Question

A 25.00 mL sample of a solution of nitric acid, HNO3, is titrated with a 0.1570 M solution of sodium hydroxide. The titration reaches the end point when 31.77 mL of the NaOH solution has been added. What is the molarity of the nitric acid solution?

Answer 1

Answer: -

0.1995 M

Explanation: -

Volume of HNO₃ = 25.00 mL = $\frac{25.00}{1000}$ = 0.025 L

Strength of NaOH = 0.1570 M

Volume of NaOH = 31.77 mL = $\frac{31.77}{1000}$ = 0.03177 L

Since the end point is reached, we use the formula

Strength of HNO₃ x Volume of HNO₃ = Strength of NaOH x Volume of NaOH

Strength of HNO₃ x 0.025 L = 0.1570 M x 0.03177 L

Strength of HNO₃ = $\frac{0.1570 M x 0.03177 L}{0.025 L}$

= 0.1995 M

Thus when a 25.00 mL sample of a solution of nitric acid, HNO3, is titrated with a 0.1570 M solution of sodium hydroxide and the titration reaches the end point when 31.77 mL of the NaOH solution has been added, the molarity of the nitric acid solution would be 0.1995 M