Question

What is the pH for a 3.75*10^-3 molar solution of butyric acid in water?

Ka for butyric acid = 1.51*10^-5

*please show steps! thanks

the answer given by my teacher is: pH=3.33

Answer 1

                   CH₃CH₂CH₂COOH + H₂O <===> CH₃CH₂CH₂COO^- + H₃O⁺

initial                3.75 x 10^-3                                          0                 0

equilibroum     (3.75 x 10^-3 - x)                                     x                 x

K_a = [CH₃CH₂CH₂COO^-][H₃O⁺] / [CH₃CH₂CH₂COOH]

1.51 x 10^-5 = x² / (3.75 x 10^-3 - x)                   *assume 3.75 x 10^-3 >> x

x² = 5.66 x 10^-8

x = 2.37 x 10^-4 M = [H₃O⁺]

pH = -log[H⁺] = -log(2.37 x 10^-4)

pH = 3.62