Please choose the write answer

Question

Calculate the percent ionization of weak acid ina 0.0010 M solution

Ka= 9.2*10^-7

a)0.959

b)0.303

c)3.03

d)30.3

Answer 1

First, the weak acid dissociation reaction and relevant ICE table is helpful to express:

HA_(aq) + H₂O_(l) ⇔ H₃O⁺_(aq) + A^-_(aq)

I 0.0010 - 0 0

C -x - x x

E 0.0010 - x - x x

The Ka relation can then be written and set equal to the given Ka value:

Ka = [H₃O⁺_(aq) x A^-_(aq)] / HA_(aq)

Ka = [x] ⋅ [x] / [0.0010 - x] = x² / (0.0010 - x) ≅ x² / (0.0010)

9.2 x 10^-7 ≅ x² / (0.0010)

x = 3.03 x 10^-5

Now we can set up our percent dissociation expression:

% dissociation = (x / HA) ⋅ 100%

% dissociation = [(3.03 x 10^-5) / (0.0010)] ⋅ 100%

% dissociation = 3.03% => So the answer is choice C

**An important step to remember here that makes the math much easier is to assume "x" is much lesser than the original HA concentration of 0.0010 M. This is a good assumption given weak acids dissociate very little. This then allows us to solve the Ka expression without using the quadratic formula.**