Question

please consider that all of the parts are related to each other and are about solubility of INSOLUBLE SALTS. I need brief and right answers please consider that while answering my questions.

a) Will any changes be observed in a saturated solution of BaSO4 if some barium chloride solution is added? Write a balanced equation to support your answer. Explain.

b) The measured volume of saturated Ca(IO3)2 solution (filtrate W) was believed to be 1.0 ml and this volume was used in calculations, but due to improper use of a pipette, the actual volume delivered and used for titration was lower. Will the calculated solubility product constant of Ca(IO3)2 increase, decrease or remain unchanged. Justify your answer.

c) Why does the saturated solution of calcium iodate have to be filtered before titration? Hint: think about the equilibrium....

d) A water sample collected from a household faucet in Park City was known to contain Ag+ . Sodium iodide solution was used to detect Ag+ in this water sample. The Ksp for silver iodide (AgI) is 1.5 x 10-16. If the [ I- ] concentration has to reach 0.067 M in this water sample before a precipitate can be detected, what is the detection limit for the silver ion in mol/L as well as in ppm?

Answer 1

a) When BaSO₄ is added to BaCl₂ some of the BaCl₂ dissolves and increases the Ba²⁺ concentration and the concentration of SO₂ decreses since the SO₂ precipitates with the Ba²⁺ formed

BaSO₄ = Ba²⁺ + SO4^2-

b) Ca(IO₃)₂ disassociates into Ca²⁺ and (IO₃)^2-

Ksp = [ Ca²⁺][(IO₃)^2-]

If the measured volume changes the concentration is calculated will be wrong. If the volume is less than the actual then solubility product will increase as the concentration of the terms will be higher.

c) To determine the concentrations of the Ca2+ and (IO3)2- calcium iodate must be filtered out as it will form a reversible reaction leading to the formation of calcium iodate which will result in error

d) Ksp = [Ag⁺][I^-]

1.5 x 10^-16 = [Ag⁺] x 0.067

[Ag⁺] = 2.23 x 10^-15 M

to convert from mol/L to ppm

ppm = mol/L x 1000 x MW

2.23 x 10^-15 x 1000 x 108 = 2.408 x 10^-10 ppm