Question

Calculate the Molar solubility at 25 degrees clelcius of Cr(OH)3 In:

Water:

1.50 M Cr(NO3)3:

A solution buffered to 10.50:

(please include all of the steps)

Answer 1

K_sp of Cr(OH)₃ is 7*10^-31.

1) Consider the dissociation of Cr(OH)₃ in water; write down the dissociation equation:

Cr(OH)₃ (aq) <======> Cr³⁺ (aq) + 3 OH^- (aq) …..(1)

                                           s                   3s

Let s be the molar solubility (in moles/L) of Cr(OH)₃ in water; as per the molar equation, s mole/L Cr(OH)₃ must give s mole/L Cr³⁺ and 3s mole/L OH^-. The equilibrium constant is

K_sp = [Cr³⁺][OH^-]³ = (s).(3s)³ = 27s⁴

===> 7*10^-31 = 27s⁴

===> s = ⁴√2.5925*10^-321.2689*10^-8 ≈ 1.27*10^-8

The molar solubility of Cr(OH)₃ in water = 1.27*10^-8 mole/L (ans).

2) This is an example of common ion effect; Cr(NO₃)₃ is completely soluble in water as below:

Cr(NO₃)₃ (aq) <====> Cr³⁺ (aq) + 3 NO₃^- (aq)

1.50 M = 1.50 mole/L Cr(NO₃)₃ will furnish 1.5 mole/L Cr³⁺ as per the above dissociation equation.

Let the molar solubility of Cr(OH)₃ be s’mole/L; therefore, s’ mole/L Cr(OH)₃ will furnish 3s’ mole/L OH^- as per equation (1) above.

Therefore,

7*10^-31 = [Cr³⁺][OH^-] = (1.50).(3            s’)³ ( we will assume [Cr³⁺] from dissociation of Cr(OH)₃ is extremely low).

===> 27s’³ = 4.6667*10^-31

===> s’³ = 1.7284*10^-32

===> s’ = 2.5855*10^-11 ≈ 2.58*10^-11

The molar solubility of Cr(OH)₃ in presence of Cr(NO₃)₃ is 2.58*10^-11 M; note that the molar solubility has gone down from part (1) above. This is known as common ion effect. An ion in common with the electrolyte will suppress the dissociation of the electrolyte (ans).

3) pH of the solution is 10.50; we know that pH + pOH = 14.

Therefore, pOH = 14 – pH = 14 – 10.50 = 3.50

Again, pOH = -log [OH^-]

====> [OH^-] = 10^-pOH = 10^-3.50 = 3.1623*10^-4

Move back to equation (1) above and note that

K_sp = [Cr³⁺][OH^-] = (x).(3.1623*10^-4)³ where x is the molar solubility of Cr(OH)₃; Also note that we will ignore the contribution of OH^- from Cr(OH)₃ since the buffer supplies a much higher concentration of OH^-.

Therefore,

7*10^-31 = x.(3.1326*10^-11)

====> x = 2.2136*10^-20 ≈ 2.21*10^-20

The molar solubility of Cr(OH)₃ in buffered solution = 2.21*10^-20 M (ans).