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

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years