Transcript Slide 1

Calculate the pH of a solution that is 0.150 M in Benzoic
acid (HC7H5O2) and 0.050 M in potassium benzoate
(KC7H5O2). The pKa of benzoic acid is 4.20)
This is a buffer
solution since it
contains a weak acid
and its conjugate
base (a salt of the
acid).
What is the equilibrium chemical equation?
KC7H5O2 dissociates in water to K+ and C7H5O2
-
The acid establishes an equilibrium.
HC7H5O2 + H2O ↔ H3O+ + C7H5O2-
Since this is a buffer solution, the Henderson-Hasselbalch equation
can be used: pH = pKa + Log{[conj. Base])/[Acid]}
0.050
pH = 4.20 + Log
= 3.72
0.150
Calculate the pH of a solution that is 0.150 M in hydrazine (N2H4) and 0.050
M in its conjugate acid (N2H5Cl ). The Kb of hydrazine is 1.7 x 10-6.
What the chemistry?
N2H5Cl → N2H5+ + Cl-
Dissociate the salt:
Write equilibrium expression for the weak base: N2H4 + H2O ↔ N2H5+ + OH-
[N2H5+ ][OH1- ]
Kb =
[N2H5Cl]
I. 0.150
C. -x
E. 0.150 – x
[0.050 + x][x]
1.7 x 10 =
[0.150 - x]
0.050
+x
0
+x
0.050 + x
+x
-6
X = [OH1-] = 5.1 x 10-6
pOH = 5.29 and pH = 8.71
Or with the Henderson Hasselbach equation:
pH = pK a + log
Base
Conj.Acid

-14
pH = -log 10
1.7x10-6

+ log
0.150
= 8.71
0.050