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 KMnO4 is required to reach the equivalence point. a. What is the concentration of the Sn2+ solution?b. Find the concentration of the Sn2+(aq) in mol/L: (give your answer to 3 decimal places)

1.00 M

Explanation:

Sn^2+ reacts with KMNO4 as follows;

5Sn^2+(aq) + 2MnO4^-(aq) + 16H^+(aq) > 5Sn^4+(aq) + 2Mn^+(aq) + 8H2O(l)

The number of moles of MnO4^- reacted = 42.1/1000 L × 0.145 mol/L

= 0.0061 moles

If 5 moles of Sn^2+ reacts with 2 moles of MnO4^-

x moles of Sn^2+ reacts with 0.0061 moles of MnO4^-

x= 5 × 0.0061/2

x= 0.015 moles

Since the volume of the Sn^2+ solution is 15.00mL or 0.015 L

number of moles = concentration × volume

Concentration = number of moles/volume

Concentration= 0.015 moles/0.015 L

Concentration = 1 M

they both have a y in them