What volume (in milliliters) of 0.140 M HClO₃ are required to neutralize 60.0 mL of 0.160 M LiOH? _____mL
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Answer:
Approx.
19
⋅
m
L
. We know that
Concentration
=
Amount of substance in moles
Volume of solution
Explanation:
Moles of HCl
=
30.4
×
10
−
3
L
×
0.152
⋅
m
o
l
⋅
L
−
1
=
4.62
×
10
−
3
⋅
m
o
l
.
We need an equivalent quantity of sodium hydroxide;
0.250
⋅
m
o
l
⋅
L
−
1
sodium hydroxide is available.
So,
4.62
×
10
−
3
⋅
m
o
l
0.250
⋅
m
o
l
⋅
L
−
1
×
10
3
⋅
m
L
⋅
L
−
1
=
?
?
m
L
What I should have done at the beginning is to write the stoichiometric equation:
N
a
O
H
(
a
q
)
+
H
C
l
(
a
q
)
→
H
2
O
(
l
)
+
N
a
C
l
(
a
q
)
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