1. The equation for the reaction of bromine and methanoic acid is

a. Suggest a way of keeping the concentration of methanoic acid virtually constant

b. The result of an experiment in which the concentration of bromine was monitored through out a reaction are shown in the table

Time ( s )	(/mol dm-3
0	0.01
25	0.009
50	0.008
75	0.007
100	0.0065
180	0.0050
240	0.0045
360	0.0030
420	0.0025
480	0.0020

Plot a graph of againts time show that this reaction is first order by working out three values for half life.

2. The half life of radioactive iodine 131 is 8.0 days what fraction of the initial amount of iodine 131 should be present in a patient after 24 days if none were to be eliminated through natural body process.

3. Explain why no external indicator is needed when titrating solution containing ethamidioated ions with potassium magnate (VII).

4. Calculated the volume of 0.02 mol dm^-3 potassium magnate (VII) solution need to completely oxidize 25.0cm³ if:

a. 0.01 moldm^-3 iron (II) sulphate solution

b. 0.2 moldm^-3 hydrogen peroxide solution

c. 0.1 moldm^-3 ethamidioate solution.

5. A 1,340 gr sample of iron are dissolved in dilute sulphuric acid to produce a solution of iron (II) sulphate this was titrated with 0.02 moldm^-3 potassium magnate (VII) solution and it was found that 28.75 cm³ was required. Calculated the percentage of iron in the iron ore.

Answer

1. Known : v olume bleach 2 cm³

0.05 moldm^-3

2.

Nt = 16.375 gr

3. Because the potassium manganate (VII) has ability of self indicating. So, there’s no needed to its solution.

4.

a. Known :

Concentration of KMnO₄ = 0.02 mol dm^-3

Volume = 25 cm³ = 25 x 10^-3 dm³

Concentration of FeSO₄ = 0.01 mol dm^-3

Solution :

Amount of KMnO₄ = 0.02 mol dm^-3 x 25 x 10^-3 = 5 x 10^-4 mol

From redox equation :

MnO₄^-_(aq) + 8H⁺_(aq) + 5Fe²⁺_(aq) à 5Fe³⁺_(aq) + 4H₂O_(l) + Mn²⁺_(aq)

1 mol of MnO₄^- will react with 5 mol of Fe³⁺

So, the amount of Potassium manganate (VII) = 5 . 10^-4 mol x 5 = 25 . 10^-4 mol = 2.5 x 10^-3 mol

4B. Known :

Concentration of KMnO₄ = 0.02 mol dm^-3

Volume = 25 cm³ = 25 x 10^-3 dm³

Concentration of H₂O₂ = 0.2 mol dm^-3

Solution :

Amount of KMnO₄ = 0.02 mol dm^-3 x 25 x 10^-3 dm³ = 5 . 10^-4 mol

From redox equations :

MnO₄^-_(aq) + 8H⁺_(aq) + 5O^-_(aq) à Mn²⁺_(aq) + 5/2O_2(aq) + 4H₂O_(aq)

1 mol of MnO₄^- will reacts with 5/2 mol of O₂

So, the amount of Potassium Manganate (VII) = 5 . 10^-4 mol x 5/2 = 1.25 x 10^-3 mol

4C

Known :

Concentration of KMnO₄ = 0.02 mol dm^-3

Volume = 25 cm³ = 25 x 10^-3 dm³

Concentration of FeSO₄ = 0.01 mol dm^-3

Solution :

Amount of KMnO₄ = 0.02 mol dm^-3 x 25 x 10^-3 = 5 x 10^-4 mol

From redox equation :

MnO₄^-_(aq) + 8H⁺_(aq) + 5C₂H₄O₂^-_(aq) à Mn²⁺_(aq) + 5/2 C₂H₄O_{2 (aq)} + 4H₂O_(aq)

1 mol of MnO₄^- will react with 5 mol of Mn²⁺

So, the amount of Potassium manganate (VII) = 5 . 10^-4 mol x 5 = 25 . 10^-4 mol = 2.5 x 10^-3 mol