Finding pH from Ka

Question

If  of butanoic acid, ₄₈₂ is dissolved in enough water to make  of solution, what is the resulting pH?

Ka of butanoic acid = 1.5 x 10^-5

Answer 1

HA ==> H⁺+ A^-

^{Ka = [H+][A-]/[HA]}

8.2 g/88 g/mol = 0.093 mol/L

Ka =1.5x10^-5 = (x)(x)/0.093-x. Assume x is small and neglect it

x² = 1.4x10^-6

x = .0012 = [H⁺]

pH = -log [H⁺] = 2.93