Asample of impure tin of mass 0.540 g is dissolved in strong acid to give a solution of sn2+. the solution is then titrated with a 0.0448 m solution of no3−, which is reduced to no(g). the equivalence point is reached upon the addition of 3.39×10−2 l of the no3− solution. find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents.

The sample has 50% tin

Explanation:

Step 1: Determine the (half) reactions

Sn(s) → Sn2+(aq)

NO3-(aq) →NO(g)

Step 2: Balance the Oyxgen atoms by adding H+

4H+ NO3- → NO + 2H2O(l)

4H+(aq) + NO3-(aq) -→ NO(g) + 2H2O(l)

Step 3: Balance the charges by adding electrons

Sn(s) → Sn2+(aq) +2e-

4H+(aq) + NO3-(aq) +3e-→ NO(g) + 2H2O(l)

Step 4: To balance the 2 equations we should multiply the first reaction by 3; and multiply the second reaction by 2

3 Sn(s) → 3 Sn2+(aq) +6e-

8H+(aq) + 2 NO3-(aq) +6e-→ 2 NO(g) + 4H2O(l)

3Sn + 8H+ +2NO3- → 3Sn2+ + 2NO +4H2O

Step 5: Calculate moles of NO3-

Moles of NO3- = 0.0448 M * 0.0339 = 0.00152 moles

Step 6: Calculate moles of Sn

As we can see in the balanced equation, for 2 moles of NO3- Consumed, we need 3 moles of Sn to consume.

This means that for 0.00152 moles of NO3- we need 0.00152 * (3/2) = 0.00228 moles of Sn

Step 7: Calculate mass of Sn

mass of Sn = moles of Sn * Molar mass of Sn

mass of Sn = 0.00228 moles * 118.71g/mole = 0.27 grams

Step 8: Calculate the mass %

(0.27g / 0.54g) * 100 % = 50%

The sample has 50% tin

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.