Exemplar Problems

Problem 5 : A solution contains 0.1 M NH₄⁺ and 0.05 M NH₃. Calculate the pH of this solution. Given K_b for NH₃ is $$1.8 \times 10^{-5}) M.

Solution :

We need to consider the hydrolysis of NH₄⁺ in the presence of NH₃ :

$$[NH_4^+ + H_2O \right leftharpoons NH_3 + H_3O^+]$$

Given K_b for NH₄⁺ is $$1.8 \times 10^{-5}$$, we can set up an ICE table:

Initial: $$[NH_4^+] = 0.1 M$$ $$[NH_3] = 0.05 M$$ $$[H_3O^+] = 1 \times 10^{-7} M$$

Change: $$[NH_4^+] decreases by (-x)$$ $$[NH_3] increases by (x)$$ $$[H_3O^+] increases by (x)$$

Equilibrium: $$[NH_4^+] - (x) M$$ $$[NH_3] + x\ \text{M}$$ $$[H_3O^+] + (x) M$$

Using the K_b expression: $$[K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = \frac{(x)(x)}{0.1 - x}]$$

Since (x) is small compared to 0.1, we can approximate (0.1 - x) as (0.1):

$$[1.8 \times 10^{-5} = \frac{x^2}{0.1}]$$

Solve for (x): $$[x^2 = 1.8 \times 10^{-5} \times 0.1]$$ $$[x^2 = 1.8 \times 10^{-6}]$$ $$[x = \sqrt{1.8 \times 10^{-6}}]$$ $$[x \approx 1.34 \times 10^{-3}]$$

Now, we can calculate the H₃O⁺ concentration and the pH: $$[H_3O^+] = x \approx 1.34 \times 10^{-3}] M$$ $$[pH = -\log[H_3O^+] = -\log(1.34 \times 10^{-3})]$$ $$[pH \approx 2.87]$$

The pH of the solution is approximately 2.87.