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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?

Respuesta :

joseaaronlara

Answer:

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

[Hâ‚‚SOâ‚„] = ?

volume of Hâ‚‚SOâ‚„ = 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 Hâ‚‚SOâ‚„ and KOH

Hâ‚‚SOâ‚„ = (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 Hâ‚‚SOâ‚„

                   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 Hâ‚‚SOâ‚„

Molarity = 0.0019/0.015

Molarity = 0.127

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