# 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?

Jun 7, 2016

${K}_{s p}$ $P b {\left(I {O}_{3}\right)}_{2}$ $=$ ??

#### Explanation:

We can represent the solubility of $P b {\left(I {O}_{3}\right)}_{2}$ as:

$P b {\left(I {O}_{3}\right)}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s P {b}^{2 +} + 2 I {O}_{3}^{-}$

${K}_{s p}$ $=$ $\left[P {b}^{2 +}\right] {\left[I {O}_{3}^{-}\right]}^{2}$

And if we let $S = \text{solubility of lead iodate}$, then,

${K}_{s p}$ $=$ $\left(S\right) {\left(2 S\right)}^{2}$ $=$ $4 {S}^{3}$.

So now we work out the solubility of $P b {\left(I {O}_{3}\right)}_{2}$.

We are given ${S}_{\text{mass}} = 0.76 \cdot g \cdot {L}^{-} 1$

${S}_{\text{molar}} = \frac{0.76 \cdot g}{557.00 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{1 \cdot L}$

$=$ $1.37 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

And thus ${K}_{s p} = 4 \times {\left(1.37 \times {10}^{-} 3\right)}^{3}$ $=$ ??

Lead iodate is thus quite an insoluble beast.

See here and and here for other examples.