{-6} \text{ M}

\text{pOH} = -\log(7.45 \times 10^{-6}) \approx 5.13

\text{pH} = 14 - 5.13 \approx 8.87

The pH is greater than 7, confirming the solution is basic.

Why This Works

The key principle: salts of weak acids and strong bases give basic solutions. This is because the conjugate base of a weak acid is a relatively strong base — it has significant affinity for protons from water.

Conversely: salts of strong acids and weak bases (e.g., NH₄Cl) give acidic solutions — the cation (NH₄⁺) donates a proton to water, generating H₃O⁺.

Salts of strong acid + strong base (e.g., NaCl): neutral (neither ion hydrolyses).

Salts of weak acid + weak base (e.g., CH₃COONH₄): depends on relative $K_a$ and $K_b$ values.

The relationship $K_h = K_w / K_a$ shows that the weaker the parent acid, the stronger the conjugate base, and the more extensive the hydrolysis — a larger $K_h$ means a higher pH for the same concentration.

Alternative Method

A quick memory rule for salt hydrolysis: “The weaker parent wins.” If the acid parent is weaker than the base parent, the salt solution is basic. If the base parent is weaker, the salt solution is acidic.

For sodium acetate: acid parent = acetic acid (weak, $K_a = 1.8 \times 10^{-5}$ ); base parent = NaOH (strong). Weak acid “wins” → basic solution.

Common Mistake

Students sometimes say “sodium acetate dissolves in water to give a neutral solution because it’s a salt.” This is wrong — not all salt solutions are neutral. Only salts of strong acids with strong bases (like NaCl, KBr, NaNO₃) give neutral solutions. The nature of the parent acid and base always determines whether the salt solution is acidic, basic, or neutral.

In JEE Main, salt hydrolysis questions often give you the $K_a$ and $K_b$ of the parent acid/base and ask for the pH of a salt solution. The key formulas to know: pH of salt of weak acid + strong base = $\frac{1}{2}(pK_w + pK_a + \log C)$ , and pH of salt of strong acid + weak base = $\frac{1}{2}(pK_w - pK_b - \log C)$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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