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. What is the pH of the solution after the addition of 30.0 mL of HNO₃?

Answer 1

Writing the reaction taking place, we have...

B + HNO₃ ==> BH⁺ + NO₃^-

Moles of weak base originally present = 0.050 L x 0.315 mol/L = 0.01575 moles

Moles of HNO3 added = 0.030 L x 0.340 mol/L = 0.0102 moles HNO3

Setting up an ICE table we can see what our final concentrations are...

B + HNO₃ ==> BH⁺ + NO₃^-

0.01574...0.0102.....0..........0........Initial

-0.0102...-0.0102....+0.0102.........Change

0.00554.....0............+0.0102........Equilibrium (note: we aren't interested in the NO₃^- as it's a spectator)

Final concentrations at equilibrium are:

[B] = 0.00554 M

[BH⁺] = 0.0102 M

This creates a BUFFER because we have a weak base (B) and the conjugate acid (BH⁺)

The pH of a basic buffer can be calculated from the Henderson Hasselbaclh equation;

pOH = pKb + log [conj.acid]/[base]

pKb = -log Kb = 5.12

pOH = 5.12 + log [0.0102]/[0.00554] = 5.12 + 0.27

pOH = 5.39

pH = 14 - pOH

pH = 8.61