what volume of pure HNO3 of sp. Great will be required to neutralize 0.892 mole of KOH??
Answers
Explanation:
Potassium hydroxide,
KOH
, will react with hydrochloric acid,
HCl
, to produce aqueous potassium chloride and water.
The balanced chemical equation that describes this neutralization reaction looks like this
KOH
(
a
q
)
+
HCl
(
a
q
)
→
KOH
(
a
q
)
+
H
2
O
(
l
)
So, the reaction consumes potassium hydroxide and hydrochloric acid in a
1
:
1
mole ratio, which implies that the number of moles of potassium hydroxide needed to completely neutralize the hydrochloric acid solution will be equal to the number of moles of hydrochloric acid present in said solution.
As you know, molarity tells you the number of moles of solute present in
1 L
=
10
3
mL
of solution.
This implies that the hydrochloric acid solution contains
500
.
mL
⋅
1.0 moles HCl
10
3
mL
=
0.50 moles HCl
Therefore, you can say that a complete neutralization requires
0.50
moles of potassium hydroxide. The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the solution.