A volume of 60.0 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 17.2 mL of 1.50 M H2SO4 was needed? The equation is 2KOH (aq) + H2SO4 (aq) →K2SO4 (aq) + 2H2O (l)

Question

Howe

Chemistry

2 July, 14:40

A volume of 60.0 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 17.2 mL of 1.50 M H2SO4 was needed? The equation is 2KOH (aq) + H2SO4 (aq) →K2SO4 (aq) + 2H2O (l)

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Answers (1)

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Larson · Answer 1

The molarity of KOH = 0.86M

Explanation:

From the equation of reaction

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

nA = 1, nB=2

From the question

CA = 1.5M, VA = 17.2ml, VB = 60ml, CB=?

Applying (CAVA) / (CBVB) = nA/nB

(1.5*17.2) / (CB*60) = 1/2

Simplify

CB = (1.5*17.2*2) / (60*1)

CB = 0.86M