The solubility of lead (II) Iodate, #Pb(IO_3)_2#, is 0.76 g/L at 25*C. How do you calculate the Value of Ksp at this same temperature?

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Answer 1 · 2016-06-07T08:12:26

#K_(sp)# #Pb(IO_3)_2# #=# #??#

We can represent the solubility of #Pb(IO_3)_2# as:

#Pb(IO_3)_2(s) rightleftharpoons Pb^(2+) + 2IO_3^-#

#K_(sp)# #=# #[Pb^(2+)][IO_3^-]^2#

And if we let #S="solubility of lead iodate"#, then,

#K_(sp)# #=# #(S)(2S)^2# #=# #4S^3#.

So now we work out the solubility of #Pb(IO_3)_2#.

We are given #S_("mass")=0.76*g*L^-1#

#S_("molar")=(0.76*g)/(557.00*g*mol^-1)xx1/(1*L)#

#=# #1.37xx10^-3*mol*L^-1#

And thus #K_(sp)=4xx(1.37xx10^-3)^3# #=# #??#

Lead iodate is thus quite an insoluble beast.

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