A 2.5 g impure sample containing weak monoacidic base (Molecular weight = 45) is dissolved in 100 ml — Ionic Equilibrium Chemistry Question
Question
A 2.5 g impure sample containing weak monoacidic base (Molecular weight = 45) is dissolved in 100 ml water and titrated with 0.5 M HCl at $25^\circ\text{C}$. When $1/5^{\text{th}}$ of the base was neutralized, the pH was found to be 9 and at equivalent point, pH of solution is 4.5.
💡 Solution & Explanation
At $1/5^{\text{th}}$ neutralization, the ratio $[\text{Salt}]/[\text{Base}] = 1/4$. Using $\text{pOH} = pK_b + \log(1/4)$, with $\text{pH}=9 \implies \text{pOH}=5$, gives $pK_b \approx 5.6$ and $K_b \approx 10^{-5.6}$, which is $> 10^{-6}$ (A is False). At equivalence, $\text{pH} = 7 - \frac{1}{2}pK_b - \frac{1}{2}\log C$. Substituting $\text{pH}=4.5$ and $pK_b=5.6$ gives $\log C = -0.6 \implies C = 0.25\text{ M}$ (B is True). Let $V$ be the volume of HCl. Equivalence concentration $C = \frac{0.5 V}{0.1 + V} = 0.25 \implies V = 0.1\text{ L} = 100\text{ ml}$ (C is True). Moles of base = $0.5 \times 0.1 = 0.05\text{ mol}$. Mass of base = $0.05 \times 45 = 2.25\text{ g}$. Mass \% = $(2.25 / 2.5) \times 100 = 90\%$ (D is False). Therefore, correct answer is B,C.