See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Kb = 10-5, pKb = 5 BOH + HCl  BCl + H2O At half neutralization, 50% of the base is converted to its salt, with strong acid HCl, it forms a basic buffer.     b salt pOH pK log base       salt pOH 5 log base   [salt] = [base] pOH = 5 pH = 14 – pOH = 9