Step 1: Understanding the Question:
A solution that "turns blue litmus red" is acidic, meaning its pH is less than 7. We need to determine which of the given salts undergoes hydrolysis to produce an acidic aqueous solution.
Step 2: Key Formula or Approach:
The acidity or basicity of an aqueous salt solution depends on the relative strengths of the parent acid and base that neutralized to form the salt.
Strong Acid + Weak Base = Acidic Solution.
Weak Acid + Strong Base = Basic Solution.
Weak Acid + Weak Base = Depends on $K_a$ and $K_b$ values (If $K_a > K_b$, it is acidic).
Step 3: Detailed Explanation:
Let's systematically analyze the parent acid and base for each given salt:
(A) $\text{NH}_4\text{CN}$: Formed from HCN (a very weak acid) and $\text{NH}_4\text{OH}$ (weak base). Since the $K_b$ of $\text{NH}_4\text{OH}$ is greater than the $K_a$ of HCN, the resulting solution is slightly basic.
(B) $\text{NH}_4\text{F}$: Formed from HF (weak acid) and $\text{NH}_4\text{OH}$ (weak base). The $K_a$ of HF ($6.8 \times 10^{-4}$) is significantly greater than the $K_b$ of $\text{NH}_4\text{OH}$ ($1.8 \times 10^{-5}$). Therefore, the hydrolysis of the ammonium ion dominates, making the solution slightly acidic.
(C) $\text{CH}_3\text{COONa}$: Formed from $\text{CH}_3\text{COOH}$ (weak acid) and NaOH (strong base). This salt undergoes anionic hydrolysis to form a basic solution.
(D) $\text{CH}_3\text{COONH}_4$: Formed from $\text{CH}_3\text{COOH}$ (weak acid) and $\text{NH}_4\text{OH}$ (weak base). Since $K_a \approx K_b$ for these two specific parent compounds, the resulting solution is almost perfectly neutral.
Because $\text{NH}_4\text{F}$ is the only slightly acidic solution among the choices, it is the one that will turn blue litmus red.
Step 4: Final Answer:
The salt that turns blue litmus red is $\text{NH}_4\text{F}$, which corresponds to option (B).