Question

The K, of a monoprotic weak acid is 2.30 x 10°. What is the percent ionization of a 0.166 M solution of this acid? Number 11.77

Answer 1

Let us give equation for ionization constant

K_a = ( [H⁺] [Ac¯] ) / [HAc]

Let us keep Ac^- is anion

From the above equation we can calculate the [H⁺]:

2.3 x 10^-3 = [(x) (x)] / 0.166

x = 0.01953  M

To get ionization we have to divide the [H⁺] by the concentration, then multiply by 100:

(0.01953 M / 0.166 M) x 100 = 11.77 % dissociated

Hence the monoprotoc acid was 11.77 % dissociated