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 is the pH of the solution at the equivalence point?

Answer 1

At the equivalence point, the moles of base is equal to the moles of acid, and so you have the salt with no acid or base remaining. To find the pH, one then needs to look at the hydrolysis of the salt that is formed.

B + H⁺ ==> BH⁺

moles B = 30.0 ml x 1 L / 1000 ml x 0251 mols / L = 0.00753 mols B

At the equivalence point, 0.00753 mols H⁺ have been added, and 0.00753 mols BH⁺ have been formed:

B + H⁺ ==> BH⁺

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

-0.00753....-0.00753...+0.00753...Change

0...................0...........0.00753.....Equilibrium

To find final volume, we need to find volume of HI that contains 0.00753 mols.

0.00753 mols HI x 1 L / 0.150 mols = 0.0502 L = 50.2 mls

Final volume = 30.0 mls + 50.2 mls = 80.2 mls = 0.0802 L

Final [BH⁺] = 0.00753 mols / 0.0802 L = 0.0939 M BH⁺

Now look at hydrolysis of this salt:

BH⁺ + H₂O ==> B + H₃O⁺

Ka for BH⁺ = 1x10^-14 / Kb = 1x10^-14 / 1.3x10^-10 = 7.69x10^-5

Ka = [B][H₃O⁺] / [BH⁺]

7.69x10^-5 = (x)(x) / 0.0939 - x (ignore x in denominator to avoid using quadratic)

7.69x10^-5 = x² / 0.0939

x² = 7.22x10^-6

x = 2.69x10^-3 M = [H₃O⁺] (note: this is less than 3% of the 0.0939 so above assumption was valid)

pH = -log [H₃O⁺] = -log 2.69x10^-3

pH = 2.57