Presentation on theme: "E NVIRONMENTAL CHEMISTRY E 12. water and soil. W ATER AND SOIL Solve problems relating to the removal of heavy- metal ions, phosphates and nitrates from."— Presentation transcript:
W ATER 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).
C HEMICAL PRECIPITATION Heavy metal ions can be removed by reacting them with anions, such as OH -, Cl -, PO 4 3- and S 2-, 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: Pb 2+ (aq) + 2Cl - (aq) → PbCl 2 (s) Cr 3+ (aq) + PO 4 3- (aq) → CrPO 4 3 (s)
C HEMICAL PRECIPITATION : HOW MUCH ? The solubility product constant allows us to calculate how much of the anion needs to be added to precipitate out a metal ion. This is possible because dissolution of a salt is an equilibrium process as a dynamic equilibrium is set up between the salt and its aqueous ions as shown below: MX(s) aM + (aq) + bX - (aq) (metal M = metal ion and X = non metal ion) The equilibrium expression for this heterogenous system would be K c = [M + (aq)] a [X − (aq)] b [MX(s)] is a constant as it is a solid!
S OLUBILITY PRODUCT K sp only changes with temperature. Its unit is mol 2 dm -6. The K sp value of an ionic compound is a measure of how soluble it is in water. In general: low K sp = low solubility high K sp = high solubility solubiity product constants A saturated solution is when the product of the concentrations of the aqueous ions according to the solubility product expression = K sp at that temperature. Amount of ion above the solubility product is precipitated.
S OLUBILITY PRODUCT CALCULATIONS Calculating ion concentration of metal ion in a saturated solution when K sp is given If MX(s) M + (aq) + X - (aq) then… K sp = [M + (aq)] x [X − (aq)] or as [M + (aq)] = [X − (aq)] then K sp = [M + (aq)] 2 and [M + (aq)] = K sp
S OLUBILITY PRODUCT CALCULATIONS Calculating ion concentration of metal ion in a saturated solution when K sp is given If MX 2 (s) M 2+ (aq) + 2X - (aq) then … K sp = [M 2+ (aq)] x [X − (aq)] 2 as [X − (aq)] = 2 x [M 2+ (aq)] then K sp = [M 2+ (aq)] x 2 [M 2+ (aq)] 2 and K sp = 4[M 2+ (aq)] 3 and [M 2+ (aq)] = (K sp /4) 1/3
S OLUBILITY PRODUCT CALCULATIONS Calculating ion concentration of metal ion in a saturated solution when K sp is given If MX 3 (s) M 3+ (aq) + 3X - (aq) then… K sp = [M 3+ (aq)] x [X − (aq)] 3 as [X − (aq)] = 3 x [M 3+ (aq)] then K sp = [M 3+ (aq)] x 3 [M 3+ (aq)] 3 and K sp = 27 [M 3+ (aq)] 4 and [M 3+ (aq)] = (K sp /27) 1/4
C OMMON ION EFFECT To precipitate out a salt, the concentrations of the aqueous ions needs to be greater than the solubility product.!!! The equilibrium shifts to the left producing more insoluble salt (s) and decreasing the concentration of the ions. In practical terms, if a metal ion, e.g. Cr 3+, needs to be removed from a solution, then another solution with the same non-metal ion, e.g. OH -, as a chromium salt with a low K sp, e.g. Cr(OH) 3, needs to be added. The OH - is considered the ion common to both the chromium salt with low solubility, Cr(OH) 3 and the solution added, e.g. NaOH.
C ATION EXCHANGE CAPACITY - CEC (1) cations such as K +, Ca 2+ and Mg 2+ 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. CEC is the amount of exchangeable cations a soil can hold, usually in clay measured in millequivalent (mg) of H + usually per 100 g of soil the higher the CEC the more fertile the soil.
C EC (2) plants need to absorb cations they do this through nutrient exchange with the soil at their root hairs; exchange between H + and K + or Mg 2+ or Ca 2+. the amount of cations the soil can exchange depends on amount of cations it adsorbs in the first place. most important factor that affects the amount of cations a soil can adsorb is the amount of clay or humus/SOM. if cations are not adsorbed they are easily washed away (=leached) e.g. in sandy soils. measurements of CEC
C EC (3) clay has a layer structure which has an overall negative charge. this negative charge attracts cations to surface of the clay sheets to balance out the negative charge. cations can be exchanged for hydrogen ions, H + (aq), at the roots of plants. this exchange is facilitated by the large surface area of clay.
C EC (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 Fe atoms has a greater CEC than a clay with many aluminium atoms.
C EC (5) - HUMUS Humus/SOM contains weak organic acids, RCOOH, which also can exchange cations as shown below and therefore contribute to the CEC of a soil. RCOOH (humus) + K + (aq) RCOOK (humus) + H + (aq) At the roots of plants during cation exchange the reverse reaction occurs.
E FFECT OF P H 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 - H + Mg 2+ (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 - + H 2 O
E FFECT OF P H ON AVAILABILITY (1) Ionic nutrients are available if they are aqueous and adsorbed onto clay or SOM. If the ions are insoluble they are unavailable as nutrients; they are also unavailable if they are aqueous but not adsorbed onto clay or humus as they are then likely to be leached from the soil. The best availability of nutrients is between pH 6 and 6.5; around neutral.
P H AND AVAILABILITY ionlow pHneutralhigh pH Ca 2+ /Mg 2+ low availability as H + ions displace Ca 2+ /Mg 2+ from clay surfaces and Ca 2+ /Mg 2 are washed away. maximum availability not available as they form insoluble carbonates or phosphates Fe 3+ /Al 3+ maximum availabilityform insoluble hydroxides PO 4 3- most available apart from Fe 3+ /Al 3+ which form insoluble phosphates maximum availability forms insoluble phosphates with Ca 2+
P H AND AVAILABILITY ionlow pHneutralhigh pH NO 3 - NO 3 - is reduced to NH 4 + which is not available to plants Half-equation of reduction: NO 3 - +10H + + 8e - → NH 4 + +3H 2 O Less nitrogen available to plants maximum availability maximum availability of nitrate – some NH 4 + lost as NH 3 (g) at higher pH: NH 4 + (aq) + OH - (aq) → H 2 O (l) + NH 3 (g) K+K+ washed away at low pH as displaced by H + maximum availability Cu 2+ /Zn 2+ maximum availability unavailable as forms insoluble hydroxides
C HEMICAL 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