Question
7.41 ml of a solution of Fe2+ (aq) is titrated with 19.3 ml of 0.367 M...
7.41 ml of a solution of Fe2+ (aq) is titrated with 19.3 ml of 0.367 M KMnO4 in acidic solution. The balanced Redox equation is: MnO4 (aq) + 5 Fe2+(aq) + 8 H(aq) --> 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(1) Calculate the concentration, in M, of Fe2(aq) in the sample. Express your answer to the appropriate number of significant figures.
Answers
Here:
M(MnO4-)=0.367 M
V(MnO4-)=19.3 mL
V(Fe2+)=7.41 mL
According to balanced reaction:
5*number of mol of MnO4- =1*number of mol of Fe2+
5*M(MnO4-)*V(MnO4-) =1*M(Fe2+)*V(Fe2+)
5*0.367*19.3 = 1*M(Fe2+)*7.41
M(Fe2+) = 4.7794 M
Answer: 4.78 M
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