Question

A buffer solution is prepared that is 0.50 m in propanoic acid and 0.40 m in sodium propanoate with a solution volume of 1.00 liters. (ka for propanoic acid = 1.3 x 10-5). what is the ph of the solution when 0.060 mol of naoh(s) is added to the solution? assume no change in solution volume

Answer 1

                CH3CH2COOH + H2O ⇄ H3O+ + CH3CH2COO-
By adding 0.060 mol of NaOH(s), this adds 0.060 mol of OH-. 0.060 mol of H3O gets neutralized by this OH, making the equilibrium shift right. 0.060 mol of CH3CH2COOH are "removed" and 0.060 mol of CH3CH2COO are "created".

The new concentrations of propionic acid and the propionic ion are 0.44 and 0.46 respectively. Therefore, since

Ka = [CH3CH2COO-][H3O+]/[CH3CH2COOH] = 1.3 x 10^-5

we get

[H3O+] = [CH3CH2COOH] / [CH3CH2COO-] * 1.3 x 10^-5
[H3O+] = 0.44 / 0.46* 1.3 x 10^-5
[H3O+] = 1.243 x 10^-5

hence

pH = -log[H3O+] = -log(1.243 x 10^-5) = 4.91

The answer to the question is 4.91.