Consider the titration of 50.0 mL of 0.310 M weak base B (Kb = 7.5 x 10⁻⁶) with 0.340 M HNO₃. What is the pH of the solution after the addition of 90.0 mL of HNO₃?

Answer 1

B + HNO₃ ==> BH⁺ + NO₃^- .. reaction of base B with acid HNO₃

moles B initially present = 50.0 ml x 1 L / 1000 ml x 0.325 mol / L = 0.0155 mols

moles of HNO₃ added = 90.0 ml x 0.340 mol / L = 0.0306 mols

Since there are more mols of acid (HNO₃) than of base (B), all of B will be converted to BH⁺ and there will still be excess HNO₃ (a strong acid) left over. The final volume = 50 ml + 90 ml = 140 ml = 0.140 L

[BH⁺] = 0.0155 mol / 0.140 L = 0.111 M

[H⁺] = 0.0306 mol - 0.0155 mol = 0.0151 mol / 0.140 L = 0.108 M

The Ka for BH⁺ = 1x10^-14 / 7.5x10^-6 = 1.33x10^-9. As such, it will not contribute significantly to the final [H⁺], and thus will not contribute significantly to the final pH. The final pH will be determined by the [H⁺] from HNO₃.

pH = -log [H⁺] = -log 0.108 M

pH = 0.967