Q.Ammonium acetate, $\text{CH}_3\text{COONH}_4$, is the salt of a weak acid and a weak base with almost equal $K_a$ and $K_b$ values ($1.8 \times 10^{-5}$ each). Using the relation $\text{pH} = 7 + \tfrac{1}{2}(\text{p}K_a - \text{p}K_b)$, show that its aqueous solution is very nearly neutral, and explain in words why this differs from the hydrolysis of $\text{CH}_3\text{COONa}$.

Sodium formate is the salt of a weak acid and a strong base, so its solution is basic (hydrolysis of the formate anion); the standard salt-hydrolysis formula for exactly this weak-acid/strong-base case gives $pH$ in terms of $pK_a$ of the acid and the salt's concentration.

Sodium formate, $HCOONa$, dissociates completely to give $Na^+$ and $HCOO^-$ (formate ion, the conjugate base of the weak acid formic acid, $HCOOH$). The formate ion undergoes hydrolysis in water: $HCOO^- + H_2O \rightleftharpoons HCOOH + OH^-$ making the solution basic (pH $>7$).

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HCOONa is basic, C6H5NH3Cl is acidic, and KCN is basic in aqueous solution.

• HCOONa is the salt of a weak acid (HCOOH, formic acid) and a strong base (NaOH). The formate ion hydrolyses to give a basic solution.
• C6H5NH3Cl (anilinium chloride) is the salt of a weak base (aniline, C6H5NH2) and a strong acid (HCl). The anilinium ion hydrolyses to give an acidic solution. …

A salt hydrolyses only if it comes from a weak acid and/or a weak base; sodium chloride (strong acid + strong base) is the only option among the four whose ions do not react with water.

Salt hydrolysis is the reaction of a salt's constituent ion(s) with water to regenerate the (weak) parent acid or base, altering the solution's pH. Sodium chloride is the salt of a strong acid (HCl) and a strong base (NaOH); neither $Na^+$ nor $Cl^-$ has any appreciable tendency to react with water (their conjugate acid/base, NaOH and HCl, are both fully dissociated strong electrolytes), so NaCl solutions remain neutral — it does not hydrolyse. Sodium formate (from weak formic acid + strong NaOH) hydrolyses to give a basic solution, since the formate ion accepts a proton from …

Methyl orange is used because the HCl-Na2CO3 equivalence point is acidic, not alkaline.

When HCl (a strong acid) is titrated against $Na_2CO_3$ (a salt that behaves as a weak base in water), the overall reaction is:

$Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$

At complete neutralisation, the solution contains $CO_2$ dissolved as carbonic acid ($H_2CO_3$), which makes the equivalence-point pH mildly acidic (around 3.5-4), unlike a strong acid-strong base titration where the equivalence point is neutral (pH 7).

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