Water+(HL)

=Water HL=

Joe Yang Topic 12.1 Solve the problem relating to the removal of heavy-metal ions, phosphates and nitrates from water by chemical precipitation.

From topic 6.2, we know the three different stages of treatment. The primary treatment, secondary treatment and tertiary treatment. Different treatment have it's specific job on solving different kind of wastes in water. The primary treatment is use to filter out large structure of waste like rubbish and dead animals. Secondary treatment is basically removing the organic material and bacterias from the water. The tertiary treatment is to remove the heavy metal from the water by the process of chemical precipitation =Removal of heavy metal ions and phosphates from water by chemical precipitation.=

From the previous chapter, we know that heavy metal and phosphates can be remove from water by the tertiary treatment by chemical precipitation. But, not all the salt formed by the precipitation of heavy metal are insoluble in water. So not all the heavy metal can be removed from the water using this method.

=The equilibrium solubility product= If a metal M and a non metal X formed a salt MX, the chemical equation will be M+ + X- <-> MX, which MX is a solid. and from the formular, we can find out the equilibrium constant by multiplying [M+] and [X-], the expression of this equilibrium will be Ksp =[M+][X-] where Ksp stands for solubility product. The solubility product affects the solubility of salt in water. Very insoluble salt has a relatively small value of Ksp,

An example will be AgCl, the solubility product for AgCl is 1.80x10^-10 To reduce the level of heavy metal in water, we need to decrease the amount of Ag+ dissolved in water. Therefore we need to calculate the amount of Ag+ dissolved in water. From Ksp=[M+][X-] we know that Ksp=[Ag+][Cl-] and [Ag+]=[Cl-] so [Ag+]^2 = Ksp= 1.80x10^-10 at 298 K and we can get the result that [Ag+]=1.34x10^-5 mol/dm3 From the result, we can calculate the mass of silver dissolve in one litre of water by multiply the molar mass of silver which is 107.87 to the concentration of Ag+. 107.87x1.34x10^-5=1.45x10^-3g so 1.45x10^-3g of Ag+ ion will be dissolve in one litre of water at 298 K. Although the amount of Ag looks small but it is a significant amount which can harm the human body.

=Common Ion effect= From the previous image, we can see that by adding OH- ion into the equilibrium will result on the increase of production of NH3 in order to balance the equilibrium. In the water treatment case, we are maximizing the amount of heavy metal precipitated.

The common ion effect is basically changing one of the solutes to decrease or increase the concentration of the other solute in the solution. In this example, if we increase the concentration of Cl- from 1.80x10^-10 mol/dm3 to 1.00 mol/dm3 then from the equation [Ag+]=[Cl-]=1.80x10^-10, we can get the result that after increase the concentration of Cl-, we will then decrease the concentration of Ag+ from 1.34x10^-5 to 1.80x10^-10. Then we can calculate the mass of Ag+ after we increase the concentration of Cl-. 107.87x1.80x10^-10= 1.94x10^-8g of Ag+ dissolved in one litre of water when the concentration of Cl- increases. So this method can help us to reduce the concentration of heavy metal in water which can make the water we're drinking cleaner.

=Phosphate and heavy metal ions= For phosphate and heavy metal found in the water, the principle of removing it is the same as the previous one. An example of this will be removing chromium ions (Cr3+) in waste water by adding hydroxide ions (OH-). The chemical equation of this reaction will be: Cr3+(aq) + 3OH-(aq) <->Cr(OH)3(s) And the Ksp which is solubility product is 1.00x10^-33 at 298K. From the value they give us, we can calculate the concentration of Cr3+ ion in one liter of water. Ksp= [Cr3+][OH-]^3 from the expression we can know that [Cr3+]=3[OH-] therefore the concentration of Cr3+ will be 1.00x10^-33=27[Cr3+]^4 which is 5.93x10^-9 mol/dm3 Then we multiply the molar mass of Cr which is 51.99. 5.93x10^-9x51.99=3.08x10^-7 g of Cr+ions dissolve in one litre of water. Due to the common ion effect, we can reduce this number to a smaller extent by increase the concentration of hydroxide ion. If we increase the concentration of hydroxide ions to 1.00 mol/dm3, then the mass of Cr+ ions dissolve in water will be decrease from 3.08x10^-7 g to 5.199x10^-32.

The method of precipitation can help us remove a significant amount of heavy metal in the water, however there are some heavy metal ions re-dissolve when the concentration of hydroxide ion is too high. For example zinc will react hydroxide to become [Zn(OH)4]2-(aq) which means the zinc we are trying to remove react back to an aqeous solution, so we cannot precitipate zinc out from the water.

When facing the heavy metal like Hg2+, Zn2+ and Cd2+, we add in bubbling hydrogen sulphide(H2S). The hydrogen sulphide will react with the heavy metal to form a metal sulfide and hydrogen ions. Example: Zn2+(aq) + H2S(g) --> ZnS(s) + 2H+ (aq)

References: 1. Neuss,G. (2007). Chemistry Course Companion. In G. Neuss, Chemistry Course Companion. Oxford University Press

2. Water.(2009). Water Treatment. Retrived October 13th 2010. []

3. Emmellin Tung(n.d.). Common ion effect. Retrived October 13th 2010. []

4.Water Treatment(n.d.). Retirved October 14th 2010. []

5.