Environmental chemistry E 12. water and soil

Water and soil Solve problems relating to the removal of heavy-metal ions, phosphates and nitrates from water by chemical precipitation. State what is meant by the term cation-exchange capacity (CEC) and outline its importance. Discuss the effects of soil pH on cation-exchange capacity and availability of nutrients. Describe the chemical functions of soil organic matter (SOM).

Chemical precipitation Heavy metal ions can be removed by reacting them with aqueous anions, such as OH-, Cl- , PO43- and S2-, that will make a salt with a low solubility; the heavy metal ions can then be filtered or precipitated out. Examples of such precipitation reactions are: Pb2+ (aq) + 2Cl- (aq) → PbCl2 (s) Cr3+ (aq) + PO43- (aq) → CrPO43 (s)

Chemical precipitation: how much? The solubility product constant is an idea that allows us to calculate how much, for instance, of an anion needs to be added to precipitate out a dissolved hazardous metal cation. We can also calculate to what level the concentration of the cation will be lowered. This calculation is possible because when a salt dissolves to form a saturated solution in water a dynamic equilibrium is set up between the solid salt and its aqueous ions as shown below: MX(s)  aM+ (aq) + bX- (aq) The equilibrium expression for this heterogenous system would be Kc = [M+(aq)]a [X− (aq)]b

Solubility product Ksp only changes with temperature. Its unit depends on the expression; could be mol2 dm-6. The Ksp value of an ionic compound is a measure of how soluble it is in water. In general: low Ksp = low solubility high Ksp = high solubility solubiity product constants The solubility product constant only applies to saturated solutions because only then is there a solid with which an equilibrium is set up. If the solution is not saturated there will be only aqueous ions. Amount of ion above the solubility product is precipitated.

Solubility product expression calculations Solubility product constant At saturation a solution is in a state of equilibrium and the expression exists. The solubility product can be calculated by finding (e.g. through titration) or measuring (probes) the concentration of one of the ions in a saturated solution of the ionic compound and then using the solubility product expression.

Solubility product expression calculations Concentration or solubility (s) of an aqueous metal ion in a saturated solution if Ksp is given. If MX(s)  M+ (aq) + X- (aq) then… Ksp = [M+(aq)] x [X−(aq)] or as [M+(aq)] = [X−(aq)]   then Ksp = [M+(aq)]2 and [M+(aq)] = Ksp

Solubility product expression calculations If MX2(s)  M2+ (aq) + 2X- (aq) then … Ksp = [M2+(aq)] x [X− (aq)]2 as [X− (aq)] = 2 x [M2+(aq)]   then Ksp = [M2+(aq)] x 2 [M2+(aq)]2 and Ksp = 4[M2+(aq)]3 and [M2+(aq)] = (Ksp/4)1/3

Solubility product expression calculations If MX3 (s)  M3+ (aq) + 3X- (aq) then… Ksp = [M3+(aq)] x [X− (aq)]3 as [X− (aq)] = 3 x [M3+(aq)]   then Ksp = [M3+(aq)] x 3 [M3+(aq)]3 and Ksp = 27 [M3+(aq)]4 and [M3+(aq)] = (Ksp/27)1/4

Solubility product expression calculations The solubility of a compound if Ksp is given For this we need to know the solubility or concentration of the metal ion and then use the ratio of the metal ion to the compound. In all three examples in the table above the solubility of the metal ion is also equal to the solubility of the salt MX as they are in a 1:1 ratio. Therefore, if the ratio of M+ : MX is 1 : 1 the solubility of a salt compound, s, at a given temperature can be calculated from its solubility product in exactly the same way as the calculation of the metal ion concentration as shown in the table below.

Common ion effect To precipitate out a salt, the product of the concentrations of its aqueous ions needs to be greater than the solubility product constant. In that case the equilibrium shifts to the left producing more insoluble salt (s) and decreasing the concentration of the aqueous ions. In practical terms, if a metal ion, e.g. Cr3+, needs to be removed from a solution, then another solution with the same non-metal ion, e.g. OH-, as a chromium compound with a low Ksp, e.g. Cr(OH)3, needs to be added. Another common ions are phosphates and carbonates as many phosphate salts and carbonates have low solubility. The common ion is added as part of a compound with high solubility e.g. a sodium salt. The OH- is considered the common ion as it is in both the chromium compound with low solubility, Cr(OH)3 and in the compound that is in the solution that has been added, e.g. NaOH.

Common ion effect The effect of the addition of the common ion is the lowering of the concentration and precipitation of most Cr3+ (aq). We can calculate the new concentration of the hazardous ions after the common ion has been added. When the common ion solution has been added to the saturated solution of the hazardous ion the concentration of the ions in the saturated solution are ignored and the concentration of the common ion is used in the calculation.

