The Ka of a monopratic weak acid is .00410. what is the percent ionization of a .152M solution of this acid?

Answer 1

Let the monoprotic acid be HA

HA ==> H⁺ + A^-

Ka = [H⁺][A^-]/[HA]

0.00410 = (x)(x)/0.152-x ... for now, I'll assume x is small relative to 0.152 M so will ignore it

0.00410 = x²/0.152

x² = 0.0006232

x = 0.0249 and this is about 16% of the value 0.152 so the above assumption was invalid. Will now repeat the calculation leaving x in the denominator and use a quadratic to solve.

0.00410 = x²/0.152-x

x² + 0.00410x + 0.00062 = 0

x = 0.0229 M = [H⁺] = [A^-]

% ionization = 0.0229 M/0.152 M = 0.151 x 100% = 15.1 % ionized