Question

If the Ka of a monoprotic weak acid is 3.8 × 10-6, what is the pH of a 0.11 M solution of this acid?

Answer 1

costruct the ICE table

no need to cinsider H2O concentration since it is very large

            HA +         H2O       <----->      H3O+ + A-

I            0.11                                          0              0

C           -x                                              +x            +x

E            0.11-x                                      +x              +x

write the disociation expression

Ka = [H3O+][A-] / [HA]

3.8 × 10^-6 = [x] [x] / [0.11-x]

3.8 × 10^-6 = x² / [0.11-x]

x² + 3.8 × 10^-6x - 0.418 * 10^-6 = 0

solve the quadratic equation

x = 0.000644 = [H3O+]

pH = -log[H3O+]

= -log [0.000644]

= 3.19