A volume of 40.0 mL of Iron (II) Sulfate is oxidized to Iron (III) by 20.0 mL of 0.100 M Potassium dichromate solution. What is the concentration of the Iron (III) Sulfate Solution?
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This is a classic chemistry problem that deals with Molarity.
Just use this equation M1V1 = M2V2
M1 is molarity for the potassium dichromate = 0.100M
V1 is the volume of original Iron (II) sulfate before oxidation = 40.0 mL
V2 is the final volume after the Iron(II) sulfate is oxidized to Iron (III) by 20 ml Potassium dichromate. So the
final volume is 40ml + 20ml = 60 ml
Solve for M2
M2 = M1V1/V2
= (0.100M) ( 40 ml) / 60 ml
= 4 M ml/ 60 ml
= 0.067 M
Thank you for your time and this helped tremendiously.
Hello, sure and if you have anymore questions. Pls do not hesitate to ask.
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