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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 21.2 mL of 1.50 M H2SO4 was needed? The equation is

2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Express your answer with the appropriate units.

Respuesta :

joseaaronlara

Answer:

The answer to your question:  0.7 M

Explanation:

Data

V of KOH = 90 ml

[KOH] = ?

V H2SO4 = 21.2 ml

[H2SO4] = 1.5 M

                       2KOH(aq)  +  H₂SO₄(aq)   →   K₂SO₄(aq)  +  2H₂O(l)

Molarity = moles / volume

moles of H₂SO₄ = (1.5) (21.2)

                           = 31.8

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

                   x                           --------------  31.8 mol of H₂SO₄

                    x = (31.8)(2) / 1

                    x = 63.8 moles of KOH

Molarity = 63.8 / 90

             = 0.7 M

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