5Fe(2+ charge) + MnO4(-1 charge) +8H(1+ charge) -> 5Fe(3+ charge) + Mn(2+ charge) + 4H2O
In a titration experiment based on the equation above, 25ml of an Acidified Fe(2+ charge) solution requires 14 ml of standard .050 M MnO4(1- charge) solution to reach the equivalence point. What is the M of the Original Fe(2+ charge) solution?
1. Calculate mols MnO4^-. mols = L x M.
2. Convert mols MnO4^- to mols Fe^+2 using the balanced equation you have written.
3. Convert mols Fe^+2 to M using L x M = mols. You know L and you know mols so solve for M.
Post your work if you get stuck.
I come up with 1.4M for the answer but the paper says the answer is .14M. I know it is only minor but i want to find my mistake. heres my work: (.14L)(.5M)= .07mol/1mol=.07x5mol=.35mol/.25L=1.4M
Its 0.014 L (14 mL) and 0.05 M (not 0.5 M) to give 0.0007 mols MnO4^-
0.0007 mols x 5/0.025 L = 0.14 M.
25 mL is 0.025 L.
To determine the M (molarity) of the original Fe(2+) solution, we can use the concept of stoichiometry and the volume and molarity of the MnO4(-1) solution used in the titration.
Given:
Volume of MnO4(-1) solution used in the titration = 14 mL
Molarity of MnO4(-1) solution = 0.050 M
From the balanced chemical equation:
5 Fe(2+) + MnO4(-1) + 8 H(+) → 5 Fe(3+) + Mn(2+) + 4 H2O
According to the stoichiometry of the balanced equation, the ratio of the moles of Fe(2+) to MnO4(-1) is 5:1.
Since the volume of the MnO4(-1) solution used in the titration is given in mL, we need to convert it to liters by dividing it by 1000:
Volume of MnO4(-1) solution used in the titration = 14 mL = 14/1000 L = 0.014 L
Using the equation:
Moles of MnO4(-1) = Molarity × Volume (in L)
Moles of MnO4(-1) = 0.050 M × 0.014 L = 0.0007 moles
From the stoichiometry of the balanced equation, we know that the ratio of moles of Fe(2+) to moles of MnO4(-1) is 5:1.
Therefore, the moles of Fe(2+) in the original solution would also be 0.0007 moles.
To find the Molarity of the original Fe(2+) solution, we can use the formula:
Molarity = Moles / Volume (in L)
The volume of the original Fe(2+) solution is given as 25 mL. Convert it to liters by dividing it by 1000:
Volume of Fe(2+) solution = 25 mL = 25/1000 L = 0.025 L
Molarity = 0.0007 moles / 0.025 L = 0.028 M
Therefore, the Molarity of the original Fe(2+) solution is 0.028 M.