Question

A certain weak acid, HA, has a Ka value of 5.5×10−7.

Part A

Calculate the percent ionization of HA in a 0.10 M solution.

Part B

Calculate the percent ionization of HA in a 0.010 M solution.

Answer 1

A)

HA dissociates as:

HA -----> H+ + A-

0.1 0 0

0.1-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.5*10^-7)*0.1) = 2.345*10^-4

since c is much greater than x, our assumption is correct

so, x = 2.345*10^-4 M

% dissociation = (x*100)/c

= 2.345*10^-4*100/0.1

= 0.2345 %

Answer: 0.234 %

B)

HA dissociates as:

HA -----> H+ + A-

1*10^-2 0 0

1*10^-2-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.5*10^-7)*1*10^-2) = 7.416*10^-5

since c is much greater than x, our assumption is correct

so, x = 7.416*10^-5 M

% dissociation = (x*100)/c

= 7.416*10^-5*100/0.01

= 0.7416 %

Answer: 0.742 %