Verified Answer :
- Identify the 1:1 stoichiometry of the neutralization reaction \(\text{HNO}_3 + \text{NaOH} \to \text{NaNO}_3 + \text{H}_2\text{O}\).
- Calculate the moles of \(\text{NaOH}\) using the volume and concentration: \(n_{base} = 0.150 \text{ mol/L} \times 0.025 \text{ L} = 0.00375 \text{ mol}\).
- Set the moles of \(\text{HNO}_3\) equal to the moles of \(\text{NaOH}\) due to the 1:1 ratio, so \(n_{acid} = 0.00375 \text{ mol}\).
- Compute the concentration of the \(\text{HNO}_3\) solution: \(C_{acid} = 0.00375 \text{ mol} / 0.085 \text{ L} \approx \boxed{0.0441 \text{ mol/L}}\).
Explanation
- Analysis of the Reaction and Data
The problem involves a neutralization reaction between nitric acid (\(\text{HNO}_3\)) and sodium hydroxide (\(\text{NaOH}\)). The balanced chemical equation for this reaction is:
\[\text{HNO}_3(aq) + \text{NaOH}(aq) \to \text{NaNO}_3(aq) + \text{H}_2\text{O}(l)\]
The stoichiometry of the reaction is 1:1, meaning one mole of nitric acid reacts exactly with one mole of sodium hydroxide. We are given:
- Volume of nitric acid (\(V_{acid}\)) = \(85 \text{ mL} = 0.085 \text{ L}\)
- Volume of sodium hydroxide (\(V_{base}\)) = \(25.00 \text{ mL} = 0.025 \text{ L}\)
- Concentration of sodium hydroxide (\(C_{base}\)) = \(0.150 \text{ mol/L}\)
Our goal is to find the molar concentration of the nitric acid solution (\(C_{acid}\)).
- Calculating the Number of Moles
First, calculate the number of moles of sodium hydroxide used in the titration using the definition of molarity: \(n = C \times V\).
\[n_{base} = 0.150 \text{ mol/L} \times 0.025 \text{ L} = 0.00375 \text{ mol}\]
Since the stoichiometry of the reaction is 1:1, the number of moles of nitric acid that reacted must be equal to the number of moles of sodium hydroxide:
\[n_{acid} = n_{base} = 0.00375 \text{ mol}\]
- Calculating the Concentration of Nitric Acid
Now that we know the moles of nitric acid (\(n_{acid} = 0.00375 \text{ mol}\)) and the volume of the nitric acid solution (\(V_{acid} = 0.085 \text{ L}\)), we can determine the concentration (\(C_{acid}\)) using the formula \(C = n/V\):
\[C_{acid} = \frac{n_{acid}}{V_{acid}} = \frac{0.00375 \text{ mol}}{0.085 \text{ L}} \approx 0.0441176 \text{ mol/L}\]
Rounding to three significant figures consistent with the provided data, we get:
\[C_{acid} \approx \boxed{0.0441 \text{ mol/L}}\]
Examples
Titrations are a fundamental technique in analytical chemistry used to determine the unknown concentration of a substance. For example, a quality control lab in a manufacturing plant might use titration to verify that the concentration of an acidic cleaning solution matches the required specifications for industrial safety and performance. By neutralizing a small, measured volume of the acid with a known base concentration, the lab can precisely calculate the acid's concentration and ensure the manufacturing process remains consistent and safe.