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A10.3 Solubility Product

For a metal hydroxide that is only very slightly (sparingly) soluble, when the solid is in a saturated solution, the following equilibrium exists with exchange between ions in the solid and in the solution:

                                     M(OH)n (s)\textsf{M(OH)}_\textsf{n}\textsf{ (s)}M(OH)n​ (s)
  ⇌ Mnn+(aq) + nOH−(aq)\textsf{Mn}{^\textsf{n+}} \textsf{(aq) + nOH}{^{-}} \textsf{(aq)}Mnn+(aq) + nOH−(aq)
 

At this point, the addition of any more hydroxide ions will cause a little more of the metal hydroxide to precipitate.
Since this is an equilibrium reaction we can write :
 Kc =[Mn+][OH−]n[M(OH)n]\textsf{Kc =} \frac{\textsf{[M}^\textsf{n+}]\textsf{[OH}^{-}]^\textsf{n}}{\textsf{[M(OH)}_\textsf{n}\textsf{]}}Kc =[M(OH)n​][Mn+][OH−]n​
     where [  ] represents the concentration in mol dm−3\textsf {mol dm}^{-3}mol dm−3
 
Since M(OH)n is a solid, its concentration is a constant. So we can now make use of a new constant, called the Solubility Product, Ksp\textsf K_\textsf{sp}Ksp​
 
                                       Ksp\textsf{K}_\textsf{sp}Ksp​
=  [Mn+][OH−]n\textsf{[M}^\textsf{n+}\textsf{][OH}^{-}\textsf{]}^\textsf{n}[Mn+][OH−]n
 
Since the substance is barely soluble. Ksp\textsf K_{\textsf{sp}}Ksp​
 will have a very low value.
 Worked example 🖊
 
Calculations using the of solubility product
(These are most difficult for a 1:2 like cadmium hydroxide)
The Ksp\textsf{K}_\textsf{sp}Ksp​
 of cadmium hydroxide = 2.5×10-14mol3dm−9 \textsf{2.5}\times \textsf{10} ^\textsf{-14} \textsf{mol}^{3} \textsf{dm}^{-9}2.5×10-14mol3dm−9
  at 298K 
1.Calculate the solubility of Cd(OH)2\textsf{Cd(OH)}_\textsf{2}Cd(OH)2​
 in g dm-3\textsf{g dm}^\textsf{-3}g dm-3
 in pure water

 Let the solubility Cd(OH)2\textsf{Cd(OH)}_\textsf{2}Cd(OH)2​
 = s g dm-3\textsf{= s g dm}^\textsf{-3}= s g dm-3
 
 this means that in a saturated solution  [Cd2+] = s\textsf{[Cd}^\textsf{2+} \textsf{] = s}[Cd2+] = s
 and  [OH-] = 2s\textsf{[OH} ^\textsf{-}\textsf{] = 2s}[OH-] = 2s

 Ksp=[Cd2+] [OH-] 2 = s ×(2s)2 = 4s3\textsf{K}_\textsf{sp} = \textsf{[Cd}^\textsf{2+} \textsf{] }\textsf{[OH} ^\textsf{-}\textsf{] }^\textsf{2}\textsf{ = s }\times\textsf{(2s)}^\textsf{2}\textsf{ = 4s}^{3}Ksp​=[Cd2+] [OH-] 2 = s ×(2s)2 = 4s3
 sos3 = 14(2.5×10-14), s = 2.92×10−5\textsf{s}^\textsf{3}\textsf{ = }\frac{1}{4}  \textsf{(2.5}\times\textsf{10}^\textsf{-14}\textsf{), s = 2.92}\times\textsf{10}^{-5}s3 = 41​(2.5×10-14), s = 2.92×10−5

