The ph of a 0.175 m aqueous solution of a weak acid is 3.52. what is ka for this acid?

Chemistry

1 answer:

Naddika [18.5K]3 years ago

3 0

Hey there!

pH = - log [ H⁺ ] = 3.52

[ H⁺ ] = 10^-pH

[ H⁺] = 10^ ( -3.52 )

[H⁺] = 3.02*10⁻⁴ M

[HA] =0.175 M

Therefore:

Ka = [ H⁺]* [A⁻] / [ HA]

Ka = (3.02*10⁻⁴)² / 0.175

Ka = 9.1204*10⁻⁸ / 0.175

Ka = 5.2*10⁻⁷

Hope that helps!

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