Question

Consider a 1.0 L buffer containing 0.109 mol L-1 HOCl and 0.095 mol L-1 OCl-. What is the pH of the solution after adding 7.2 x 10-3 mol of NaOH? Express your answer to 2 decimal places.

Answer 1

Ka of HOCl = 3.5*10^-8

mol of NaOH added = 0.0072 mol

HOCl will react with OH- to form OCl-

Before Reaction:

mol of OCl- = 0.095 M *1.0 L

mol of OCl- = 0.095 mol

mol of HOCl = 0.109 M *1.0 L

mol of HOCl = 0.109 mol

after reaction,

mol of OCl- = mol present initially + mol added

mol of OCl- = (0.095 + 0.0072) mol

mol of OCl- = 0.1022 mol

mol of HOCl = mol present initially - mol added

mol of HOCl = (0.109 - 0.0072) mol

mol of HOCl = 0.1018 mol

Ka = 3.5*10^-8

pKa = - log (Ka)

= - log(3.5*10^-8)

= 7.456

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

use:

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

= 7.456+ log {0.1022/0.1018}

= 7.458

Answer: 7.46