What is Ksp in chemistry?

1 Answer
Jun 5, 2016

Answer:

#K_(sp)# is the so-called solubility product, that quantifies the solubility of a salt in water.

Explanation:

Consider a sparingly soluble salt, #MX#, in water.

We can represent its solubility in water in the following way:

#MX(s) rightleftharpoonsM^+ + X^-#

As for any equilibrium, we can write (and quantify) this equilibrium:

#([M^+][X^-])/([MX(s)])# #=# #K_(sp)#

But #[MX(s)]# is meaningless, as you cannot have the concentration of a solid, so we are left the solubility expression:

#K_(sp) = [M^+][X^-]#

#K_(sp)# have been measured for a great variety of insoluble and sparingly soluble salts . Why? Because suppose you were trying to isolate precious metal salts, i.e. those of gold, or rhodium, or iridium. You don't want to throw precious metals away. Likewise, if you had lead, or cadmium, or mercury salts, you don't want to throw these metals away, for the reason that you might poison the waterways.

#K_(sp),"lead chloride "=1.62xx10^-5# at #25# #""^@C#. A temperature is specified because a hot solution can normally hold more solute than a cold one.

#PbCl_2(s) rightleftharpoons Pb^(2+) + 2Cl^-#

And, #K_(sp)=[Pb^(2+)][Cl^-]^2=1.62xx10^-5#.

If we say #[Pb^(2+)]=S#, then #K_(sp)=(S)(2S)^2#.

i.e. #K_(sp)=4S^3#.

And thus #S# #=# #""^3sqrt{{(1.62xx10^-5)/(4)}# #=# #??*mol*L^-1#.

I leave it to you to solve for the solubility of lead chloride in water in #g*L^-1# under standard conditions.