Question

A 1.32L buffer solution consists of 0.333M propanoic acid and 0.148M sodium propanoate. Calculate the pH of the solution following the addition of 0.070 mol HCI. Assume that any contribution of the HCI to the volume of the solution is negligible. The Ka of propanoic acid is 1.34x10^-5 .

please answer will rate

Answer 1

mol of HCl added = 0.07 mol

C2H5COO- will react with H+ to form C2H5COOH

Before Reaction:

mol of C2H5COO- = 0.148 M *1.32 L

mol of C2H5COO- = 0.1954 mol

mol of C2H5COOH = 0.333 M *1.32 L

mol of C2H5COOH = 0.4396 mol

after reaction,

mol of C2H5COO- = mol present initially - mol added

mol of C2H5COO- = (0.1954 - 0.07) mol

mol of C2H5COO- = 0.1254 mol

mol of C2H5COOH = mol present initially + mol added

mol of C2H5COOH = (0.4396 + 0.07) mol

mol of C2H5COOH = 0.5096 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 {0.1254/0.5096}

= 4.264

Answer: 4.26