Consider the titration of 50.0 mL of 0.320 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

Consider the reaction taking place:

B + HNO₃ ==> BH⁺ + NO₃^-

moles B present = 50.0 ml x 1 L / 1000 ml x 0.320 mol/L = 0.0160 mols B

moles HNO3 present = 90.0 ml x 1 L / 1000 ml x 0.340 mol/L = 0.0306 mols HNO3

After reaction, moles B = 0

After reaction, moles HNO3 = 0.0306 - 0.016 = 0.0146 mols

After reaction, moles BH⁺ = 0.0160 mols

NOTE: this solution does NOT make a buffer since there is no weak acid or weak base along with the conjugate pair. We have excess HNO₃ (a strong acid) in a final volume of 140 ml (0.140 L)

Final [HNO₃] = 0.0146 mols / 0.140 L = 0.104 M

pH = -log [H⁺] = - log 0.104

pH = 0.98