An aqueous solution of nitric acid is standardized by titration with a 0.106 M solution of calcium hydroxide. If 13.1 mL of base are required to neutralize 20.7 mL of the acid, what is the molarity of the nitric acid solution?

0.134M

Explanation:

First, we begin by writing a balanced equation for the reaction. This is illustrated below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + 2H2O

From the equation:

nA (mole of the acid) = 2

nB (mole of the base) = 1

Data obtained from the question include:

Ma (Molarity of acid) =?

Va (volume of acid) = 20.7 mL

Mb (Molarity of base) = 0.106 M

Vb (volume of base) = 13.1 mL

Using MaVa/MbVb = nA/nB, we can calculate the molarity of the acid as follow:

MaVa/MbVb = nA/nB

Ma x 20.7/ 0.106 x 13.1 = 2/1

Cross multiply to express in linear form as shown below:

Ma x 20.7 = 0.106 x 13.1 x 2

Divide both side by 20.7

Ma = (0.106 x 13.1 x 2) / 20.7

Ma = 0.134M

Therefore, the molarity of the acid is 0.134M

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