Question

Calculate the pH of a 0.50 M HIO. The Ka of hypoiodic acid, HIO, is 2.3x10–11.0.305.325.479.474.80

Answer 1

Answer:

pH = 5.47

Explanation:

The equilibrium that takes place is:

HIO ↔ H⁺ + IO⁻

Ka = $\frac{[H+][IO-]}{[HIO]}$ = 2.3 * 10⁻¹¹

At equilibrium:

• [HIO] = 0.5 M - x

• [H⁺] = x

• [IO⁻] = x

Replacing those values in the equation for Ka and solving for x:

$Ka=\frac{x^2}{0.5-x}=2.3*10^{-11} \\x^2=(2.3*10^{-11})(0.5-x)\\x^2=1.15*10^{-11}-2.3*10^{-11}x\\x^2+2.3*10^{-11}x-1.15*10^{-11}=0\\x=3.39*10^{-6}$

Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47