Question

How many milliliters of a 5.17 M H2SO4 solution are needed to neutralize 70.00 mL of a 0.899 M KOH solution?

Answer 1

take the balanced equation for the neutralization reaction
H2SO4 + 2 KOH ---> K2SO4 + 2 H2O

Balanced equation says 1 mole H2SO4 needs to neutralize 2 moles
KOH ;mole ratio H2SO4:KOH =1:2

let us find out moles of KOH in the70.00 mL [which is equal to 0.07 Liter] of a 0.899 M KOH solution

# moles of KOH= molarity* volume in Liters = 0.899 mole/L* 0.07 L
= 0.06293 moles

SO moles of H2SO4 needed = [1/2]*0.06293 moles = 0.031465 moles
[since mole ratio is 1:2]

Now convert this moles to volume ,using molarity of H2SO4 [5.17 M]

volume of H2SO4 = # moles of H2SO4/molarity
=0.031465 moles/5.17 mols/L = 0.006086 Liters

volume of H2SO4 in ML = 6.086 ml
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