(8.d) pH = 1.46

Since [HSO₄^-] < 100 Ka then (0.10 - x) is not = 0.10.

Therefore you must use the quadratic equation.

Ka = {[H⁺][SO₄^-2]}/[HSO₄^-]

1.2 x 10^-2 = {(x)(x)}/(0.10 - x )

x² = 1.2 x 10^-2(0.10 - x)

x² = 1.2 x 10^-2x - 1.2 x 10^-3 = 0 Use the quadratic formula

a = 1

b = 1.2 x 10^-2

c = 1.2 x 10^-3

x = {-1.2 x 10^-2 + 0.07}/2

x = 0.029 OR.... x = - 0.81 (not valid)

pH = - 1og [H⁺]

pH = - log(0.029)

pH = 1.54