Question

A solution of 0.133 M KOH is used to titrate 15.0 mL of a H2SO4 solution. Write out the balanced chemical equation and what is the molarity of the H2SO4 solution if 28.2 mL of the KOH solution is required?

Answer 1

The answer to your question is a) H₂SO₄ + 2KOH ⇒ K₂SO₄ + 2H₂O

b) Molarity = 0.127

Step-by-step explanation:

Data

[KOH] = 0.133 M

volume of KOH = 28.2 ml

[H₂SO₄] = ?

volume of H₂SO₄ = 15 ml

1) Balanced chemical reaction

H₂SO₄ + 2KOH ⇒ K₂SO₄ + 2H₂O

Reactants Elements Products

2 K 2

1 S 1

4 H 4

6 O 6

2) Calculate the molar mass of H₂SO₄ and KOH

H₂SO₄ = (1 x 2) + (32 x 1) + (16 x 4) = 2 + 32 + 64 = 98 g

KOH = (39 x 1) + (16 x 1) + (1 x 1) = 39 + 16 + 1 = 56 g

- Calculate the number of moles of KOH

Molarity = moles / volume

-Solve for moles

moles = Molarity x volume

-Substitution

moles = 0.133 x 28.2/1000

= 0.0038

- Calculate the moles of H₂SO₄

1 mol of H₂SO₄ ------------------ 2 moles of KOH

x ------------------ 0.0038 moles

x = (0.0038 x 1)/2

x = 0.0019 moles of H₂SO₄

-Calculate the molarity of H₂SO₄

Molarity = 0.0019/0.015

Molarity = 0.127