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?

1 Answer
Jun 7, 2016

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

Explanation:

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.