Answer :
To find the concentration of HNO3 in both scenarios, we can use the concept of neutralization and stoichiometry. The balanced chemical equation for the reaction between HNO3 (nitric acid) and NaOH (sodium hydroxide) is:
[tex]\text{HNO}_3 + \text{NaOH} \rightarrow \text{NaNO}_3 + \text{H}_2\text{O}[/tex]
This equation shows that HNO3 and NaOH react in a 1:1 mole ratio.
Next, let's tackle each scenario one by one.
Scenario 1:
For the reaction of 50.0 mL of HNO3 with 120.0 mL of 2.50 M NaOH:
Calculate the moles of NaOH:
- Moles of NaOH = Volume (L) × Molarity
- [tex]0.120 \text{ L} \times 2.50 \text{ M} = 0.300 \text{ moles of NaOH}[/tex]
Determine moles of HNO3 needed:
- Because of the 1:1 mole ratio, moles of NaOH = moles of HNO3 needed = 0.300 moles.
Calculate the concentration of HNO3:
- Molarity of HNO3 = Moles / Volume (L)
- [tex]\frac{0.300 \text{ moles}}{0.050 \text{ L}} = 6.00 \text{ M}[/tex]
For this scenario, the concentration of HNO3 is 6.00 M (Option 4).
Scenario 2:
For the reaction of 25.0 mL of HNO3 with 20.0 mL of 3.00 M KOH:
Balanced equation for HNO3 and KOH:
- [tex]\text{HNO}_3 + \text{KOH} \rightarrow \text{KNO}_3 + \text{H}_2\text{O}[/tex]
Calculate the moles of KOH:
- [tex]0.020 \text{ L} \times 3.00 \text{ M} = 0.060 \text{ moles of KOH}[/tex]
Determine moles of HNO3 needed:
- Again, using a 1:1 mole ratio, moles of KOH = moles of HNO3 needed = 0.060 moles.
Calculate the concentration of HNO3:
- [tex]\frac{0.060 \text{ moles}}{0.025 \text{ L}} = 2.4 \text{ M}[/tex]
For this scenario, the concentration of HNO3 is 2.4 M (Option 2).
By using the mole concept and stoichiometry, we've determined the concentrations for both scenarios clearly.
