Question

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 1

0.300 M

Step-by-step explanation:

• Cr₂O₇²⁻ (aq) + Sn²⁺ (aq) → Sn⁴⁺(aq) + Cr³⁺(aq)

The balanced equation is:

• Cr₂O₇²⁻ (aq) + 3Sn²⁺ + 14H⁺ (aq) → 3Sn⁴⁺(aq) + 2Cr³⁺(aq) + 7H₂O

Now we calculate how many Cr₂O₇²⁻ moles reacted, using the given volume and concentration:

• 12.50 mL * 0.0800 M = 1 mmol Cr₂O₇²⁻

Then we convert Cr₂O₇²⁻ moles into Sn²⁺ moles, using the stoichiometric coefficients of the balanced reaction:

• 1 mmol Cr₂O₇²⁻ *
  `(3molSn^(+2))/(1molCr_2O_7^(2-))`= 3 mmol Sn²⁺

Finally we calculate the concentration of the Sn²⁺ species:

• 3 mmol Sn²⁺ / 10.0 mL = 0.300 M