Jadaahnya0 Jadaahnya0
  • 19-11-2020
  • Chemistry
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what is the ka for a 0.25 m solution of a monoprotic weak acid with a ph of 4.65

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maacastrobr
maacastrobr maacastrobr
  • 26-11-2020

Answer:

ka = 2x10⁻⁹

Explanation:

Using the pH we can calculate the molar concentration of H⁺, [H⁺]

  • pH = -log[H⁺] = 4.65
  • [tex]10^{-4.65}[/tex] =  [H⁺] = 2.24 x 10⁻⁵ M

Then we use the expression of Ka for the equilibrium of a weak monoprotic acid:

HA ↔ H⁺ + A⁻

Where:

ka = [tex]\frac{[H^+][A^-]}{[HA]}[/tex]

ka =  [tex]\frac{(2.24*10^{-5})(2.24*10^{-5})}{0.25-2.24*10^{-5}}[/tex] =  2x10⁻⁹

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