Ionic Equilibrium

The reaction for the base can be written as BOH + HCl → BCl + $H_2O$ $B^{+}+H_2O$ ⇌ BOH + $H^{+}$ Initial concentration : c → 0 , 0 Equilibrium concentration : c - cα → cα , cα Volume of HCl used can be obtained as $M_1V_1(BOH)$ = $M_2V_2(HCl)$ $\frac{2}{5} \times 2.5$ = $\frac{2}{15} \times V_2$ $V_2$ = 7.5 ml Similarly, concentration of salt can be obtained as $M_1V_1(\text{salt})$ = $M_2V_2(BOH)$ $M_1$ × (2.5 + 7.5) = 2.5 × $\frac{2}{5} M_1$ = 0.1 M We know that $K_a \times K_b$ = $K_w$ $K_a$ = $\frac{c \alpha^2}{1-\alpha}$ = $\frac{K_w}{K_b}$. Solving, we get a = 0.27. $[H^{+}]$ = cα = 2.7 × $10^{-2}$ M