Question

A certain weak acid, HA, has a Ka value of 8.7×10⁻⁷.

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)

ionization = [H+]/M*100%

get [H+]

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 = 0.1 M; then

x^2 + (8.7*10^-7)x - 0.1*(8.7*10^-7) = 0

solve for x

x =2.945*10^-4

substitute

[H+] = 0 + 2.945*10^-4 = 2.945*10^-4M

% ionization = (2.945*10^-4) / (0.10) * 100% = 0.2945% ---> 0.29%

B)

similarly.

if M = 0.01 M; then

x^2 + (8.7*10^-7)x - 0.01*(8.7*10^-7) = 0

solve for x

x =9.2183*10^-5

substitute

[H+] = 0 + 9.2183*10^-5 = 9.2183*10^-5 M

% ionization = (9.2183*10^-5) / (0.010) * 100% = 0.92%