How much water should be added to 55.0 mL of 0.201 M HCl to dilute the solution to 0.155 M?
To solve this problem, we can use the formula for dilution:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
In this case, we are given:
C1 = 0.201 M (initial concentration)
V1 = 55.0 mL (initial volume)
C2 = 0.155 M (final concentration)
We need to find V2 (final volume), which represents the volume of water that needs to be added.
Rearranging the formula, we have:
V2 = (C1 * V1) / C2
Now let's substitute the values into the equation:
V2 = (0.201 M * 55.0 mL) / 0.155 M
Calculating this expression, we find:
V2 = 71.03 mL
Therefore, you need to add 71.03 mL of water to the 55.0 mL of 0.201 M HCl to dilute the solution to 0.155 M.