Exemplar Problems
Problem 3 : Determine the pH of a 0.02 M NH4OH solution. Given Kb for NH4OH is $$(1.8 \times 10^{-5}) M.$$
Solution :
Given Kb for NH4OH is $$(1.8 \times 10^{-5}) M,$$ we can set up an ICE table for the dissociation of NH4OH:
Initial: $$[NH_4OH] = 0.02 M$$ $$[NH_4^+] = 0 M$$ $$[OH^-] = 0 M$$
Change: $$[NH_4OH] decreases by (-x)$$ $$[NH_4^+] increases by (x)$$ $$[OH^-] increases by (x)$$
Equilibrium: $$[NH_4OH] - (x) M$$ $$[NH_4^+] + (x) M$$ $$[OH^-] + (x) M$$
Using the Kb expression: $$[K_b = \frac{[NH_4^+][OH^-]}{[NH_4OH]}]$$
Substitute the equilibrium concentrations: $$[1.8 \times 10^{-5} = \frac{(x)(x)}{0.02 - x}]$$
Since (x) is small compared to 0.02, we can approximate $$(0.02 - x) as (0.02):$$
$$[1.8 \times 10^{-5} = \frac{x^2}{0.02}]$$
Solve for (x): $$[x^2 = 1.8 \times 10^{-5} \times 0.02]$$ $$[x^2 = 3.6 \times 10^{-7}]$$ $$[x = \sqrt{3.6 \times 10^{-7}}]$$ $$[x \approx 6 \times 10^{-4}]$$
Now, we can calculate the OH- concentration: $$([OH^-] = x \approx 6 \times 10^{-4}) M$$
To find the (pOH): $$[pOH = -\log[OH^-] = -\log(6 \times 10^{-4})]$$ $$[pOH \approx 3.22]$$
To find the pH: $$[pH = 14 - pOH = 14 - 3.22]$$ $$[pH \approx 10.78]$$
BETA
