A 40.0 ml solution of weak base, BOH is titrated with 0.1 N - solution. The pH of the solution is found to be 10.0 and 9.0 after adding 5.0 ml and 20.0 ml of the acid, respectively. The dissociation c

Question

A 40.0 ml solution of weak base, BOH is titrated with 0.1 N - $HCl$ solution. The pH of the solution is found to be 10.0 and 9.0 after adding 5.0 ml and 20.0 ml of the acid, respectively. The dissociation constant of the base is (log 2 = 0.3)

Prashant Kotian · Accepted Answer

Correct answer: A. Let n be the initial millimoles of BOH. Addition of reacts with BOH: BOH + H+ -> B+ + . 1) At 5 mL addition: millimoles of B+ formed = 5 × 0.1 = 0.5 mmol. Remaining BOH = n - 0.5. Since pH = 10.0 => pOH = 14.0 - 10.0 = 4.0. pOH = pKb + log([B+] / [BOH]) => 4.0 = pKb + log(0.5 / (n - 0.5)). 2) At 20 mL addition: millimoles of B+ formed = 20 × 0.1 = 2.0 mmol. Remaining BOH = n - 2.0. Since pH = 9.0 => pOH = 14.0 - 9.0 = 5.0. pOH = pKb + log(2.0 / (n - 2.0)). Subtracting equation 1 from equation 2: 1.0 = log[(2.0 × (n - 0.5)) / (0.5 × (n - 2.0))] => 10 = 4 × (n - 0.5) / (n - 2.0) => 10n - 20 = 4n - 2 => 6n = 18 => n = 3.0 mmol. Substituting: 4.0 = pKb + log(0.5 / 2.5) = pKb - 0.70 => pKb = 4.70. Kb = 10^-4.70 = 2.0 × 10^-5.

💡 Solution & Explanation

Practice 22,000+ questions like this