The base imidazole has a of at 25°C. In what volumes should 0.02 M – HCl and 0.02 M imidazole be mix — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Given $K_b = 10^{-7}$, $pK_b = 7$, meaning $pK_a = 7$. For a buffer with $\text{pH} = 7$, we need $[\text{Salt}] = [\text{Base}]$. Let $V$ be the volume of imidazole. Volume of HCl $= 120 - V$. The strong acid reacts with the base to form salt. Moles of salt formed $= (120 - V) \times 0.02$. Moles of unreacted base $= V \times 0.02 - (120 - V) \times 0.02 = (2V - 120) \times 0.02$. Setting them equal: $120 - V = 2V - 120 \implies 3V = 240 \implies V = 80\text{ ml}$ (imidazole). Thus, volume of HCl is $40\text{ ml}$.