Question

Calculate the pH of a mixture that contains 0.19 M of HNO3 and 0.27 M of HC6H5O.

Answer 1

HC6H5O is a weak acid, so it will dissociate only partially

HNO3 is a strong acid, so expect 100% dissocition

initially

[HNO3] = 0.19 M

[HC6H5O] = 0.27 M

[H+] = 0

[NO3-] = 0

[C6H5O-] = 0

after ionization

[HNO3] = 0

[HC6H5O] = 0.27 - x

[H+] = 0.19 + x

[NO3-] = 0.19

[C6H5O-] = x

where "x" is the unkown dissociation of the weak acid

Ka = [H+][C6H5O-] / [HC6H5O]

Ka = HC6H5O 1.3*10^–10

1.3*10^–10 = (0.19 + x)(x) / (0.27 - x)

solve for x

(1.3*10^–10)(0.27) = 0.19x + x^2

x^2 + 0.19x - 3.51*10^-11 = 0

x = 1.85*10^-10

[H+] = 0.19 + x = 0.19 + 1.85*10^-10 =0. 19

pH = -log(0.19) = 0.7212