Avolume 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.

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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