A basic buffer of pH is manufactured by dissolving of solid ammonium sulphate and of ammonia into en — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Given $pH = 9.26$, the $pOH = 14 - 9.26 = 4.74$. The Henderson equation gives $pOH = pK_b + \log\frac{[Salt]}{[Base]}$. Since $pOH = pK_b$, $\log\frac{[NH_4^+]}{[NH_3]} = 0$, meaning $[NH_4^+] = [NH_3] = \frac{0.1}{0.1\text{L}} = 1.0 \text{ M}$. $x \text{ moles}$ of $(NH_4)_2SO_4$ in $100 \text{ mL}$ gives an $[NH_4^+]$ of $\frac{2x}{0.1} \text{ M}$. Equating them: $\frac{2x}{0.1} = 1.0 \Rightarrow 2x = 0.1 \Rightarrow x = 0.05$. Therefore, $x \times 100 = 5$.