Question

Calculate the percent dissociation of each of the following solutions. (a) 0.154 M propanoic acid (HC3H5O2, Ka = 1.3 x10^-5) (part A solved, 0.92%dissociation) (b) a mixture containing 0.154 M HC3H5O2 and 0.154 M NaC3H5O2

Answer 1

a)

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 = 0.154 M; then

x^2 + (1.3*10^-5)x - 0.154 *(1.3*10^-5) = 0

solve for x

x =0.00141

substitute

[H+] = 0 + 0.00141= 0.00141M

[A-] = 0 + 0.00141= 0.00141M

%ion = [A-]/[HA] * 100% = 0.00141 / 0.154 * 100 = 0.915 % --> 0.92%

b)

this is a buffer

pH = pKa + log(NaC3H5O2/HC3H5O2 )

pH = -log(1.3*10^-5) + log(0.154 /0.154 )

pH = 4.886

[A-] = 10^-pH = 10^-4.886

[HA] = 0.154

% ion = (10^-4.886)/(0.154) * 100% = 0.00844 %