concentration of Sn2+?

Question

A sample of pure tin metal is dissolved in nitric acid to produce 15.00 mL of solution containing Sn²⁺.  When this tin solution is titrated, a total of 42.1 mL of 0.145 mol/L KMnO₄is required to reach the equivalence point. 

b.    What is the concentration of the Sn²⁺ solution?

 Find the concentration of the Sn2+_(aq) in mol/L: (give your answer to 3 decimal places)

Answer 1

When Sn2+ is titrated with KMnO4 in HNO4 solution, the chemical process can be described with reaction:

5 Sn(NO3)2 + 2 KMnO4 + 16 HNO3 = 5 Sn(NO3)4 + 2 Mn(NO3)2 + 2 KNO3 + 8 H2O

Sn changes it's oxidation number from +2 to +4, loosing 2 electrons

Mn changes it's oxidation number from +7 to +2, gaining 5 electrons

If x is Sn2+ concentration than equation will be:

V(Sn2+) * C(Sn2+) / 5 = V(KMnO4) * C(KMnO4) / 2

Replacing with actual values will give:

15.00 * x / 5 = 42.1 * 1.145 / 2

x = 1.017 mol/L