To determine the volume of the 4.71 M iron(II) bromide stock solution needed to prepare 325 mL of a 2.00 M working solution, we will use the dilution formula:
[tex]\[ C_1 V_1 = C_2 V_2 \][/tex]
where:
- [tex]\( C_1 \)[/tex] is the concentration of the stock solution (4.71 M),
- [tex]\( V_1 \)[/tex] is the volume of the stock solution that we need to find,
- [tex]\( C_2 \)[/tex] is the concentration of the working solution (2.00 M),
- [tex]\( V_2 \)[/tex] is the volume of the working solution (325 mL).
Rearranging the dilution formula to solve for [tex]\( V_1 \)[/tex]:
[tex]\[ V_1 = \frac{C_2 \cdot V_2}{C_1} \][/tex]
Now, substituting the known values into the formula:
[tex]\[ V_1 = \frac{2.00 \, \text{M} \times 325 \, \text{mL}}{4.71 \, \text{M}} \][/tex]
Performing the multiplication and division:
[tex]\[ V_1 = \frac{650 \, \text{M} \cdot \text{mL}}{4.71 \, \text{M}} \][/tex]
[tex]\[ V_1 \approx 138.004245625 \, \text{mL} \][/tex]
Rounding to three significant digits:
[tex]\[ V_1 \approx 138 \, \text{mL} \][/tex]
So, the chemist should pour out approximately:
[tex]\[ 138 \, \text{mL} \][/tex]
of the 4.71 M iron(II) bromide stock solution to prepare 325 mL of a 2.00 M working solution.
