The Ksp of FeCO3 is 3.13×10−11, the Ksp of Fe(OH)2 is 4.87×10−17, and the Ksp of FeF2 is 2.36×10−6. If a solution contained CO2−3, OH−, and F− ions, each at a concentration of 0.1 M, and the Fe2+ ion concentration is slowly increased from a trace amount, which of the salts would precipitate first?

Fe(OH)₂ will precipitate first

Explanation:

Ksp formulas for FeCO₃, Fe(OH)₂ and FeF₂ are:

FeCO₃:

FeCO₃(s) ⇄ Fe²⁺ + CO₃²⁻

Ksp = [Fe²⁺] [CO₃²⁻]. Replacing with the Ksp and CO₃²⁻ concentration values: 3.13x10⁻¹¹ = [Fe²⁺] [0.1M]

3.13x10⁻¹⁰ =  [Fe²⁺]

That means, the system reaches the equilibrium when   [Fe²⁺]  = 3.13x10⁻¹⁰, for higher concentrations the FeCO₃ begins precipitation.

For the other salts:

Fe(OH)₂:

Fe(OH)₂(s) ⇄ Fe²⁺ + 2OH⁻

Ksp = [Fe²⁺] [OH⁻]². Replacing with the Ksp and OH⁻ concentration values: 4.87x10⁻¹⁷ = [Fe²⁺] [0.1M]²

4.87x10⁻¹⁵ =  [Fe²⁺]

FeF₂:

FeF₂(s) ⇄ Fe²⁺ + 2F⁻

Ksp = [Fe²⁺] [F⁻]². Replacing with the Ksp and F⁻ concentration values: 2.36x10⁻⁶ = [Fe²⁺] [0.1M]²

2.36x10⁻⁴ =  [Fe²⁺]

As you can see, the Fe(OH)₂ begins the precipitation with the lower Fe²⁺ concentration, thus, Fe(OH)₂ will precipitate first.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole