Question

Ammonium hexachloroplalinate (IV), (NH₄)₂ (PtCI₆) is one of the few sparingly soluble ammonium salts. It’s K_SP K_SP at 20⁰c is 5.6X10^-6. Calculate its solubility in grams per litre of solution.

Short answer

• Solubility is defined as the maximum amount of a substance that will dissolve in a given amount of solvent at a specific temperature.
• The solubility of most solids and liquids increases with increasing temperature.

Step-by-step solution

Step(1):- Answer of the question

Given:- K_SP at 20⁰c is 5.6X10^-6 .

The solubility of(NH₄)₂ (PtCI₆) is 4.9gm/lit.

Step(2):-  Calculation of molar solubility

Calculation of molar solubility of (NH₄)₂ (PtCI₆)

The equilibrium equation is-

${\left(N{H}_4\right)}_2\left(PtC{I}_6\right)\left(s\right)\rightleftarrows N{H}_4^{+}\left(aq\right)+PtC{I}_6^{-2}\left(aq\right)$

So, the solubility product is-

$K_{SP}={\left[N{H}_4^{+}\right]}^{2}X\left[PtC{I}_6^{2-}\right]K_{SP}={\left[N{H}_4^{+}\right]}^{2}X\left[PtC{I}_6^{2-}\right]$

If S mole of(NH₄)₂ (PtCI₆) is dissolved In water molar concentrations of NH₄⁺ and PtCI₆^2- will be:-

[NH₄⁺ ] = 2S and [ PtCI₆^2- ] = S

$\begin{array}{rcl}{K}_{SP}& =& {\left(2S\right)}^{2}XS\\ K_{SP}& =& 4S^{2}XS\\ K_{SP}& =& 2S^3\\ 4S^3& =& K_{SP}\\ 4S^3& =& 5.6X{10}^{-6}\\ S^3& =& 1.4X{10}^{-6}\\ S& =& 1.1X{10}^{-2}\end{array}$

Step(3):- Calculation of soubility

Calculation of soubility of (NH₄)₂ (PtCI₆)

The molar mass of (NH₄)₂ (PtCI₆) is 443.87 gm/mol

Gram solubility = molar solubility X molar mass

= ( 1.1X10^-2mol/lit) X(443.87gm/mol)

= 4.9gm/lit

Step(4):- Conclusion

Gram solubility represents the solubility of (NH₄)₂ (PtCI₆) in the unit of gm/mol .

The solubility of (NH₄)₂ (PtCI₆) is 4.9gm/lit