What is the concentration of Fe(III) with a stock concentration of .00025 M in a solution of 1 ml FeCl3 and 9 ml of HCl?
2 ml of FeCl3 and 8 ml of HCl?
3 ml of FeCl3 and 7 ml of HCl?
4 ml of FeCl3 and 6 ml of HCl?
5 ml of FeCl3 and 5 ml of HCl?
I got
.000025
.00005
.000075
.00010
.000125
is this right?
Was it right?
To determine the concentration of Fe(III) in the given solutions, you need to consider the dilution that occurs when mixing the stock solution (0.00025 M) with the given volumes of FeCl3 and HCl.
The dilution formula you can use is:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
Now, let's calculate the concentrations for each solution:
For the first solution (1 ml FeCl3 and 9 ml HCl):
C1 = 0.00025 M
V1 = 1 ml + 9 ml = 10 ml
C2 = ?
V2 = 10 ml
Using the dilution formula: C1V1 = C2V2
(0.00025 M)(10 ml) = C2(10 ml)
C2 = (0.00025 M)(10 ml) / (10 ml)
C2 = 0.00025 M
Therefore, the concentration of Fe(III) in the first solution is 0.00025 M.
Now, let's calculate the concentrations for the other solutions:
For the second solution (2 ml FeCl3 and 8 ml HCl):
C1 = 0.00025 M
V1 = 2 ml + 8 ml = 10 ml
C2 = ?
V2 = 10 ml
Using the dilution formula: C1V1 = C2V2
(0.00025 M)(10 ml) = C2(10 ml)
C2 = (0.00025 M)(10 ml) / (10 ml)
C2 = 0.00025 M
Therefore, the concentration of Fe(III) in the second solution is also 0.00025 M.
You can follow the same procedure for the remaining solutions:
For the third solution (3 ml FeCl3 and 7 ml HCl):
C2 = 0.00035 M
For the fourth solution (4 ml FeCl3 and 6 ml HCl):
C2 = 0.0004 M
For the fifth solution (5 ml FeCl3 and 5 ml HCl):
C2 = 0.0005 M
Hence, the correct concentrations for the solutions are:
1 ml FeCl3 and 9 ml HCl: 0.00025 M
2 ml FeCl3 and 8 ml HCl: 0.00025 M
3 ml FeCl3 and 7 ml HCl: 0.00035 M
4 ml FeCl3 and 6 ml HCl: 0.0004 M
5 ml FeCl3 and 5 ml HCl: 0.0005 M
Based on the information you provided, your calculated concentrations are correct.