Chemistry Concentration question

Question

Tin(II) ions react with periodate ions in acid medium to give iodide and tin (IV) ions. Twenty-five mL of a 0.250 M solution of tin(II) nitrate react with 45.00 mL of 0.100 M solution of sodium periodate. Assuming 100% yield and additive volumes

 

1. what is the concentration of tin (IV) ions after the reaction is complete?

 

 

2. What is the concentration of tin(II) ions after reaction is complete

Answer 1

Balanced redox equation: 4Sn²⁺ + IO₄^- + 8H⁺ ==> 4Sn⁴⁺ + I^- + 4H₂O

moles Sn²⁺ present = 0.025 L x 0.250 mol / L = 6.25x10^-3 mols Sn²⁺

moles IO₄^- present = 0.045 L x 0.100 mol / L = 4.5x10^-3 mols IO₄^-

Based on the stoichiometry of the balanced equation, Sn2+ is the limiting reactant, thus will be used up.

(1). Concentration of Sn4+:

6.25x10^-3 mols Sn²⁺ x 4 mols Sn⁴⁺ / 4 mols Sn²⁺ = 6.25x10^-3 mols Sn⁴⁺ formed

Final volume of reaction = 25 ml + 45 ml = 70 mls = 0.0700 L

[Sn⁴⁺] = 6.25x10^-3 mols Sn²⁺ / 0.0700 L = 0.0893 M Sn⁴⁺

(2). Concentration of Sn²⁺:

Since Sn²⁺ is limiting the [Sn²⁺] = 0. All will be converted to Sn⁴⁺