Question

A 1.49 L buffer solution consists of 0.286 M propanoic acid and 0.189 M sodium propanoate. Calculate the pH of the solution following the addition of 0.066 mol HCL. Assume that any contribution of the HCl to the volume of the solution is negligible. The K, of propanoic acid is 1.34 x 10-5.

Answer 1

mol of HCl added = 0.066 mol

C2H5COO- will react with H+ to form C2H5COOH

Before Reaction:

mol of C2H5COO- = 0.189 M *1.49 L

mol of C2H5COO- = 0.2816 mol

mol of C2H5COOH = 0.286 M *1.49 L

mol of C2H5COOH = 0.4261 mol

after reaction,

mol of C2H5COO- = mol present initially - mol added

mol of C2H5COO- = (0.2816 - 0.066) mol

mol of C2H5COO- = 0.2156 mol

mol of C2H5COOH = mol present initially + mol added

mol of C2H5COOH = (0.4261 + 0.066) mol

mol of C2H5COOH = 0.4921 mol

Ka = 1.34*10^-5

pKa = - log (Ka)

= - log(1.34*10^-5)

= 4.873

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.873+ log {0.2156/0.4921}

= 4.514

Answer: 4.51