Question

The molarity of an aqueous solution of sodium hydroxide ( NaOH ) is determined by titration against a 0.132 M nitric acid ( HNO3 ) solution. If 31.9 mL of the base are required to neutralize 29.2 mL of nitric acid , what is the molarity of the sodium hydroxide solution? Give your answer to three significant figures.

Answer 1

Nitric acid and sodium hydroxide react as follow

NaOH + HNO₃   H₂O + NaNO₃

In the above neutrilisation reaction NaOH and HNO₃ react in equimolar proportion

no. of mole = malarity volume of solution in liter

no. of mole of nitric acid = 0.132 M 0.0292 L = 0.0038544 mole

In the neutrilisation reaction NaOH and HNO₃ react in equimolar proportion.

therefore no .of mole of sodium hydroxide = 0.0038544 mole

volume of sodium hydroxide = 31.9 ml = 0.0319 L

molarity = no.of mole / volume of solution in liter

= 0.0038544/0.0319 = 0.120M

Molarity of NaOH = 0.120M