Question

How many milliliters of 0.120 M NaOH are required to tritrate 50.0 mL of 0.0998 M hydrochloric acid to the equivale point the Ka of hypochlorous acid is 3.0×10^-8. What is the pH after the NaOH is added?

Answer 1

The reaction is

HA + NaOH ----> NaA + H2O

50x 0.0998 v x 0.120 initial milimoles

0 0 50x0.0998 at the equivalence point

At the equivalence point

millimoles of acid = millimoles of base

Thus    50x 0.0998 = v x 0.120

hence V = volume of base = 50x 0.0998 /0.120 =41.58 mL

At the equivlence point , the solution contains only the salt NaA, neither acid nor base.

Thus the pH depends on the salt which is a salt of strong base and weak acid.

Thus it undergoes anionic hydrolysis and the solution is basic in solution.

A^-  + H2O -------> HA + OH-

The pH of such a salt is given by

pH = 1/2 [pkw + pka + log C] where C is the concentration of the salt.

At equivalence the volume of solution = 50 + 41.58mL= 91.58mL and its mmoles are =50x0.0998

Thus molar concentration of salt = 50x 0.0998 /91.58 and Ka is given as 3.0x 10^-8

Hence pH of solution = 1/2 [ 14 + (8-log 3.0) + log ( 50x 0.0998 /91.58 )]

= 10.129