how many milliliters of 0.10 M HCl must be added to 50.0 mL of 0.40 HCl to give a final solution that has a molarity of 0.25 M?
work with the moles.
x ml of .1M HCl has .0001x moles HCl
50ml of .40M HCl contains .05*.4 = .02M HCl
(x+50)ml of .25M HCl contains (x+50)/1000 * .25 = .00025x + .0125 M of HCl
so, the moles need to add up:
.0001x + .02 = .00025x + .0125
x = 50
check:
since equal amounts of .1M and .4M acid are mixed, the concentration will be the average of .1 and .4, or .25
To find the amount of 0.10 M HCl that must be added, we can use the formula:
M1V1 = M2V2
Where:
M1 = initial molarity of HCl solution
V1 = initial volume of HCl solution
M2 = final molarity of HCl solution
V2 = final volume of HCl solution
Given:
M1 = 0.40 M (initial molarity)
V1 = 50.0 mL (initial volume)
M2 = 0.25 M (final molarity)
V2 = unknown
Rearranging the formula, we have:
V2 = (M1 * V1) / M2
Substituting the given values:
V2 = (0.40 M * 50.0 mL) / 0.25 M
V2 = (20 mL * 50) / 0.25
V2 = (1000 mL) / 0.25
V2 = 4000 mL
Therefore, 4000 mL (or 4.0 L) of 0.10 M HCl must be added to 50.0 mL of 0.40 M HCl to obtain a final solution with a molarity of 0.25 M.
To solve this problem, we can use the concept of dilution.
The formula for dilution is:
M1V1 = M2V2
Where
M1 is the initial molarity of the solution
V1 is the initial volume of the solution
M2 is the final desired molarity of the solution
V2 is the final desired volume of the solution
We have the following values:
M1 = 0.40 M (initial molarity of 50.0 mL of 0.40 M HCl)
V1 = 50.0 mL (initial volume of 0.40 M HCl)
M2 = 0.25 M (final desired molarity)
V2 = ?
Using the formula, we can rearrange it to solve for V2:
V2 = (M1V1) / M2
Substituting the known values:
V2 = (0.40 M) * (50.0 mL) / (0.25 M)
Calculating:
V2 = 80.0 mL
Therefore, you would need to add 80.0 mL of 0.10 M HCl to 50.0 mL of 0.40 M HCl to obtain a final solution with a molarity of 0.25 M.