Question

your lab has yet another analysis to perform. the following was submitted: 0.018 M sodium sulfate and 0.036 M sodium chromate. you have elected to utilize a solution of 0.012 M barium nitrate to test the feasibility of seperation of the ions by selective precipitation.

a) determine what concentration of the barium solution will initiate precipitation for each salt.

b)which barium salt will form first?

c) what is the concentration of the less soluble salt when the most soluble one starts to precipitate?

d) would you be successful in your effort to seperate sulfate and chromate? If so, indicate to what degree.

Answer 1

(a) We know the solubility product (Ksp) of BaSO₄ and BaCrO₄.

BaSO₄ <=> Ba²⁺ + SO₄^2-

Ksp = [Ba²⁺][SO₄^2-] = 1.08x10^-10 .....................(1)

BaCrO₄ <=> Ba²⁺ + CrO₄^2-

Ksp = [Ba²⁺][SO₄^2-] = 1.17x10^-10 . ....................(2)

Given, concentration of SO₄^2- = [SO₄^2-] = 0.018 M [ Na₂SO₄ concentration = 0.018 M]

concentration of CrO₄^2- = [CrO₄^2-] = 0.036 M [ Na₂CrO₄ concentration = 0.036 M]

We know that the precipitation starts when the Ksp is greater than the ionic product (K_i). Now, we have to find the concentration of Barium ion in each case.

(1) [Ba²⁺] = Ksp/[SO₄^2-] = 1.08x10^-10 / 0.018 = 60x10^-10 M = 6.0x10^-9 M

(2) [Ba²⁺] = Ksp/[CrO₄^2-] = 1.17x10^-10 / 0.036 = 32.5x10^-10 M = 3.25x10^-9 M

Hence, concentration of the barium solution to initiate precipitation in each barium sulphate (BaSO₄) and barium chromate (BaCrO₄) are 6.0x10^-9 M and 3.25x10^-9 M.

(b) Barium chromate (BaCrO₄) precipitation will start first because to precipitate of barium chromate requires only

3.25x10-⁹ M of Barium ion which is less than that of Barium sulphate.

(c) Here in this case, less soluble salt completely gets precipitate out from the solution. We know the concentration when the most soluble salt start precipitate i.e. 6.0x10^-9 M concentration of Ba²⁺ will reach.

[CrO₄^2-] = Ksp/[Ba²⁺] = 1.17x10^-10/6.0x10^-9 = 0.195x10^-1 M = 1.95x10^-2 M

Hence, concentration of the less soluble salt is 1.95x10^-2 M.

(d) To find the degree of separation, we have to divide the initial concentration of CrO₄^2- to the separated concentration of CrO₄^2-.

Degree of separation = [CrO₄^2-]_sep/[CrO₄^2-]in

= 1.95x10^-2 M/0.036 M = 0.542 (in percentage 54.2%)

Hence, the separation degree is 54.2%.