Question

What is the pH of a 0.13 M solution of hypoiodous acid?

Answer 1

H⁺ + IO^-

(0.13 - X) X X

Ka = 2.3 x 10^-11

Ka = (H⁺) x (IO^-) / (HIO)

Ka = X² / (0.13 - X)

0.13Ka - XKa = X²

2.99x10^-12 - 2.3x10^-11X = X²

X² + 2.3x10^-11X - 2.99x10^-12=0

X = 1.73x10^-6

(H⁺) = 1.73x10^-6

pH = -Lg(H⁺)

pH = -Lg(1.73x10^-6)

pH = 5.76