Question

Suppose a solution is prepared by dissolving 15.0g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of OH- ions in the solution after the reaction has gone to completion. Assume that there is no volume change when adding the grams of NaOH.

Answer 1

The reaction is:
NaOH (aq) + HNO3 (aq) ? NaNO3 (aq) + H2O
The molecular mass of NaOH is 40 g/mole so:
(15 g) / (40 g/mol) = 0.375 moles of NaOH
are dissolved. Since the volume of the solution is 0.150 L (150 mL) then the initial and final concentration of Na+ is:
(0.375 moles) / (0.150 L) = 2.50 mol/L (M)
because Na+ is a "spectator ion" of the reaction.
The initial number of moles of nitric acid is:
(0.150 L) (0.250 M) = 0.0375 moles.
Since there is a 1:1 stoichiometry of all reactants and products in the reaction, then OH- is present in excess, and all the nitric acid will react. Therefore the final concentration of nitrate ion will equal the initial concentration of nitric acid; NO3- = 0.250 M.
Finally, there will be 0.375 moles - 0.0375 moles = 0.3375 moles of OH- left over from the reaction, also because of the 1:1 stoichiometry of the reagents. The final concentration of OH- is then:
(0.3375 moles) / (0.150 L) = 2.25 M OH-.
The resultant solution is basic (alkaline).