In a redox titration 12.50 mL of 0.0800 mol/L K2Cr2O7 (aq) was used in acidic solution to oxidize Sn2 (aq) ions to Sn4 (aq) ions. The volume of K2Cr2O7 (aq) used was just sufficient to oxidize all the Sn2 (aq) in 10.0 mL of the solution. Calculate the concentration of the Sn2 (aq) ions in the solution according to the following unbalanced equation.
Cr2O72- (aq) + Sn2+ (aq) → Snº+(aq) + Cr3+(aq)
Answer:
0.300 M
Explanation:
The balanced equation is:
Now we calculate how many Cr₂O₇²⁻ moles reacted, using the given volume and concentration:
Then we convert Cr₂O₇²⁻ moles into Sn²⁺ moles, using the stoichiometric coefficients of the balanced reaction:
Finally we calculate the concentration of the Sn²⁺ species: