# chemistry

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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?

• chemistry - ,

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.

• chemistry - ,

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

• chemistry - ,

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.

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