                                                                                                                                                       
 So the solubility of Cd(OH) Cd(OH)2 = 2.92×10-5mol dm−3\textsf{Cd(OH)}_\textsf{2}\textsf{ = 2.92}\times\textsf{10}^\textsf{-5}\textsf{mol dm}^{-3}Cd(OH)2​ = 2.92×10-5mol dm−3
 and in g dm-3\textsf{g dm}^\textsf{-3}g dm-3
  = 2.92×10-6×molar mass = 2.92×10-6×[112.4 +2(16.00 + 1.01)]= 2.92×10−6×146.42 = 4.27×10−4g dm−3\textsf{= 2.92}\times\textsf{10}^{\textsf{-6}}\times\textsf{molar mass } \\  \textsf{= 2.92}\times \textsf{10}^\textsf{-6}\times\textsf{[112.4 +2(16.00 + 1.01)]} \\ \textsf{= 2.92}\times\textsf{10}^{-6}\times\textsf{146.42 = 4.27}\times\textsf{10}^{-4}\textsf{g dm}^{-3} = 2.92×10-6×molar mass = 2.92×10-6×[112.4 +2(16.00 + 1.01)]= 2.92×10−6×146.42 = 4.27×10−4g dm−3
  
​
2.Calculate the solubility of Cd(OH)2\textsf{Cd(OH)}_\textsf{2}Cd(OH)2​
 in g dm-3\textsf{g dm}^\textsf{-3}g dm-3
  in a 0.10 mol dm-3\textsf{0.10 mol dm}^\textsf{-3}0.10 mol dm-3
  solution of NaOH\textsf{NaOH}NaOH
 

This time the concentration of hydroxide ions from the NaOH will be affected the way the Cd(OH)2\textsf{Cd(OH)}_\textsf{2}Cd(OH)2​
 dissolves

As above:  
Let the solubility of Cd(OH)2= s g dm-3\textsf{Cd(OH)}_\textsf{2}\textsf{= s g dm}^\textsf{-3}Cd(OH)2​= s g dm-3
  
this means that in a saturated solution  [Cd2+] = s\textsf{[Cd}^\textsf{2+} \textsf{] = s}[Cd2+] = s
  but  [OH-] = 0.10\textsf{[OH} ^\textsf{-}\textsf{] = 0.10}[OH-] = 0.10
 Ksp=[Cd2+ ][OH−]2= s×(0.10)2 = 0.01s\textsf{K}_\textsf{sp} =  \textsf{[Cd}{^\textsf{2+}}\textsf{ ][OH}{^{-}}\textsf{]}{^\textsf{2}}  \textsf{=  s} \times \textsf{(0.10)} {^\textsf{2}}  \textsf{ =  0.01s}Ksp​=[Cd2+ ][OH−]2= s×(0.10)2 = 0.01s
 so 2.5×10-14 = 0.01s\textsf{2.5}\times \textsf{10} {^\textsf{-14}}  \textsf{ =  0.01s}2.5×10-14 = 0.01s
 

and in g dm-3= 2.6×10−12×146.42 = 3.80×10−10 g dm-3\textsf{g dm} {^\textsf{-3}}\textsf{= 2.6}\times\textsf{10}^{-\textsf{12}}\times \textsf{146.42 = 3.80}\times\textsf{10}^{-\textsf{10}}\textsf{ g dm} {^\textsf{-3}}g dm-3= 2.6×10−12×146.42 = 3.80×10−10 g dm-3
 
                                                                        
From the calculations above, you can see that the solubility of cadmium ions decreases from  2.6 x10 −6{^{-6}}−6
  to  2.6 x10 −12{^{-12}}−12
 by adding the sodium hydroxide.
This explains why the metal ions are precipitated by increasing the pH of the solution.
The calculations are much simpler for 1:1 compounds like barium sulphate.
You can follow the same steps as in the example above.

The Ksp\textsf K_\textsf{sp}Ksp​
​=BaSO4=\textsf{BaSO}_\textsf4=BaSO4​
 at 298\textsf{298}298
K is 1.1×10-10\textsf{1.1}\times\textsf{10}{^\textsf{-10}}1.1×10-10
 
​[Ba2+]\textsf{[Ba}{^\textsf{2+}}\textsf{]}[Ba2+]
  = SO42-\textsf{= SO}_\textsf4{^\textsf{2-}}= SO4​2-
 = s =\textsf{= s =}= s =
 (1.1×10-10)\sqrt{\textsf{(1.1}\times\textsf{10}}{^\textsf{-10}\textsf)}(1.1×10​-10)
 
and from this s = 1.05×10-5mol dm−3\textsf{1.05}\times\textsf{10}{^\textsf{-5}}\textsf{mol dm}^{-\textsf{3}}1.05×10-5mol dm−3
 
A10.2 Removal of Heavy Metals
A10.4 Polydentate Ligands and Chelation
Last modified 2yr ago