Ethanoic acid has pKa = 4.76. A solution has a concentration of 0.10 mol/L…

$K_a = 10^{-\text{p}K_a}$. For a weak acid HA at concentration $c$, $K_a = \dfrac{x^2}{c - x}$, so $x = [\text{H}^+] = \dfrac{-K_a + \sqrt{K_a^2 + 4K_a c}}{2}$, and $\text{pH} = -\log x$. The approximation $x \approx \sqrt{K_a c}$ holds only when the acid is ionised by a few per cent. Withdrawing groups make an acid stronger, and the effect falls with distance. In a titration, moles of base $= c \times V$.