an aqueous solution of nitric acid is standardized by titration with a 0.168 m solution of calcium hydroxide. if 12.9 ml of base are required to neutralize 22.5 ml of the acid, what is the molarity of the nitric acid solution?



Answer :

The molarity of the nitric acid solution is 0.193M if nitric acid is standardized by titration with a 0.168 M solution of calcium hydroxide.

First, we write a balanced equation for the reaction as follows;

[tex]2HNO_{3} + Ca (OH)_{2} --- > Ca(NO_{3}) _{2} + 2H_{2}O[/tex]

Therefore from this equation;

Moles of acid (nA) = 2

Moles of base (nB) = 1

As the volume of acid (Va) is 22.5 ml and the volume of base (Vb) is 12.9 ml and the molarity of the base (Mb) is 0.168 M, we can calculate the molarity of the acid by using the following formula;

MaVa/MbVb = nA/nB

Substituting the values;

Ma × 22.5 / 0.168 × 12.9 = 2 / 1

Ma × 22.5 = 2 × 0.168 × 12.9

Ma × 22.5 = 4.3344

Ma = 4.3344 / 22.5

Ma = 0.193M

Therefore, the molarity of the nitric acid solution is calculated to be 0.193M.

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