Question

The balanced equation for the neutralization reaction of aqueous H2SO4 with aqueous KOH is shown. H2SO4(aq)+2KOH(aq)⟶2H2O(l)+K2SO4(aq)...

The balanced equation for the neutralization reaction of aqueous H2SO4 with aqueous KOH is shown. H2SO4(aq)+2KOH(aq)⟶2H2O(l)+K2SO4(aq) What volume of 0.220 M KOH is needed to react completely with 13.0 mL of 0.230 M H2SO4 ?

Number of moles of H2SO4 = molarity * volume of solution in L

Number of moles of H2SO4 = 0.230 * 0.0130 L = 0.00299 mole

From the balanced equation we can say that

1 mole of H2SO4 requires 2 mole of KOH so

0.00299 mole of H2SO4 will require

= 0.00299 mole of H2SO4 *( 2 mole of KOH / 1 mole of H2SO4)

= 0.00598 mole of KOH

molarity of KOH = number of moles of KOH / volume of solution in L

0.220 = 0.00598 / volume of solution in L

volume of solution in L = 0.00598 / 0.220 = 0.0272 L

1 L = 1000 mL

0.0272 L = 27.2 mL

Therefore, the volume of KOH required would be 27.2 mL

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