Question

0.950 L of 0.420 M H2SO4 is mixed with 0.900 L of 0.280 M KOH. What concentration of sulfuric acid remains after neutralization?

Answer 1

In first step we calculate mol of both H2SO4 and KOH and then by using balanced chemical equation we can get limiting reactant. Excess reactant gives idea about calculation of the final concentration of H2SO4 in the solution.

Solution:

n H2SO4 = volume in L x molarity

=0.950*0.420

=0.399 mol

n KOH

=0.900*0.280

=0.252 mol

Reaction:

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

Mol ratio of H2SO4 : KOH is 1 : 2

Limiting reactant :

Calculation of moles of KOH needed to react with 0.399 mol H2SO4

n KOH required = 0.399 mol H2SO4 x 2 mol KOH / 1 mol H2SO4

=0.798 mol KOH

But actually there are only 0.252 moles of KOH are present. So KOH is limiting reactant and H2SO4 is excess reactant.

Calculation of excess moles of H2SO4

n H2SO4 (excess)

= Original moles – ( moles of H2SO4 needed to reacts with 0.252 mol KOH

=0.399 – ( 0.252 mol KOH x 1 mol H2SO4/ 2 mol KOH)

=0.399-0.252/2 mol H2SO4

=0.273 mol H2SO4

After neutralization concentration of H2SO4 = mol of H2SO4/ total volume of the solution.

= 0.273 / (0.420+0.900 ) L

= 0.21 M

Molarity of the H2SO4 in the final solution would be 0.21 M