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?

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

                                                          b) Molarity = 0.127

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

the answer is d. 74.02