Question

A 25.0-mL sample of 0.150 M butanoic acid is titrated with a 0.150 M NaOH solution. What is the pH before any base is added? The Ka of butanoic acid is 1.5 × 10-5.

Answer 1

HA ⇄ H⁺ + A⁻ 
so:
$\frac{[H^+][A^-]}{[HA]} = 1.5 x 10^{-5}$
and now:
$\frac{(x)(x)}{(0.150-x)}$ = 1.5 x 10⁻⁵
x is considered very small compared to 0.15
x² = 2.25 x 10⁻⁶
x = 1.5 x 10⁻³
So [H⁺] = 1.5 x 10⁻³
pH = - log [H⁺] = - log (1.5 x 10⁻³) = 2.83