For the precise calculation of in a basic buffer solution formed by a weak base () and its conjugate — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The equilibrium is $BOH \rightleftharpoons B^+ + OH^-$. The dissociation constant $K_b = \frac{[B^+][OH^-]}{[BOH]}$. Rearranging for $[OH^-]$ gives $[OH^-] = K_b \times \frac{[BOH]}{[B^+]}$. Taking the negative logarithm of both sides yields $pOH = -\log K_b - \log\frac{[Base]}{[Salt]}$, which algebraically inverts to the standard Henderson-Hasselbalch form: $pOH = pK_b + \log\frac{[Salt]}{[Base]}$.