Answer :
In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.
What is Neutralization?
The interaction of H+ ions and OH- ions produces water in a neutralization reaction, which occurs when an acid and a base react to form water and salt. A pH of 7 results from the neutralization of a potent acid and a potent base.
Calculation:
The balanced reaction is
H₂SO₄ + 2 KOH ⇒ 2 H₂O + K₂SO₄
The molarity M is the number of moles of solute that are dissolved in a given volume, expressed as:
number of moles = molarity × volume
For H₂SO₄
35.00 mL= 0.035 L (being 1,000 mL= 1 L)
Molarity= 0.737 M
Then,
number of moles= 0.737 M × 0.035 L
number of moles= 0.0258
So you must neutralize 0.0258 moles of H₂SO₄. Now you can apply the following rule of three: if by stoichiometry 1 mole of H₂SO₄ is neutralized with 2 moles of KOH, 0.0258 moles of H₂SO₄ are neutralized with moles of KOH =?
moles of KOH = 0.0516,
Then 0.0516 moles of KOH are needed.
Molarity= 0.827 M
number of moles= 0.0516
volume=?
Replacing in the definition of molarity: 0.827M = 0.0516 moles/ volume.
= 0.0516moles/0.827M = volume
volume = 0.0624 L= 62.4 mL.
Hence, 62.4 mL of a 0.827 m KOH solution is required for neutralization.
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