Answer to Question #58952 in General Chemistry for Tasha Brown

Question

An analytical chemist is titrating
232.3mL
of a
0.3600M
solution of hydrazoic acid
HN3
with a
1.100M
solution of
KOH
. The
pKa
of hydrazoic acid is
4.72
. Calculate the pH of the acid solution after the chemist has added
41.23mL
of the
KOH
solution to it.

Answer 1

HN3 + KOH → KN3 + H2O
You will produce a salt of the weak acid (KN3) and there will be some unreacted HN3 in solution . This is a typical buffer. The pH is determined using the Henderson - Hasselbalch equation:
Mol HN3 in 232.3 mL of 0.3600 M HN3 solution = 232.3/1000*0.360 = 0.0841 mol HN3
Mol KOH in 41.23 mL of 1.100 M solution = 41.23/1000*1.100 = 0.0454 mol KOH
The reaction will produce 0.0454 mol KN3 and there will be 0.0841 - 0.0454 = 0.0387 mol HN3 remaining. These are dissolved in 232.2+41.23 = 273.43 mL = 0.2734 L solution
Calculate the molarity of each component:
HN3 = 0.0387 / 0.2734 = 0.1416 M HN3
KN3 = 0.0454 /0.2734 = 0.1661 M KN3
Substitute into the H-H equation:
pH = pKa + log ([salt] / [acid])
pH = 4.72 + log (0.1661 / 0.1416)
pH = 4.72 + log 1.17
pH = 4.72 + 0.23
pH = 4.95