Question

A certain weak acid, HA, has a Ka value of 6.9×10−7.

Part A Calculate the percent ionization of HA in a 0.10 M solution. Express your answer as a percent using two significant figures.

Part B Calculate the percent ionization of HA in a 0.010 M solution. Express your answer as a percent using two significant figures.

Answer 1

a)

First, assume the acid:

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = M;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

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

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.1 M; then

x^2 + (6.9*10^-7)x - 0.1*(6.9*10^-7) = 0

solve for x

x =2.623*10^-4

now, % ionization

%ionization = [A-]/[HA]initial * 100% = (2.623*10^-4) / (0.1) * 100 = 0.2623%

b)

apply same logic

substitute in Ka

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

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.1 M; then

x^2 + (6.9*10^-7)x - 0.01*(6.9*10^-7) = 0

solve for x

x =8.27*10^-5

now, % ionization

%ionization = [A-]/[HA]initial * 100% = (8.27*10^-5) / (0.01) * 100 = 0.827%