Question

certain weak acid, HA, has a Ka value of 6.6×10⁻⁷. Calculate the percent ionization of HA in a 0.10 M solution.

Step-by-step mathematics when breaking apart the Ka equation would be appreciated.

Answer 1

First, assume the acid:

HF

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 = 0.3 M; then

x^2 + (6.6*10^-7)x - 0.01*(6.6*10^-7) = 0

solve for x

x =8.09*10^-5

substitute

[H+] = 0 + 8.09*10^-5 = 8.09*10^-5M

[A-] = 0 + 8.09*10^-5= 8.09*10^-5 M

% ionization = [A-]/[HA]initial * 100% = 8.09*10^-5 / (0.01) * 100 = 0.809 %