Question

A certain weak acid, HA , has a Ka value of 6.6×10⁻⁷.

Part A

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

Express your answer to two significant figures and include the appropriate units.

Part B

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

Express your answer to two significant figures, and include the appropriate units.

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 ((6.6*10^-7)*0.1) = 2.569*10^-4

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

so, x = 2.569*10^-4 M

% dissociation = (x*100)/c

= 2.569*10^-4*100/0.1

= 0.257 %

Answer: 0.26 %

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 ((6.6*10^-7)*1*10^-2) = 8.124*10^-5

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

so, x = 8.124*10^-5 M

% dissociation = (x*100)/c

= 8.124*10^-5*100/0.01

= 0.812 %

Answer: 0.81 %