Question
How to dilute 5L of 1M HCl to achieve a concentration of 0.25M
Answers
GPT-4o mini
To dilute a solution, you can use the dilution equation:
\[
C_1V_1 = C_2V_2
\]
Where:
- \(C_1\) is the initial concentration (1 M).
- \(V_1\) is the volume of the concentrated solution you need to use (unknown).
- \(C_2\) is the final concentration (0.25 M).
- \(V_2\) is the final volume after dilution.
In this case, you want to dilute 5 L of 1 M HCl to get a 0.25 M solution. First, decide on the final volume you want to use. If you want to keep the final volume at 5 L (which you don't have to do), we can calculate using that.
If you want to prepare a 5 L solution at a 0.25 M concentration, you would first find the total moles needed for the final solution:
\[
\text{Moles of solute} = C_2 \times V_2 = 0.25 \, \text{mol/L} \times 5 \, \text{L} = 1.25 \, \text{mol}
\]
Now, we know that the initial concentration is 1 M:
\[
\text{Moles of solute} = C_1 \times V_1 \implies V_1 = \frac{\text{Moles of solute}}{C_1} = \frac{1.25 \, \text{mol}}{1 \, \text{mol/L}} = 1.25 \, \text{L}
\]
So, you will need 1.25 L of the 1 M HCl solution for this dilution.
Next, you need to find out how much water to add:
\[
\text{Final Volume} - \text{Volume of HCl} = \text{Volume of Water}
\]
\[
5 \, \text{L} - 1.25 \, \text{L} = 3.75 \, \text{L}
\]
Therefore, to make a 0.25 M solution in 5 L, you should:
1. Measure 1.25 L of the 1 M HCl solution.
2. Add enough water to reach a total volume of 5 L which would be 3.75 L of water.
If you wish to prepare a greater final volume, adjust \(V_2\) in the equations accordingly.
\[
C_1V_1 = C_2V_2
\]
Where:
- \(C_1\) is the initial concentration (1 M).
- \(V_1\) is the volume of the concentrated solution you need to use (unknown).
- \(C_2\) is the final concentration (0.25 M).
- \(V_2\) is the final volume after dilution.
In this case, you want to dilute 5 L of 1 M HCl to get a 0.25 M solution. First, decide on the final volume you want to use. If you want to keep the final volume at 5 L (which you don't have to do), we can calculate using that.
If you want to prepare a 5 L solution at a 0.25 M concentration, you would first find the total moles needed for the final solution:
\[
\text{Moles of solute} = C_2 \times V_2 = 0.25 \, \text{mol/L} \times 5 \, \text{L} = 1.25 \, \text{mol}
\]
Now, we know that the initial concentration is 1 M:
\[
\text{Moles of solute} = C_1 \times V_1 \implies V_1 = \frac{\text{Moles of solute}}{C_1} = \frac{1.25 \, \text{mol}}{1 \, \text{mol/L}} = 1.25 \, \text{L}
\]
So, you will need 1.25 L of the 1 M HCl solution for this dilution.
Next, you need to find out how much water to add:
\[
\text{Final Volume} - \text{Volume of HCl} = \text{Volume of Water}
\]
\[
5 \, \text{L} - 1.25 \, \text{L} = 3.75 \, \text{L}
\]
Therefore, to make a 0.25 M solution in 5 L, you should:
1. Measure 1.25 L of the 1 M HCl solution.
2. Add enough water to reach a total volume of 5 L which would be 3.75 L of water.
If you wish to prepare a greater final volume, adjust \(V_2\) in the equations accordingly.