25 mL dissolution of hydrochloric acid 0.100 M is needed to neutralize 55.00 mL dissolution of sodium hydroxide. Calculate the molarity of sodium hydroxide dissolution. plz help me lol
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Explanation:
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
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