Question

Using simplifying assumptions calculate the equilibrium pH of a solution made by adding acetic acid to water to give a concentration of 10^-2 at 25C (Ignore Activity)

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 = 10^-2 M; then

x^2 + (1.8*10^-5)x - (10^-2)*(1.8*10^-5) = 0

solve for x

x =4.15*10^-4

substitute

[H+] = 0 + 4.15*10^-4 = 4.15*10^-4M

[A-] = 0 + 4.15*10^-4= 4.15*10^-4M

pH = -log(H+) = -log(4.15*10^-4) = 3.38

Ussing simplification will give us an error:;

x^2 + Kax - M*Ka = 0

if M = 10^-2 M; then

Ka = x*x/(M-x)

asume M>> x so

Ka = x^2 / M

x= sqrt(M*Ka) = sqrt((10^-2)(1.8*10^-5)) =0.0004242 M

pH = -log(0.0004242) = 3.372