Question

A volume of 90.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 16.2 mL of 1.50 M H2SO4 was needed? Watch your sig figs and include your unit in your final answer.

Answer 1

1M H₂SO₄=2N H₂SO₄ as H₂SO₄ is dibasic acid.

So, by unitary method 1.5M H₂SO₄ = 3 N H₂SO₄ where M= molarity and N= normality

1N = 1 gram-equivalence in 1000ml solution

3 N = 3 gram-equivalence 1000 ml solution

So, no of gram-equivalence in 16.2 ml solution of 3N H₂SO₄ = (16.2*3)/1000 no of gram-equivalence

Say the KOH is 'P' N i.e 'P' gram-equivalent KOH is dissolved in 1000 ml solution.

So, the gram-equivalence number of 90 ml solution of 'P' N solution is =(90*P)/1000 no. of gram-equivalence

Now, at the end point of the titration gram-equivalence number of KOH = gram-equivalence number of H₂SO₄

(90*P)/1000 = (16.2*3)/1000

P= 0.54 gram equivalence no.

so, the KOH was a 0.54 N solution and we know as KOH is monoacidic base its normality = molarity.Thus the molarity of the KOH was 0.54 M.