The reaction between nitric acid and sodium hydroxide is shown below. What is the concentration of HNO, if it takes 25.0 mL of 0.493 M NaOH to react with a 10.0 mL sample of HNO3? Write your answer to three significant figures, include units, and show your work for full credit (15 points)
HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(1)
The Correct Answer and Explanation is :
To find the concentration of HNO₃, we can use the concept of stoichiometry based on the balanced chemical equation:
HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H₂O(l)
Step 1: Write down the known values:
- Volume of NaOH (V_NaOH) = 25.0 mL = 0.0250 L
- Concentration of NaOH (C_NaOH) = 0.493 M
- Volume of HNO₃ (V_HNO3) = 10.0 mL = 0.0100 L
- We need to find the concentration of HNO₃ (C_HNO3).
Step 2: Calculate the moles of NaOH:
The number of moles of NaOH is calculated using the equation:
[
\text{moles of NaOH} = \text{concentration of NaOH} \times \text{volume of NaOH}
]
Substitute the known values:
[
\text{moles of NaOH} = 0.493 \, \text{M} \times 0.0250 \, \text{L} = 0.012325 \, \text{moles}
]
Step 3: Use stoichiometry to find the moles of HNO₃:
From the balanced chemical equation, we can see that the mole ratio between NaOH and HNO₃ is 1:1. Therefore, the moles of HNO₃ will be the same as the moles of NaOH:
[
\text{moles of HNO₃} = 0.012325 \, \text{moles}
]
Step 4: Calculate the concentration of HNO₃:
The concentration of HNO₃ can be calculated using the formula:
[
\text{concentration of HNO₃} = \frac{\text{moles of HNO₃}}{\text{volume of HNO₃}}
]
Substitute the values:
[
\text{concentration of HNO₃} = \frac{0.012325 \, \text{moles}}{0.0100 \, \text{L}} = 1.2325 \, \text{M}
]
Step 5: Round the answer:
Since we are asked to round the answer to three significant figures, the concentration of HNO₃ is:
[
\boxed{1.23 \, \text{M}}
]
Conclusion:
The concentration of HNO₃ is 1.23 M.