What is the pH of a 0.520 M solution of NaCN (Ka of HCN is 4.9 × 10⁻¹⁰)?

Answer 1

Hi Matthieu!

In order to solve this question, we need to view it from the perspective of base hydrolysis.

First, let's find the base hydrolysis constant:

k_w = k_a*k_b => k_b = k_w*k_a^-1

<=> 10^-14/(4.9x10^-10)

k_b = 2.04x10^-5

Now, let's set up our equilibrium reaction:

CN^- + H₂O <---------> HCN + OH^-

I 0.520M

Δ -x +x +x

Eq 0.520-x x x

Let's use our equilibrium expression to find [OH^-]:

K_b = [HCN][OH^-]/[CN^-] = (x)(x)/(0.520-x) = x²/(0.520-x) = 2.04x10^-5

x² = 2.04x10^-5(0.520-x)

x² = 1.06x10^-5 - (2.04x10^-5)x

x² + (2.04x10^-5)x - 1.06x10^-5 = 0

x = [OH^-] = 1.62x10^-3 M

Finally, we can calculate the pH of the solution:

pH + pOH = 14 => pH = 14 - pOH

=> pOH = -log[OH^-]

pH = 14 + log[OH^-]

<=> 14 + log(1.62x10^-3)

pH = 11.2

I hope this helps!

Cheers