Consider the titration of 30.0 mL of 0.251 M weak base B (Kb = 1.3 x 10⁻¹⁰) with 0.150 M HI. What would be the pH of the solution after the addition of 20.0 mL of HI?

Answer 1

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Let's look at what we have in the present problem. We have a weak base titrated with a strong acid.

B + H⁺ ==> BH⁺ (the I^- is a spectator ion)

Initial mols B = 30.0 ml x 1 L / 1000 ml x 0.251 mol / L = 0.00753 mols

Initial mols H⁺ = 20.0 ml x 1 L / 1000 ml x 0.150 mol / L = 0.003 mols

Setting up an ICE table, we have ...

B + H⁺ ==> BH⁺

0.00753....0.003.......0.......Initial

-0.003.....-0.003.....+0.003..Change

0.00453......0............0.003..Equilibrium

Final volume = 30 ml + 20 ml = 50 ml = 0.050 L

Final [B] = 0.00453 mol/0.05 L = 0.0906 M

Final [BH⁺] = 0.003 mol/0.05 L = 0.06 M

This creates a buffer of a weak base (B) and the conjugate acid (BH⁺). We can use the Henderson Hasselbalch equation to find the pH:

pH = pKa + log [B] / BH⁺]

KaKb = 1x10-¹⁴ so Ka = 1x10^-14 / Kb = 1x10^-14 / 1.3x10^-10 = 7.69x10^-5

pKa = -log Ka = 4.11

pH = 4.11 + log (0.0906 / 0.06) = 4.11 + 0.18

pH = 4.29