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

pH = 5.47

Step-by-step explanation:

The equilibrium that takes place is:

HIO ↔ H⁺ + IO⁻

Ka =
`([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=(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