Question

At 25$^\circ$C, the percentage of ionization of x M acetic acid is 4.242. What is the pH of the acetic acid solution?
Given: $\log 4.242 = 0.6275$, $\log 0.04242 = -1.372$, $K_a = 1.8 \times 10^{-5}$

• A. 3.37
• B. 1.70
• C. 1.37
• D. 2.37

Hint:

To calculate pH of weak acids, use degree of ionization and $K_a$ relation: $K_a = C\alpha^2$.

Answer 1

The Correct Option is A

Given $%$ ionization = 4.242 means $\alpha = 0.04242$
Let initial concentration = $C$ M
Then $[H^+] = C \cdot \alpha = x \cdot 0.04242$
Also, $K_a = C\alpha^2 = 1.8 \times 10^{-5}$
Solve for $C$: $C = \dfrac{K_a}{\alpha^2} = \dfrac{1.8 \times 10^{-5}}{(0.04242)^2}$
$\Rightarrow C \approx 0.01$ M (approx.)
Then $[H^+] = 0.01 \cdot 0.04242 = 4.242 \times 10^{-4}$
pH = $-\log [H^+] = -\log (4.242 \times 10^{-4}) = -(\log 4.242 + \log 10^{-4}) = -(0.6275 - 4) = 3.3725 \approx 3.37$