Exemplar Problems
Problem 4 : Calculate the pH of a 0.1 M CH3COOH (acetic acid) solution. Given Ka for acetic acid is $$(1.8 \times 10^{-5}).$$
Solution :
Given Ka for acetic acid is $$(1.8 \times 10^{-5})$$, we can set up an ICE table for the dissociation of acetic acid CH3COOH:
Initial: $$[CH_3COOH] = 0.1 M$$ $$[H^+] = 0 M$$ $$[CH_3COO^-] = 0 M$$
Change: $$[CH_3COOH] decreases by (-x)$$ $$[H^+] increases by (x)$$ $$[CH_3COO^-] increases by (x)$$
Equilibrium: $$[CH_3COOH] - (x) M$$ $$[H^+] + (x) M$$ $$[CH_3COO^-] + (x) M$$
Using the Ka expression: $$[K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}]$$
Substitute the equilibrium concentrations: $$[1.8 \times 10^{-5} = \frac{(x)(x)}{0.1 - x}]$$
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 H+ concentration and the pH: $$[H^+] = x \approx 1.34 \times 10^{-3}] M$$ $$[pH = -\log[H^+] = -\log(1.34 \times 10^{-3})]$$ $$[pH \approx 2.87]$$
BETA
