Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

Example 4- B:

Relative solubility; first example

We wish to determine the order of solubility of three solid species. In other words we wish to investigate

which of the three following species that are easiest to dissolve:

AgI(s) having a Ksp-value of Ksp = 1.5 · 10-16 M^2

CuI(s) having a Ksp-value of Ksp = 5.0 · 10-12M^2

CaSO 4 (s) having a Ksp-value of Ksp = 6.1 · 10-5 M^2

These three solid species give upon dissolution all two ions following the reaction schemes:

AgI(s)  Ag+(aq) + I-(aq)

CuI(s)  Cu+(aq) + I-(aq)

CaSO 4 (s)  Ca2+(aq) + SO 4 2-(aq)

Or expressed more generally:

Solid specie  Cathion + Anion

Ksp = [Cathion] · [Anion]

I few assume that the solubility of cathion as well as anion is x moles/L at equilibrium we achieve:

[Cathion] = x mol/l

[Anion] = x mol/l

And thereby by insertion in the expression for Ksp:

Ksp = [Cathion] · [Anion] = x^2 

x = Ksp½ = solubility

Thus, form the Ksp-values from above:

AgI(s): Ksp = 1.5 · 10-16 M^2  solubility = 1.2 · 10-8 M

CuI(s): Ksp = 5.0 · 10-12 M^2  solubility = 2.2 · 10-6 M

CaSO 4 (s): Ksp = 6.1 · 10-5 M^2  solubility = 7.8 · 10-3 M

CaSO 4 is thereby must dissolvable followed by CuI and lastly AgI. In that the number of ions for the three

different ions are the same (all include 2 ions) the product of solubility have the units of (M^2 ). Therefore in

this case you may determine the order of dissolubility just by looking at the sizes of the solubility products

(Ksp).

Please note that this is only possible when the units for KSP are similar.

Equilibrium