Question

In a titration of 35.00 mL of 0.737 M H2SO4, __________ mL of a 0.827 M KOH solution is required for neutralization.

Answer 1

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.

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:

$Molarity=\frac{number of moles}{volume}$

in units of $\frac{moles}{liter}$

then the number of moles can be calculated as:

number of moles= molarity* volume

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

$moles of KOH=\frac{0.0258moles of H_{2} SO_{4}*2 moles of KOH }{1mole of H_{2} SO_{4}}$

moles of KOH= 0.0516

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

• Molarity= 0.827 M
• number of moles= 0.0516
• volume=?

Replacing in the definition of molarity:

$0.827 M=\frac{0.0516 moles}{volume}$

Solving:

$volume=\frac{0.0516 moles}{0.827 M}$

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.