In a titration of 35.00 mL of 0.737 M H2SO4, mL of a 0.827 M KOH solution is required for neutralization. A) 35.0
B) 1.12
C) 25.8
D) 62.4
E) 39.3

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.

Explanation:

The balanced reaction is

H₂SO₄  +  2 KOH  ⇒  2 H₂O  +  K₂SO₄

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction)  1 mole of H₂SO₄ is neutralized with 2 moles of KOH.

The molarity M being the number of moles of solute that are dissolved in a given volume, expressed as:

in units of

then the number of moles can be calculated as:

number of moles= molarity* volume

You have acid H₂SO₄

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 are neutralized with 2 moles of KOH, 0.0258 moles of H₂SO₄ are neutralized with how many moles of KOH?

moles of KOH= 0.0516

Then 0.0516 moles of KOH are needed. So you know:

Replacing in the definition of molarity:

Solving:

volume=0.0624 L= 62.4 mL

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.

the answer is the first option

explanation:

just googled it m8. not that hard

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