You have 37.0 mL of a 3.00 M stock solution that must be diluted to 0.500 M. How much water should you add?
Dr. Bob22 and I disagreed, but use MV=MV to solve for the volume that you need, but first convert mL to L. V=[(3.00M)(0.037L)]/(0.500M). Remember, you already have an initial volume of 37mL or 0.037L, so subtract 0.037L from the answer that you get for V to get the amount of H20 to add to the stock solution.
You're right. I disagree but the practical aspects are right.
3*37/0.500 = 222 mL is the total volume you want. In practice, that is ABOUT 222-37 = ABOUT 185 mL. In theory, however, that assumes that 185 mL + 37 mL = 222 mL but volumes are not additive unless we are adding substance A to more of substance A. For example, 50 mL H2O + 50 mL ethyl alcohol = about 95 mL and not 100 mL. I know I'm repeating myself but I think it is a mistake to foster the incorrect impression among students that volumes are additive. In many cases addition of volumes is essentially correct but in many cases that isn't so. Technically, it is never correct. :-)
To dilute the stock solution from 3.00 M to 0.500 M, you need to add water. The amount of water to be added can be calculated using the formula:
C1V1 = C2V2
Where:
C1 = initial concentration (3.00 M)
V1 = initial volume (37.0 mL)
C2 = final concentration (0.500 M)
V2 = final volume (unknown, which is what we need to calculate)
Rearranging the formula to solve for V2:
V2 = C1V1 / C2
Substituting the known values into the equation:
V2 = (3.00 M)(37.0 mL) / 0.500 M
Now, we can calculate the value of V2:
V2 ≈ 222 mL
Therefore, you should add approximately 222 mL of water to the 37.0 mL stock solution to get a final concentration of 0.500 M.