Consider the titration of 50.0 mL of 0.315 M weak base B (Kb = 7.5 x 10⁻⁶) with 0.340 M HNO₃.

Question

What is the pH of the solution after the addition of 90.0 mL of HNO₃?

Answer 1

B + HNO₃ ==> BH⁺ + NO₃^-

moles B = 0.050 L x 0.315 mol/L = 0.01575 moles

moles HNO₃ = 0.090 L x 0.340 mol/L = 0.0306 moles

HNO₃ is in excess.

After reaction moles of HNO₃ left over = 0.0306 mol - 0.01575 mol = 0.01485 moles HNO₃

Final volume = 50.0 ml + 90.0 ml = 140.0 ml = 0.140 L

Final [HNO₃] = 0.01485 mol / 0.140 L = 0.106 M

pH = -log [H⁺] = -log 0.106

pH = 0.97