Question

Calculate the equilibrium concentrations of C₂H₅COO^- and C₂H₅COOH in a solution prepare by dissolving 0.04 mole of propionic acid in 150 ml of water.

Answer 1

ka for propanoic acid = 1.34*10^-5

[C2H5COOH] = number of moles / volume in L

= 0.04 mol / 0.150 L

= 0.267 M

Lets write the dissociation equation of C2H5COOH

C2H5COOH -----> H+ + C2H5COO-

0.267 0 0

0.267-x x x

Ka = [H+][C2H5COO-]/[C2H5COOH]

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.34*10^-5)*0.267) = 1.892*10^-3

since c is much greater than x, our assumption is correct

so, x = 1.89*10^-3 M

[C2H5COO] = x = 1.89*10^-3 M

[C2H5COO-] = 0.267-x

= 0.267 - 1.89*10^-3

= 0.265 M

[C2H5COO] = 1.89*10^-3 M

[C2H5COO-] = 0.265 M