Question

A certain weak acid, HA, has a Ka value of 4.6×10−7.

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

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 ((4.6*10^-7)*0.1) = 2.145*10^-4

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

so, x = 2.145*10^-4 M

% dissociation = (x*100)/c

= 2.145*10^-4*100/0.1

= 0.2145 %

Answer: 0.21 %

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 ((4.6*10^-7)*1*10^-2) = 6.782*10^-5

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

so, x = 6.782*10^-5 M

% dissociation = (x*100)/c

= 6.782*10^-5*100/0.01

= 0.6782 %

Answer: 0.68 %

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Include units, Thank you!

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