What is the concentration of the Sn2+ solution?

Question

Part 2: Find the concentration of tin(II) chloride.

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 39.8 mL of 0.100 mol/L KMnO₄ is 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

From your previous question, the reaction taking place is...

5Sn²⁺ + 2MnO₄^- + 16H⁺ + 10e- ==> 5Sn⁴⁺ + 2Mn²⁺ + 8H₂O

mols MnO4- needed = 39.8 ml x 1 L/1000 ml x 0.100 mol/L = 0.00398 mols

mols Sn2+ present = 0.00398 mols MnO4- x 5 mols Sn2+ / 2 mol MnO4- = 0.00995 mols Sn2+

Volume of sample = 15.00 ml x 1 L / 1000 ml = 0.01500 L

[Sn²⁺] = 0.00995 mols / 0.01500 L = 0.663 mol/L