Question

a.)

PbCO3 has a low solubility in fresh water but increases in saline water. Explain this observation. Calculate the difference for the molarity of Pb2+ ion from the values given below.

Ksp = 1.46 x 10^16

(CO3)2- = 0.256

Pb2+ = 0.202

b.) Explain how colloids impact the concentration of free ions in the water column as they enter estuarine enviornment.

Answer 1

Solution :- solubility of leadcarbonate (PbCO₃) increases in saline water because in saline water lead (Pb²⁺ ) forms formes precipitate with chlorine (Cl^- ) so acc. to le-chatelier principle as the product is removed form the reaction the reaction proceed in forwerd direction i.e the equilibrium will shift toward product.

difference in molarity of Pb²⁺ . Ksp = [Pb²⁺][CO₃^2-]

so in first case CO₃^2- = 0.256 , concentration of Pb²⁺ = Ksp/CO₃^2- = 1.46*10¹⁶/0.256 = 5.7*10¹⁶

in secound case concentration of Pb²⁺ = 0.202 , difference in molarity =  5.7*10¹⁶ -  0.202 = 5.7*10¹⁶