What volume of 0.100 N HNO3 is required to neutralize 50.0 mL of a 0.150 N solution of Ba(OH)2?
Answers
Answer:
Hydrochloric acid,
HCl
, and potassium hydroxide,
KOH
, react in a
1
:
1
mole ratio to produce aqueous potassium chloride,
KCl
, and water.
HCl
(
a
q
)
+
KOH
(
a
q
)
→
KCl
(
a
q
)
+
H
2
O
(
l
)
This means that a complete neutralization requires equal numbers of moles of hydrochloric acid, i.e. of hydronium cations,
H
3
O
+
, and of potassium hydroxide, i.e. of hydroxide anions,
OH
−
.
As you know,m molarity is defined as the number of moles of solute present in
1 L
of solution.
Now, notice that the potassium hydroxide solution, which has a molarity of
0.350 M
, is
0.350
M
0.100
M
=
3.5
times more concentrated than the hydrochloric acid solution, which has a molarity of
0.100 M
. In other words, for the same volume of both solutions, the potassium hydroxide solution contains
3.5
times more moles of solute than the hydrochloric acid solution.
This means that in order to have equal numbers of moles of both solutes, you need to have a volume of hydrochloric acid solution that is
3.5
times bigger than the volume of the sodium hydroixde solution.
Since the sodium hydroixde solution has a volume of
50.0 mL
, it follows that the volume of hydrochloric acid needed will be
V
HCl
=
3.5
×
50.0 mL
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
175 mL
a
a
∣
∣
−−−−−−−−−−−