Determine the molarity of NaOH to three decimal places if 20.36 mL of this NaOH were used to titrate 0.328 g of KHP (MM 204.22 g/mol)
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Answer:
Explanation:
We need a stoichiometric equation........
C
6
H
4
(
C
O
−
2
K
+
)
(
C
O
2
H
)
potassium hydrogen phthalate, KHP
+
N
a
O
H
→
C
6
H
4
(
C
O
−
2
K
+
)
(
C
O
−
2
N
a
+
)
+
H
2
O
There is thus 1:1 stoichiometry........
Moles of KHP
=
0.6939
⋅
g
204.22
⋅
g
⋅
m
o
l
−
1
=
3.398
×
10
−
3
⋅
m
o
l
.
And since this molar quantity reacted with
26.55
⋅
m
L
of titrant...
Concentration
=
3.398
×
10
−
3
⋅
m
o
l
26.55
⋅
m
L
×
10
−
3
⋅
L
⋅
m
L
−
1
=
0.1280
⋅
m
o
l
⋅
L
−
1
.........
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