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

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!