MC test: Environmental impact - heavy metals

Multiple choice test on A.10 Environmental impact - heavy metals

Use the following 'quiz' to test your knowledge and understanding of this sub-topic. As this relates to a sub-topic on the options you may need access to the IB data booklet.

If you get an answer wrong, read through the explanation carefully to learn from your mistakes.

Which can act as chelating agents?

I. ammonia

II. EDTA4−

III. ethane-1,2-diamine

Chelating agents must contain at least two non-bonding pairs of electrons (on different atoms) that are able to form co-ordinate bonds to a metal or metal ion in two or more positions. Ammonia only contains one non-bonding pair of electrons and acts as a monodentate ligand.

 

The table gives some values for solubility products at 298 K.

Compound Formula Ksp
barium sulfate BaSO4 1.08 x 10−10
lead(II) hydroxide Pb(OH)2 1.43 x 10−20
iron(II) hydroxide Fe(OH)2 4.87 x 10−17
silver(I) cyanide AgCN 5.97 x 10−17

Which will contain the lowest concentration of metal ions (in mol dm−3) in a saturated solution of the salt at 298 K?

[Ba2+] = √(1.08 x 10−10); [Pb2+] = 3√(¼ x 1.43 x 10−20); [Fe2+] = 3√(¼ x 4.87 x 10−17) and [Ag+] = √(5.97 x 10−17) so [Ag+] has the lowest concentration in mol dm−3.

 

Which heavy metals are essential in small quantities for humans but potentially toxic in larger quantities?

I. manganese

II. cadmium

III. copper

Copper is a key constituent of the respiratory enzyme complex cytochrome c oxidase and manganese is a cofactor for a large variety of enzymes. Both copper and manganese are poisonous in larger quantities. Cadmium is non-essential and is extremely poisonous even in small amounts.

 

The Haber-Weiss process is a naturally occurring slow process that generates hydroxyl free radicals from the reaction between hydrogen peroxide and the superoxide free radical ion. Which other two species are also formed in this reaction?

The equation for the Haber-Weiss reaction is .O2−  +  H2O2  →  O2  +  OH−  +  .OH

 

Which contain at least two non-bonding pairs of electrons but cannot act as chelating agents?

I. water

II. carbon monoxide

III. cyanide ions

All three contain two non-bonding pairs of electrons. In water they are both on the oxygen atom, in carbon monoxide and cyanide ions they are on different atoms but too close to each other to occupy more than one coordination site around the metal ion so they are all only monodentate ligands.

 

Which are both correct for the exothermic reaction that occurs when a hexadentate ligand, such as EDTA4−, replaces the water molecules in an hexahydrated transition metal ion?

Compared to the reactants there are more particles in the products in which the effective energy can be distributed so the entropy increases making ΔS positive, this in turn makes ΔG negative (since ΔH is negative as it is an exothermic reaction and ΔG = ΔH − TΔS) making the reaction spontaneous .

 

The solubility product for silver(I) carbonate, Ag2CO3 is given as 8.46 x 10−12 at 298 K in Section 32 of the data booklet.
Which is a true statement about a saturated solution of silver(I) carbonate at 298 K?

Since each Ag2CO3 that dissolves produces two silver ions to one carbonate ion [Ag+(aq)] = 2[CO32−(aq)] so [Ag+(aq)] > [CO32−(aq)]. Ag2CO3 is a solid so its concentration does not appear in the equilibrium expression.
Ksp = [Ag+(aq)]2 x  [CO32−(aq)] = ½[Ag+(aq)]3 so [Ag+(aq)] = 3√(2 x 8.46 x 10−12) = 3√(1.692 x 10−11) mol dm−3. 
Ksp = [Ag+(aq)]2 x  [CO32−(aq)] = 4 x [CO32−(aq)]3 so [CO32−(aq)] = 3√(¼ x 8.46 x 10−12) = 3√(2.115 x 10−12) mol dm−3.

 

What is the function of Fenton's reagent when it is added to polluted water.

Fe2+ reacts with hydrogen peroxide to produce the hydroxyl free radical .OH (together with the hydroxide ion and Fe3+).

 

The solubility product for cadmium(II) hydroxide, Cd(OH)2 is given as 7.2 x 10−15 at 298 K in Section 32 of the data booklet.
What is the solubility of cadmium(II) hydroxide at 298 K?

Since Ksp = [Cd2+(aq)] x [OH−(aq)]2 and [OH−(aq)] = 2 x [Cd2+(aq)]  then Ksp = 4[Cd2+(aq)]3 and [Cd2+(aq)] = 3√(¼ x Ksp) mol dm−3.
The concentration of Cd2+ is the same as the molar solubility of Cd(OH)2, so the correct answer is 3√(1.8 x 10−15) mol dm−3. 

 

The solubility product for lead(II) hydroxide, Pb(OH)2 is given as 1.43 x 10−20 at 298 K in Section 32 of the data booklet. What will be the approximate solubility of lead(II) ions when calcium hydroxide has been added to a saturated solution of lead(II) hydroxide to give a solution with a pH of 11?

At pH = 11,  [OH−(aq)] = 1.0 x 10−3 mol dm−3.
Ksp = [Pb2+(aq)] x [OH−(aq)]2 = [Pb2+(aq)] x (1.0 x 10−3)2 so [Pb2+(aq)] = Ksp ÷ 10−6 = 1.43 x 10−14 mol dm−3.

 

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