Calculate the molarity of a solution of potassium hydroxide if 19.55 mL of KOH is required for the titration of a 20.20 mL sample of a 0.3467 M sulfuric acid solution.

The correct answer is 0.3582 M.

Explanation:

Based on the given information, the volume of KOH given is 19.55 ml, the volume of sulfuric acid given is 20.20 ml, and the molarity of sulfuric acid is 0.3467 M. There is a need to find the molarity of KOH.

The formula to use in the given case is,

M1V1 = M2V2

Here M1 is the molarity of KOH, V1 is the volume of KOH, M2 is the molarity of sulfuric acid, and V2 is the volume of H2SO4.

Here V1 is 19.55 ml, M2 is 0.3467 M, and V2 is 2020 ml.

Now putting the values in the equation we get,

M1 * 19.55 = 0.3467 * 20.20

M1 * 19.55 = 7.00334

M1 = 7.00334/19.55

M1 = 0.3582 M

Hence, the molarity of the given KOH is 0.3582.

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