How much water must be added to 6 mL of a 0.05M stock solution to diltue it to 0.02M ?
The answerI got is 15 mL. Is this correct? Thanks
Close but no cigar.
6 mL x 0.05 = xmL*0.02
x mL = 15.0 mL which is the final volume. Therefore, you must add 9 mL to the 6 to make 15.0. Technically, you want the final volume to be 15.0 mL; the amount you add may NOT be EXACTLY 9.0 mL since volumes are not additive.
To find the amount of water needed to dilute the solution, we can use the formula for dilution:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution
V1 = volume of the stock solution
C2 = desired concentration
V2 = final volume after dilution
Let's substitute the known values into the formula:
C1 = 0.05 M
V1 = 6 mL
C2 = 0.02 M
V2 = ?
0.05 M * 6 mL = 0.02 M * V2
0.3 = 0.02V2
Now, solve for V2:
V2 = 0.3 / 0.02
V2 = 15 mL
So, the amount of water that should be added to 6 mL of the stock solution is 15 mL. Therefore, your answer of 15 mL is correct.