batmanthatguy6199 batmanthatguy6199
  • 22-10-2019
  • Chemistry
contestada

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

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maacastrobr
maacastrobr maacastrobr
  • 22-10-2019

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:

  • [HIO] = 0.5 M - x
  • [H⁺] = x
  • [IO⁻] = x

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

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