Question

The balanced equation for the neutralization reaction of aqueous H₂SO₄ with aqueous KOH is shown.

H₂SO₄(aq)====>2H₂O (l) +K₂SO₄(aq)

What volume of 0.370 M KOH is needed to react completely with 15.5 mL of 0.135 M H₂SO₄? Assume the reaction goes to completion.

Answer 1

Solution :-

balanced reaction equation is as follows

H2SO4(aq) + 2KOH(aq) ------- > K2SO4(aq) + 2H2O(l)

volume of the 0.370 M KOH needed to react with 15.5 ml of 0.135 M H2SO4 is calculated as follows

using the molarity and volume lets calculate the moles of H2SO4

moles = molarity * volume in liter

moles of H2SO4 =0.135 mol per L * 0.0155 L =0.0020925 mol H2SO4

now using the mole ratio lets calculate moles of KOH needed to react with the 0.0020925 mol H2SO4

mole ratio of the H2SO4 to KOH is 1 : 2

0.0020925 mol H2SO4 * 2 mol KOH / 1 mol H2SO4 =0.004185 mol KOH

now lets calculate the volume of the KOH needed

volume of KOH= moles of KOH / molarity of KOH

                     = 0.004185 mol /0.370 mol per L

                     = 0.0113 L

converting volume from liter to ml then we get

0.0113 L* 1000 ml / 1 L =11.3 ml

therefore volume of KOH needed = 11.3 ml