How many ml of water must be added to 40.0 ml of 2.50 m koh to prepare a solution that is 0.212 m?
mL1 x M1 = mL2 x M2
x*0.212M = 40*2.5
Solve for x and that is the TOTAK volume required. Total-40 is what must be added.
Do you know that M and m are not the same thing? Which do you want?
Apologies, the numbers corrected are: 40.0 ml of 2.50 M KOH to prep a solution that is 0.212 M
To find out how many mL of water must be added, we need to calculate the final volume of the solution. The total volume of the solution will be the sum of the volume of the water added and the initial volume of the 2.50 M KOH.
Given:
Initial volume of KOH solution (V1) = 40.0 mL
Initial concentration of KOH solution (C1) = 2.50 M
Final concentration of the solution (C2) = 0.212 M
Now, we can use the formula:
C1V1 = C2V2
Substituting the given values:
(2.50 M)(40.0 mL) = (0.212 M)(V2)
Solving for V2 (the final volume):
V2 = (2.50 M)(40.0 mL) / 0.212 M
V2 = 1180.38 mL
Therefore, you would need to add approximately 1180.38 mL of water to 40.0 mL of the 2.50 M KOH solution to prepare a solution with a concentration of 0.212 M.