Ionic Equilibrium Part 2

Concept:A basic buffer contains a weak base and its salt with a strong acid.Its pH is given by the equation $\text{pOH} = \text{p}K_b + \log\frac{[\text{salt}]}{[\text{base}]}$.Explanation:Required $\text{pH} = 9.25$, so $\text{pOH} = 14 - 9.25 = 4.75$.Given $\text{p}K_b(\mathrm{NH_4OH}) = 4.75$.Substitute: $4.75 = 4.75 + \log\frac{[\text{salt}]}{[\text{base}]}$.Thus $\log\frac{[\text{salt}]}{[\text{base}]} = 0$ → $\frac{[\text{salt}]}{[\text{base}]} = 1$.This means equal moles of salt and base must be present after mixing.Reaction: $\mathrm{NH_4OH} + \mathrm{HCl} \rightarrow \mathrm{NH_4Cl} + \mathrm{H_2O}$.Let initial moles of base = $n_b$ and moles of HCl = $n_a$.After reaction: moles of salt formed = $n_a$, moles of base left = $n_b - n_a$.Condition $[\text{salt}]/[\text{base}] = 1$ gives $n_a = n_b - n_a$ → $n_b = 2n_a$.Check each option by calculating $n_b$ and $n_a$:• Option B: $n_b = 0.2 \times 0.5 = 0.10$ mol, $n_a = 0.1 \times 0.5 = 0.05$ mol → $n_b = 2n_a$ holds.Thus after reaction: salt = $0.05$ mol, base left = $0.05$ mol → ratio = 1, giving $\text{pOH} = \text{p}K_b = 4.75$ and $\text{pH} = 9.25$.Shortcut:For $\text{pH} = \text{p}K_a$ (or $\text{pOH} = \text{p}K_b$), the buffer requires equal moles of salt and base.From reaction stoichiometry, initial base moles must be twice the acid moles. So directly compute $n_b$ and $n_a$ for each option; only option B satisfies $n_b = 2n_a$.Answer:Option B