An amount of 0.1 mole of () is mixed with 0.08 mole of HCl and diluted to one litre. What will be th — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The strong acid HCl completely reacts with the weak base $\text{CH}_3\text{NH}_2$ to form salt ($\text{CH}_3\text{NH}_3\text{Cl}$). Moles of salt formed = 0.08. Moles of base remaining = $0.1 - 0.08 = 0.02$. In 1 L, $[\text{Salt}] = 0.08\text{ M}$ and $[\text{Base}] = 0.02\text{ M}$. $[\text{OH}^-] = K_b \times \frac{[\text{Base}]}{[\text{Salt}]} = 5 \times 10^{-4} \times \left(\frac{0.02}{0.08}\right) = 1.25 \times 10^{-4}\text{ M}$. $[\text{H}^+] = \frac{10^{-14}}{1.25 \times 10^{-4}} = 8 \times 10^{-11}\text{ M}$.