- Identify the known values for the dilution formula: \(M_i = 18.0 \text{ M}\), \(M_f = 2.50 \text{ M}\), and \(V_f = 50.0 \text{ mL}\).
- Set up the equation \(V_i = \frac{M_f V_f}{M_i}\) to find the required volume of stock solution.
- Calculate the result: \(V_i = \frac{2.50 \times 50.0}{18.0} \approx 6.94 \text{ mL}\).
- Select option (b) as the correct value: \(\boxed{6.94 \text{ mL}}\).
Explanation
- Analysis of the Dilution Problem
The goal is to determine the volume \(V_i\) of a concentrated \(18.0 \text{ M } H_2SO_4\) stock solution needed to prepare \(50.0 \text{ mL}\) of a diluted \(2.50 \text{ M } H_2SO_4\) solution. We are given the dilution formula \(M_i V_i = M_f V_f\), where:
- \(M_i\) is the molarity of the stock solution (\(18.0 \text{ M}\)).
- \(V_i\) is the required volume of the stock solution (the unknown we need to find).
- \(M_f\) is the final molarity of the solution (\(2.50 \text{ M}\)).
- \(V_f\) is the final volume of the solution (\(50.0 \text{ mL}\)).
By substituting these values into the given equation, we can solve for \(V_i\).
- Setting up the Equation
Using the dilution formula \(M_i V_i = M_f V_f\), we substitute the known quantities into the equation:
\[18.0 \text{ M} \times V_i = 2.50 \text{ M} \times 50.0 \text{ mL}\]
To isolate \(V_i\), we divide both sides by \(18.0 \text{ M}\):
\[V_i = \frac{2.50 \text{ M} \times 50.0 \text{ mL}}{18.0 \text{ M}}\]
- Calculation and Conclusion
Now, perform the arithmetic calculation:
\[V_i = \frac{125 \text{ M} \cdot \text{mL}}{18.0 \text{ M}} \approx 6.9444... \text{ mL}\]
Rounding this result to three significant figures, which matches the precision of the given values, we find that the required volume is approximately \(6.94 \text{ mL}\). Comparing this result with the provided choices, it corresponds to option (b).
Examples
In a professional chemistry laboratory, precise dilution is a foundational skill. For instance, if a researcher needs to create a specific concentration of a reagent for an experiment from a more concentrated stock, they use the dilution equation. This ensures that the chemical reactions occur under controlled conditions, whether they are preparing buffer solutions for biological studies or specific concentrations of acids for titration. Mastering this formula helps maintain accuracy and safety in any scientific practice involving chemical solutions.