Here we need to prepare a diluted solution from a more concentrated stock solution using the formula for dilution:
[tex]\[ c_1 v_1 = c_2 v_2 \][/tex]
Where:
- [tex]\( c_1 \)[/tex] = concentration of the stock solution (in Molarity, M)
- [tex]\( v_1 \)[/tex] = volume of the stock solution needed to prepare the new solution (in Liters, L)
- [tex]\( c_2 \)[/tex] = concentration of the desired diluted solution (in Molarity, M)
- [tex]\( v_2 \)[/tex] = final volume of the desired solution (in Liters, L)
From the question, the following information is provided:
- [tex]\( c_1 \)[/tex] = 1.75 M (stock K2CrO4 solution)
- [tex]\( c_2 \)[/tex] = 0.100 M (desired K2CrO4 concentration)
- [tex]\( v_2 \)[/tex] = 2.00 L (final volume of the desired solution)
We want to find [tex]\( v_1 \)[/tex], which is the volume of the stock solution we need to use, and then determine how much water we should add to reach the final volume of 2.00 L.
First, let's rearrange the dilution formula to solve for [tex]\( v_1 \)[/tex]:
[tex]\[ v_1 = \frac{c_2 v_2}{c_1} \][/tex]
Plugging in the values:
[tex]\[ v_1 = \frac{0.100 \times 2.00}{1.75} \][/tex]
Performing the calculations:
[tex]\[ v_1 = \frac{0.200}{1.75} \][/tex]
[tex]\[ v_1 = 0.1143 \][/tex] (approximate value in Liters)
To convert the volume of the stock solution from liters to milliliters, we multiply it by 1000 since there are 1000 milliliters in a liter:
[tex]\[ v_1 \text{ in mL} = 0.1143 \times 1000 \text{ mL/L} \][/tex]
[tex]\[ v_1 \text{ in mL} = 114.29 \text{ mL} \][/tex]
Now, we have the volume of the stock solution that we need in milliliters. The next step is to figure out how much water we need to add to this to get to the final volume of the diluted solution (2.00 L).
Since the final volume needs to be 2.00 L or 2000 mL (since 1 L = 1000 mL), and we have already accounted for 114.29 mL of that volume with the stock solution, the volume of water needed will be:
[tex]\[ v_{\text{water}} = \text{Final volume} - v_1 \text{(stock solution volume)} \][/tex]
[tex]\[ v_{\text{water}} = 2000 \text{ mL} - 114.29 \text{ mL} \][/tex]
[tex]\[ v_{\text{water}} = 1885.71 \text{ mL} \][/tex]
Therefore, you would need 1885.71 mL of water to add to 114.29 mL of the 1.75 M K2CrO4 stock solution to prepare 2.00 L of a 0.100 M K2CrO4 solution.
