Salt Hydrolysis & Common Ion Effect

Explanation:

${\text{NH}}_4\text{Cl}$ is a salt of a strong acid ($\text{HCl}$) and a weak base (${\text{NH}}_3$). It undergoes cationic hydrolysis.

First, calculate the hydrolysis constant ($K_h$):
$K_h={\displaystyle \frac{K_w}{K_b}}={\displaystyle \frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}}=5.56\times {10}^{-10}$

Calculate the hydrogen ion concentration $[{\text{H}}^{+}]$:
$[{\text{H}}^{+}]=\sqrt{K_h\cdot C}=\sqrt{5.56\times {10}^{-10}\times 0.1}=\sqrt{5.56\times {10}^{-11}}$
$[{\text{H}}^{+}]\approx 7.45\times {10}^{-6}\text{M}$

Calculate the pH:
$\text{pH}=-\log (7.45\times {10}^{-6})=6-\log (7.45)$
$\text{pH}\approx \mathbf{5.13}$ (The solution is acidic, as expected for a strong acid + weak base salt).

Answer: C) ${\text{Na}}_2{\text{CO}}_3$.

Explanation: ${\text{Na}}_2{\text{CO}}_3$ is the salt of a weak acid (carbonic acid, ${\text{H}}_2{\text{CO}}_3$) and a strong base ($\text{NaOH}$). The carbonate ion undergoes anionic hydrolysis, producing hydroxide ions:
${\text{CO}}_3^{2-}+{\text{H}}_2\text{O}\rightleftharpoons {\text{HCO}}_3^{-}+{\text{OH}}^{-}$

Contrast: $\text{NaCl}$ (strong acid + strong base) is neutral; ${\text{NH}}_4\text{Cl}$ (strong acid + weak base) is acidic; ${\text{AlCl}}_3$ is acidic because the ${\text{Al}}^{3+}$ ion acts as a Lewis acid and heavily hydrolyses in water.

Explanation:

For a salt undergoing hydrolysis, the degree of hydrolysis ($h$) is given by the formula $h=\sqrt{{\displaystyle \frac{K_h}{C}}}$. As the solution is diluted, the concentration ($C$) decreases. Since $h$ is inversely proportional to the square root of $C$, the value of $h$ mathematically increases.

Conceptually, this perfectly aligns with Le Chatelier's principle: adding more solvent (dilution) disturbs the hydrolysis equilibrium. The system responds by shifting in the forward direction (toward more hydrolysis) to partially restore equilibrium, increasing the fraction of ions that hydrolyse.

Explanation:

Ammonium acetate is a salt of a weak acid (${\text{CH}}_3\text{COOH}$) and a weak base (${\text{NH}}_3$). The pH of such a salt depends strictly on the dissociation constants, not on the concentration $C$.

Since $K_a=K_b$, it follows that $\text{p}{K}_a=\text{p}{K}_b=4.74$.

The standard formula for the pH of a WA+WB salt is:
$\text{pH}=7+{\displaystyle \frac{1}{2}}(\text{p}{K}_a-\text{p}{K}_b)$

$\text{pH}=7+{\displaystyle \frac{1}{2}}(4.74-4.74)=7+0=\mathbf{7}$

The solution is approximately neutral because both the cation and the anion hydrolyse to the exact same extent, producing equal amounts of $[{\text{H}}^{+}]$ and $[{\text{OH}}^{-}]$.

Explanation:

When a strong acid like $\text{HCl}$ is added, it dissociates completely to provide external ${\text{H}}^{+}$ ions. In the buffer mixture, these incoming protons are immediately consumed by the conjugate base (acetate ion) acting as a reservoir:

${\text{CH}}_3{\text{COO}}^{-}+{\text{H}}^{+}\to {\text{CH}}_3\text{COOH}$

Because the added ${\text{H}}^{+}$ is neutralized and converted into the weak, largely undissociated acid (${\text{CH}}_3\text{COOH}$), a massive drop in pH is prevented. The pH does decrease very slightly because the ratio of $[\text{salt}]/[\text{acid}]$ in the Henderson-Hasselbalch equation goes down, but the common ion effect successfully "buffers" the drastic change that would occur in pure water.