Question

Suppose a solution is prepared by dissolving 15 g NaOH in 0.15 L of 0.25 M nitric acid. What is the final concentration of OH ions in the solution after the reaction has gone to completion?

a) 0.25 M
b) 0.5 M
c) 1.0 M
d) 1.5 M

Answer 1

To find the final concentration of OH- ions, first calculate the moles of NaOH and HNO3, then determine the excess after neutralization. The resulting concentration of OH- ions is the moles of excess NaOH divided by the total volume after the reaction.

Step-by-step explanation:

The goal is to determine the final concentration of OH- ions after NaOH is dissolved in a solution of nitric acid (HNO3). The reaction between NaOH and HNO3 is a neutralization reaction where NaOH and HNO3 react in a 1:1 molar ratio to form water and a salt (NaNO3).

First, calculate the moles of NaOH and HNO3. Moles of NaOH = mass (g) / molar mass (g/mol) = 15 g / 40.00 g/mol = 0.375 mol. The moles of HNO3 are given by the concentration multiplied by the volume in liters: 0.25 M * 0.15 L = 0.0375 mol.

Since NaOH and HNO3 react 1:1 and there is more NaOH present, all of the HNO3 will be neutralized, and excess NaOH will remain. Moles of remaining NaOH = 0.375 mol - 0.0375 mol = 0.3375 mol.

The final concentration of OH- ions is equal to the moles of remaining NaOH divided by the total volume of the solution. The volume of the solution after reaction is still 0.15 L, so the concentration of OH- ions is 0.3375 mol / 0.15 L = 2.25 M. However, no answer option matches this result, implying there may have been an error in the question details or answer options provided.