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?

Question

9369 views around the world

You can reuse this answer
Creative Commons License

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.

See here and and here for other examples.

The solubility of lead (II) Iodate, Pb(IO_3)_2Pb(IO_3)_2, is 0.76 g/L at 25*C. How do you calculate the Value of Ksp at this same temperature?