Question

Question 4 2 pts If 11 drops of a formic acid (HCOOH) solution are titrated with 10 drops of a 0.12 M NaOH solution, what is the percent ionization of the formic acid solution? (Ka for formic acid 1.78 x 104) 4.0% 1.3% 1.9% @ 2.5% Question 5 2 pts What is the pH of an acetic acid solution if 25 drops are titrated with 44 drops of a 0.45 M KOH solution? (K, for acetic acid 1.8 x 105) O 2.04 2.59 2.22 O 2.42

Answer 1

Solution:

11 drops HCOOH neutalizes with 10 drops of 0.12 M NaOH

Thus,

11 drops x [HCOOH] = 10 drops x [NaOH]

11 x [HCOOH] = 10 x 0.12 M

[HCOOH] = 10 x 0.12 M / 11 = 0.109 M

Formic acid is a weak acid, it is dissociated as:

HCOOH + H2O = HCOO- + H3O+

0.109 --------------------0-------------0 (initial concentrations)

(0.109 - X)--------------X---------------X (final cocentration)

Ka = X . X / 0.109 - X

X can be neglected from denominator in comparison to 0.109 because HCOOH is a weak acid. Then,

Ka = X^2 / 0.109

X^2 = 0.109 Ka = 0.109 x 1.78 x 10^-4

X^2 = 19.4 x 10^-6

X = 4.40 x 10^-3 = 0.0044

Thus, % ionization = ( 0.0044 / 0.109 ) x 100 = 4.0 %

Hence, option A.