Question

1. Determine ionic strength, I and mean ionic activity coefficient, γ± at 298 K for

1. 0.025 molal K₂SO₄
2. 0.01 molal Li₂CO₃
3. 0.01 molal Ca₂(PO₄)₂

2. Estimate the solubility of

1. MgF₂ (K_sp=6.4x10^-9) in an aqueous solution with I=0.003 molal.   
2. BaSO₄ (K_sp=1.08x10^-10) in an aqueous solution with I=0.001 molal.   

Answer 1

Ionic strength (I) is given by the formula,

c_i denotes the concentration of the ion "i" and Z denotes the valency of the concerned ion.

(i)

0.025 molal K₂SO₄

K₂SO₄ dissociates to giving 2 K⁺ ions and 1 SO₄^2- ions.

When we gave 0.025 molal of K₂SO₄, we get 0.05 molal of K⁺ ions a 0.25 molal of SO₄^2- ions.

Therefore, For K⁺ ions, c = 0.05 and valency is the charge on the ion and is equal to +1

For SO₄^2- ions, c = 0.025 and valency is -2

Therefore, ionic strength:

(ii)

0.01 Li₂CO₃

1 mole Li₂CO₃ dissociates to give 2 moles of Li⁺ and 1 mole of CO₃^2-

Therefore when we have 0.01 molal of Li₂CO₃, we get 0.02 molal of Li⁺ and 0.01 molal of CO₃^2-

Valency of Li⁺= 1 and Valency of CO₃^2-= -2

Therefore,

(iii)

0.01 molal Ca₃(PO₄)₂

1 mole of Ca₃(PO₄)₂ dissociates to give 3 moles of Ca²⁺ ion and 2 moles of PO₄^3- ion.

Therefore, 0.01 molal of Ca₃(PO₄)₂ gives 0.03 molal of Ca²⁺ and 0.02 molal of PO₄^3-.

Valency of Ca²⁺ = +2 and Valency of PO₄^3-= -3

Therefore, Ionic strength :

Solubilities:

(i)

We need to calculate the solubility of MgF₂ in a solution of MgF₂ with ionic strength of 0.003 molal

MgF₂ dissociates into Mg²⁺ and 2 F^-. Therefore let the concentration of Mg²⁺ be X molal and therefore the concentration of F^- will be 2X molal.

Valency of Mg²⁺ = +2 and the Valency of F^- = -1

Given: K_sp = 6.4 x 10^-9

On substituting the values of concentration and valency in the formula for Ionic strength, we get:

Therefore, concentration of the Mg²⁺ is 0.001 molal and the concentration of the F^- is 0.002 molal

K_sp is given by the formula,

where a is the number of molecules of cation formed and b is the number of molecules of anion formed.

From the given sample of MgF₂, Let X be the molal of MgF₂ that actually dissociates, therefore, concentration of Mg²⁺ formed is X and that of F^- is 2X.

a for Mg²⁺ is 1 and b for F^- is 2 as MgF₂ dissociates into 1 molecule of Mg²⁺ and 2 molecule of F^-

As, the solution already contains some dissolved MgF₂, we will have to consider this concentration as well.

as calculated above, the total Concentration of Mg²⁺= X+0.001

The total concentration of F^-= 2X+0.002

Therefore,

K_sp is in the order of magnitde of 10^-9 and therefore its safe to say X will be also in a similar order of magnitude because of which we can ignore powers of 2 and 3 in X as well as the last 2 terms, therefore we get:

X is the desired solubility of the compound

(ii)

BaSO₄

BaSO₄ dissociates into Ba²⁺ and SO₄^2-.

The given solution has an ionic strength of 0.001 molal

let the concentration of Ba²⁺ be X then the concentration of the SO₄²⁺ will also be X.

Valency of Ba²⁺ and SO₄^2- is (+2) and (-2) respectively.

a =1 and b=1 as only 1 unit of Ba²⁺ and SO₄^2- is formed.

Therefore, if we have X is the concentration of BaSO₄, the concentration of Ba²⁺ and SO₄^2- based on stoichiometry is X and X molal

Total Concentration of Ba²⁺= X + 0.00025 molal

Total Concentration of SO₄^2-= X + 0.00025 molal

K_sp= 1.08 x 10^-10

Ignoring X² and the second term, we get :

X is the required solubility.