Answer:
pH = 5.47
Explanation:
The equilibrium that takes place is:
HIO ↔ H⁺ + IO⁻
Ka = [tex]\frac{[H+][IO-]}{[HIO]}[/tex] = 2.3 * 10⁻¹¹
At equilibrium:
Replacing those values in the equation for Ka and solving for x:
[tex]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}[/tex]
Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47
