Exemplar Problems
Problem 5 : A solution contains 0.1 M NH4+ and 0.05 MNH3. Calculate the pH of this solution. Given Kb for NH4+ 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 Kb for NH4+ is $$(1.8 \times 10^{-5}) M,$$ we can set up an ICE table:
Initial: $$[NH_4^+] = 0.1 M$$ $$[NH_3] = 0.05 M$$ $$[H_3O^+] = 0 M$$
Change: $$[NH_4^+] decreases by (-x)$$ $$[NH_3] increases by (x)$$ $$[H_3O^+] increases by (x)$$
Equilibrium: $$[NH_4^+] - (x) M$$ $$[NH_3] + (x) M$$ $$[H_3O^+] + (x) M$$
Using the Kb expression: $$[K_b = \frac{[NH_3][H_3O^+]}{[NH_4^+]} = \frac{(x)(x)}{0.1 - x}]$$
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.
BETA