Common ion effect For instance at 298K the solubility product constant of cadmium sulphide is 1.0 x 10-27. Therefore the solubility of Cd2+ or [Cd2+(aq)] = 1.0 x 10-27 which is 3.16 x 10-14 mol dm-3 which is also the solubility of S2-. If a 0.5 mol dm-3 NaS solution is added to the CdS solution to precipitate out most Cd2+, we ignore the original concentrations of Cd2+ and S2- and use the concentration of the common ion to calculate the reduced solubility of Cd2+. The solubility of Cd2+ in 0.5 mol dm-3 NaS: 1.0 x 10-27 = [Cd2+(aq)] x 0.5 [Cd2+(aq)] = 1.0 x 10-27 / 0.5 = 2.0 x 10-27

Cation exchange capacity - CEC (1) CEC is the amount of exchangeable cations, such as K+, Ca2+ and Mg2+, that 100g of soil can hold CEC is an indicator of the fertility of a soil. It is the clay (mainly) and humus in a soil that give the soil its CEC. Measured in millequivalent (mg) of H+ (or singly positive ions) usually per 100 g of soil or could be 1kg. The higher the CEC value the more fertile the soil.

Cec (2) Plants need to absorb cations. Plants do this through cation exchange with the soil at their root hairs. Exchange of H+ with K+ or Ca2+ or Mg2+. The amount of cations the soil can exchange with plants depends on amount of cations it is able to absorb in the first place. Most important factor that affects the amount of cations a soil can absorb is the amount of clay or humus/SOM as the cations are adsorbed on their surface. If cations are not adsorbed by the soil particles, they are easily washed away (=leached) e.g. in sandy soils. measurements of CEC

Cec (3) Clay has a layer structure which has an overall negative charge. This negative charge attracts cations to the surface of the clay sheets to balance out the negative charge. Cations are attracted weakly onto the clay and humus. Cations can be exchanged for hydrogen ions, H+ (aq), at the roots of plants. clay- - K+ (s) + H+ (aq)  clay– - H+ (s) + K+ (aq) The K+ (aq) is now available to the plants. Exchange is facilitated by the large surface area of clay.

CEC

Cec (4) – negative charge on clay The net negative charge on clay occurs as a result of silicon atoms (oxidation number +4) being replaced by aluminium atoms (+3) or even iron atoms (+2) which do not balance out the negative charges of all the oxygen atoms. A clay with more iron atoms has a greater CEC value than a clay with many aluminium atoms.

CEC

Cec (5) - humus Humus/SOM contains weak organic acids, RCOOH. RCOOHs exchange cations with the soil which is how the cations are absorbed (really part of compound): RCOOH (humus) + K+ (aq)  RCOOK (humus) + H+ (aq) At the roots of plants the cations are exchanged again with the roots: RCOOK (humus) + H+ (aq)  RCOOH (humus) + K+ (aq)

Effect of pH on CEC Low pH lowers CEC At a low pH, H+ ions displace the exchangeable cations on the surface of the clay: clay - Mg + 2H+ (aq)  clay - 2H + Mg2+ (aq)  As a result these essential cations are not being adsorbed by the clay, lower CEC value, and are easily leached leaving the soil with fewer nutrients.   High pH increases CEC The hydroxide ions remove H+ ions from the hydroxyl group on the clay giving the clay a negative charge increasing CEC clay - OH + OH- (aq)  clay – O- + H2O

Effect of pH on availability (1) Ionic nutrients are available if they are aqueous (only way plants can absorb) and adsorbed onto clay or SOM – if not they are leached The best availability of nutrients is between pH 6 and 6.5; around neutral.

pH and availability ion low pH neutral high pH Ca2+/Mg2+ low availability as H+ ions displace Ca2+/Mg2+ from clay surfaces and Ca2+/Mg2 are washed away. maximum availability not available as they form insoluble carbonates or phosphates Fe3+/Al3+ form insoluble hydroxides PO43- most available unless there are Fe3+/Al3+ which form insoluble phosphates Fe3+ (aq)+ PO43- (aq)→ FePO4 (s) forms insoluble phosphates with Ca2+ Ca2+ (aq)+ PO43- (aq) → Ca3 (PO4)2 (s)

washed away at low pH as displaced by H+ Ph and availability ion low pH neutral high pH NO3- NO3- is reduced to NH4+ which is not available to plants Half-equation of reduction: NO3- +10H++ 8e- → NH4+ +3H2O Less nitrogen available to plants maximum availability maximum availability of nitrate – some NH4+ lost as NH3 (g) at higher pH: NH4+ (aq) + OH- (aq) → H2O (l) + NH3 (g) K+ washed away at low pH as displaced by H+ Cu2+/Zn2+ unavailable as forms insoluble hydroxides Cu2+ (aq) + OH- (aq) → Cu(OH)2 (s)

Chemical functions of SOM Contributes to cation-exchange capacity (CEC) as they form stable complexes with cations. Enhances the ability of soil to buffer changes in pH. Reduces the negative environmental effects of pesticides, heavy metals and other pollutants by binding contaminants. Binds to organic and inorganic compounds in the soil preventing nutrients from easily being washed away