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3 years ago

What is the ka for a 0.25 m solution of a monoprotic weak acid with a ph of 4.65

Chemistry

1 answer:

34kurt3 years ago

8 0

Answer:

ka = 2x10⁻⁹

Explanation:

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

pH = -log[H⁺] = 4.65

$10^{-4.65}$ = [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 = $\frac{[H^+][A^-]}{[HA]}$

ka = $\frac{(2.24*10^{-5})(2.24*10^{-5})}{0.25-2.24*10^{-5}}$ = 2x10⁻⁹

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tex]\large{\boxed{mole\:=\:\frac{grams}{relative\:molecular\:mass}}}[/tex]

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