The question requires us to calculate the volume of a 0.0500 M NaOH solution necessary to completely react with 25 mL of 0.100 M HNO3, given the chemical equation for the reaction between NaOH and HNO3.
We can solve this problem using the stoichiometry of the equation given and using the concentration and volume of the solutions to calculate the number of moles of HNO3 and NaOH.
The following chemical equation was provided:
[tex]HNO_{3(aq)}+NaOH_{(aq)}\to NaNO_{3(aq)}+H_2O_{(l)}[/tex]Since the reaction is already balanced, we don't need to adjust the coefficients of the equation. We can see from the stoichiometry that 1 mol of HNO3 reacts with 1 mol of NaOH.
Next, we must calculate the number of moles of HNO3 in 25 mL of a 0.100 M HNO3 solution. We can use the definition of molarity, as it follows:
[tex]\begin{gathered} \text{molar concentration = }\frac{number\text{ of moles (mol)}}{\text{volume (L)}}\to C=\frac{n}{V} \\ n=C\times V \end{gathered}[/tex]Thus, we calculate the number of moles of HNO3 used in the reaction:
[tex]n_{HNO_3}=(0.100\text{ mol/L)}\times(25\text{ mL}\times\frac{1L}{1000\text{ mL}})=0.00250\text{ mol of HNO}_3[/tex]Next, we need to calculate how many moles of NaOH would be necessary to react with 0.00250 moles of HNO3, using the stoichiometry of the reaction:
1 mol HNO3 --------------- 1 mol NaOH
0.00250 mol HNO3 --- x
Solving for x, we'll have that 0.00250 moles of NaOH are necessary.
Now we can convert the number of moles of NaOH calculated (0.00250 moles) to the volume of a 0.0500 M solution of NaOH. We'll use the definition of molarity to perform this calculation:
[tex]\begin{gathered} C=\frac{n}{V}\to V=\frac{n}{C} \\ V_{NaOH}=\frac{(0.00250\text{ mol)}}{(0.0500\text{ mol/L)}}=0.0500\text{ L} \end{gathered}[/tex]Therefore, the volume of NaOH necessary is 0.0500 L or 50.0 mL and the best option to answer the question is the fourth one (50.0 mL).
