What is the molar solubility of #PbBr_2# in pure water?

1 Answer
Dec 10, 2015

Answer:

You need a value for the solubility product, #K_(sp)# at a given temperature.

Explanation:

The interwebs give a value for the solubility product of lead bromide as #K_(sp)# #=# #1.86xx10^(-5)# at #20""^@C#. This is an equilibrium constant for the reaction:

#PbBr_2(s) rightleftharpoons Pb^(2+) + 2Br^-#

And so #K_(sp)# #=# #1.86xx10^(-5)# #=# #[Pb^(2+)][Br^-]^2#. We cannot speak of the concentration of a solid, which is why #PbBr_2# does not appear in the expression. If we were to call the solubility of lead bromide #S#, then #K_(sp) = Sxx(2S)^2# #=# #4S^3#.

So #S# #=# #root(3){(1.86xx10^(-5))/4}#. The answer has units #mol*L^-1# as required. Please do not trust my arithmetic.

Would you expect #K_(sp)# to increase or decrease at elevated temperature? Why or why not??

If we were to make #[Br^-]# artificially high (say by adding quantities of #NaBr#), would the solubility of the lead salt decrease or increase; why or why not?