How much water would you add to 50.0 mL of 0.700 M NaOH if you wanted the final concentration to be 0.300 M?
mL x M = mL x M
How much water would you add to 250.0 mL of 0.700 M NaOH if you wanted the final concentration to be 0.250 M ?
To determine how much water you would need to add, you can use the formula for dilution:
M1 x V1 = M2 x V2
Where:
M1 = initial concentration of the solution
V1 = initial volume of the solution
M2 = final concentration of the solution
V2 = final volume of the solution
In this case, we know:
M1 = 0.700 M (initial concentration of NaOH)
V1 = 50.0 mL (initial volume of NaOH)
M2 = 0.300 M (final concentration of NaOH)
We need to find V2 (final volume of the solution).
Rearranging the formula gives:
V2 = (M1 x V1) / M2
Substituting the given values, we can calculate:
V2 = (0.700 M x 50.0 mL) / 0.300 M
Simplifying the expression:
V2 = 11.67 mL
Therefore, you would need to add approximately 11.67 mL of water to 50.0 mL of 0.700 M NaOH in order to achieve a final concentration of 0.300 M.