Question

Using Ksp data online pick an anion that could be used to precipitate out Arsenic. Justify your choice with calculations as well as with a consideration for toxicity of the anion. Determine the metal ion concentration in the resulting saturated solution.

Answer 1

Precipitate is a solid insoluble ionic product of reaction. It is formed when ceation and anion react in the aqueous solution. The ion product Q of salt is product of concentration of ion raised to some powers. K_sp defines equilibrium concentration whereas Q defines concentrations not equilibrium concentrations.

If Q< K_sp then the solution is unsaturated and it will be soluble

If Q = K_sp then the solution is saturated and it will be in equilibrium

If Q > K_sp then the solution is supersaturated and ionic solid will precipitate.

K_sp is temperature dependent and at given temperature it is constant.

Arsenic belong to the same group in which phosphorous is present and its properties are similar to phosphorous. It also forms arsenate ion in which oxidation state of arsenic is -3.

Arsenic is presnt in two oxidation state ie, +3 and +5. So arsenic is oxidised to +3 to +5 by using some oxiding agents which are anions. To oxidise arsenic we can use ozone or oxygen and chlorine. Ozone is better than oxygen because in less time and in less concentration it will provide complete oxidation of arsenic. Arsenic(+5) is insoluble, so basically it forms precipitate.

To oxidise arsenic with ozone pH used = 7.6- 8.5, concentration= 46-62 g/L

K_sp = 4.6 X 10^-5 g/L X 1 mol of As/ 122.92 = 3.74 X 10^-7 moles of (AsO₄)^3-

Arsenate oxide or arsenate trioxide are sparingly soluble in water and they have no taste or smell. They are heavier than water.

(AsO₄)^3- --------------------> As⁵⁺ + 2O₂

We have to calculate for As then we already calculated its K_sp value

So,

K_sp = [As][O₂]²

K_sp = x X x² = x³

3.74 X 10^-7 = x³

x³ = 0.0072

Now multiply this value with atomic mass of arsenic

Metal ion concentration = 0.0072 X 74.92 = 0.539 g/L.