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RUDIKE [14]
3 years ago
8

How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?

Chemistry
1 answer:
Kipish [7]3 years ago
4 0

Answer:

V KOH = 41 mL

Explanation:

for neutralization:

  • ( V×<em>C </em>)acid = ( V×<em>C </em>)base

∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L

∴ V H2SO4 = 41 mL = 0.041 L

∴ <em>C</em> KOH = 0.0050 N = 0.0050  eq-g/L

∴ E KOH = 1 eq-g/mol

⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L

⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH

⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)

⇒ V KOH = 0.041 L

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The mass of Calcium bromide added in the flask is 29.7 g.

<h3>What is Molarity? </h3>

Molarity is defined as the ratio of number of moles od solute to the number of volume of solution in litres.

Molarity = number of moles/ volume

<h3>Calculation of Moles</h3>

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Mole is defined as the given mass of substance to the molar mass of substance.

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learn more about Molarity:

brainly.com/question/19517011

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DISCLAIMER:

The above question is incomplete. Below is the complete question

A chemist adds 410.0mL of a 0.363 M calcium bromide solution to a reaction flask. calculate the mass in grams of calcium bromide the chemist has added to the flask. round your answer to 3 significant digits.

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