A solution of weak base, BOH is titrated with solution. The pH of the solution is found to be 10.0 a — Ionic Equilibrium Chemistry Question
Question
A $40.0 \text{ ml}$ solution of weak base, BOH is titrated with $0.1 \text{ N} – \text{HCl}$ solution. The pH of the solution is found to be 10.0 and 9.0 after adding $5.0 \text{ ml}$ and $20.0 \text{ ml}$ of the acid, respectively. The dissociation constant of the base is ($\log 2 = 0.3$)
💡 Solution & Explanation
Let total mmol of base = $a$. Adding $v$ ml of 0.1 N HCl neutralizes $0.1v$ mmol of base, forming $0.1v$ mmol of salt and leaving $(a - 0.1v)$ mmol of base. $\text{pOH} = pK_b + \log(\frac{0.1v}{a - 0.1v})$. At 5 ml, $\text{pH} = 10 \implies \text{pOH} = 4$: $4 = pK_b + \log(\frac{0.5}{a - 0.5})$. At 20 ml, $\text{pH} = 9 \implies \text{pOH} = 5$: $5 = pK_b + \log(\frac{2.0}{a - 2.0})$. Subtracting equations: $1 = \log(\frac{2.0(a - 0.5)}{0.5(a - 2.0)}) = \log(\frac{4a - 2}{a - 2}) \implies 10 = \frac{4a - 2}{a - 2} \implies 10a - 20 = 4a - 2 \implies 6a = 18 \implies a = 3.0 \text{ mmol}$. Substitute $a=3.0$ into first eq: $4 = pK_b + \log(0.5 / 2.5) = pK_b - \log 5$. Given $\log 5 = 1 - 0.3 = 0.7$, $pK_b = 4.7$. $K_b = 10^{-4.7} = 10^{0.3} \times 10^{-5} = 2 \times 10^{-5}$. Therefore, correct answer is A.