Question

The pKa of the complex [Cr(H2O)6]Cl3 is 4.2. Calculate the pH of a 0.35 M solution of the complex ion. You MUST show all workings.

Answer 1

Write the balanced chemical equation for the dissociation of [Cr(H₂O)₆]Cl₃ as below.

[Cr(H₂O)₆]Cl₃ (aq) --------> H⁺ (aq) + [Cr(OH)(H₂O)₅]Cl₃^- (aq)

The acid dissociation constant is given as

K_a = [H⁺][Cr(OH)(H₂O)₅]Cl₃^-]/[Cr(H₂O)₆Cl₃] …..(1)

Let [H⁺] = x M; due to the 1:1 nature of ionization, [Cr(OH)(H₂O)₅]Cl₃^-] = x M and [Cr(H₂O)₆Cl₃] = (0.35 – x) M.

Again, pK_a = 4.2; therefore, K_a = antilog (-pK_a) = antilog (-4.2) = 6.3096*10^-6.

Since K_a is small, we will assume x << 0.35 M and hence [Cr(H₂O)₆Cl₃] ≈ 0.35 M. Therefore,

6.3096*10^-6 = x²/(0.35)

====> x² = 1.8027*10^-5

====> x = 0.0042459

Therefore, [H⁺] = 0.0042459 M and pH = -log [H⁺] = -log (0.0042459) = 2.3720 ≈ 2.372 (ans).