The solubility of cadmium oxalate, , in 0.150 M ammonia is mol/L. What is the oxalate ion concentration in the saturated solution? If the solubility product constant for cadmium oxalate is , what must be the cadmium ion concentration in the solution? Now, calculate the formation constant for the complex ion

[Cd²⁺] = 2.459x10⁻⁶M

Kf = 9.96x10⁶

Explanation:

Solubility of CdC₂O₄ is 6.1x10⁻³M and ksp is 1.5x10⁻⁸

The ksp of CdC₂O₄ is:

CdC₂O₄(s) ⇄ Cd²⁺(aq) + C₂O₄²⁻(aq)

ksp = [Cd²⁺] [C₂O₄²⁻] = 1.5x10⁻⁸

As solubility is 6.1x10⁻³M, concentration of C₂O₄²⁻ ions is 6.1x10⁻³M. Replacing:

[Cd²⁺] = 1.5x10⁻⁸ / [6.1x10⁻³M]

All Cd²⁺ in solution is 6.1x10⁻³M and exist as Cd²⁺ and as Cd(NH₃)₄²⁺. That means concentration of Cd(NH₃)₄²⁺ is:

[Cd(NH₃)₄²⁺] + [Cd²⁺] = 6.1x10⁻³M

[Cd(NH₃)₄²⁺] = 6.1x10⁻³M - 2.459x10⁻⁶M = 6.098x10⁻³M

In the same way, the whole concentration of NH₃ in solution is 0.150M, as you have 4ₓ6.098x10⁻³M = 0.024M of NH₃ producing the complex, the concentration of the free NH₃ is:

[0.150M] = [NH₃] + 0.024M

The equilibrium of the complex formation is:

Cd²⁺ + 4 NH₃ → Cd(NH₃)₄²⁺

The kf, formation constant, is defined as:

Kf = [Cd(NH₃)₄²⁺] / [Cd²⁺] [NH₃]⁴

Replacing:

Kf = [6.098x10⁻³M] / [2.459x10⁻⁶M] [0.1256M]⁴

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.