A 15.60-mL sample of aqueous sodium hydroxide is titrated to the stoichiometric point with 18.00 mL?

Question

A 15.60-mL sample of aqueous sodium hydroxide is titrated to the stoichiometric point with 18.00 mL of 0.256 M HNO3(aq).
What is the molarity of the basic solution (sodium hydroxide) BEFORE it is titrated ? what is the mass of sodium hydroxide in the basic solution?

Answer 1

The neutralization equation is NaOH + HNO3 --> NaNO3 + H2O

Therefore one mole of HNO3 neutralizes one mole of NaOH. The titration added a total of 0.018*0.256 = 0.00461 moles of HNO3. Therefore the orginal solution contained 0.00461 moles of NaOH. This was in 15.6mL (0.0156L) of solution, so the concentration was 0.00461/0.0156 or 0.295M. The mass in the solution was 0.00461*molecular mass of NaOH, or 0.00461*40 = 0.184g

Answer 2

1.
NaOH + HNO3 ===> NaNO3 + H2O

From stoichiometry

15.6 x m /1000 = 18 x 0.256 / 1000 or m = 0.295 M

Molarity of NaOH 0.295 M

2.
Mass of NaOH in 15.6 mls of 0.295 M is

15.6 x 0.295 x 40 / 1000 = 0.184 g