First, let HA represent acetic acid, CH₃COOH.

For acetic acid, CH₃COOH
K_a = 1.80 × 10^-5

from RICE table: \(\displaystyle K_{\rm a} = {\rm [H_3O^+][A^-]\over [HA]} = {x^2 \over 0.01 – x} \)

\(K_{\rm a}(0.01) - K_{\rm a}(x) = x^2 \)

\(0 = x^2 + K_{\rm a}(x) - K_{\rm a}(0.01) \)

\(\displaystyle x = {-K + \sqrt{K^2 + 4K(0.01)}\over 2}\)

\(\displaystyle x = 4.1536 \times 10^{-4} \)

using assumption: \(x = \sqrt{K_{\rm a}(0.01 )} = 4.2426 \times 10^{-4}\)
which is a percent difference of pH = 2.14 %
this is a fairly good match to the actual answer

For 0.01 M acid solution
where K_a = 1.80 × 10^-5

[H₃O⁺] = 4.15 × 10^-4 M

[OH^–] = 2.41 × 10^-11 M

pH = 3.38

pOH = 10.62