Question

1/ The molar solubility of PbI2 is 1.5 x 10-3 M.

a/ What is the molar concentration of iodide ion in a saturated PbI2 solution?

b/ Determine the solubility constant, Ksp, for lead(II) iodide

2/ Calculate the molar solubility of PBI2 in the presence of 0.10 M NaI

3/ Compare the molar solubility given in problem 1 and the molar solubility calculated in problem 2. Explain the cause of the difference.

Answer 1

Concepts and reason

The maximum quantity of a given solute soluble in a given solvent at a given temperature and pressure is termed as the solubility of the solute. When the quantity is expressed in moles, the molar solubility is obtained.

Ionic product: The product of concentration of chemically active ions in a solution forms the ionic product of the solution.

Saturated solution: A solution in which no more solute could be dissolved.

Solubility product: The ionic product of a saturated salt solution is termed as the solubility product of the salt. Hence, when the ionic product of a given solution exceeds the solubility product of the salt, the solute precipitates out from the solution.

Fundamentals

A sparingly soluble salt AB exists in the following equilibrium,

The equilibrium constant could be given as follows:

The chemical activity of AB is taken as 1,

Hence $K=K=[A*].[B] $ … solubility product

With the help of ICE table, the molar concentrations of the acid and base in the equilibrium reaction can be calculated.

ICE table:

I denotes the initial concentration of each species in the equilibrium reaction.

C denotes the change in the concentration of each species that move towards the equilibrium.

E denotes the concentration of each species at the equilibrium.

(1.a)

The equilibrium reaction would be,

It could be rewritten as follows:

‘x’ moles shall dissociate to give ‘2x’ moles of iodide ions.

There are of dissociated; hence, there shall be of ions.

The concentration of iodide ions is.

‘x’ moles shall dissociate to give ‘2x’ moles of iodide ions.

There are of dissociated; hence, there shall be of ions.

The concentration of iodide ions is.

(1.b)

The solubility product could be written as follows:

The concentration of the Pb (II) ions shall be the same as that of. Hence, the solubility product could be calculated as follows:

(2)

The ICE table could be written as follows:

The solubility product expression could be written as follows:

The equation obtained is solved for ‘x’.

The molar solubility ‘x’ is.

(3)

The solubility before the addition of NaI was, while after addition it decreased to. This is because of the common ion effect where the common ion is the iodide ion.

Ans: Part 1.a

The concentration of iodide ions in a saturated solution is