The solubility of lead (II) Iodate, $P b {(I O_{3})}_{2}$ $Pb(IO_3)_2$ , 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

$K_{s p}$ $K_(sp)$ $P b {(I O_{3})}_{2}$ $Pb(IO_3)_2$ $=$ $=$ $? ?$ $??$

We can represent the solubility of $P b {(I O_{3})}_{2}$ $Pb(IO_3)_2$ as:

$P b {(I O_{3})}_{2} (s) ⇌ P b^{2 +} + 2 I O_{3}^{-}$ $Pb(IO_3)_2(s) rightleftharpoons Pb^(2+) + 2IO_3^-$

$K_{s p}$ $K_(sp)$ $=$ $=$ $[P b^{2 +}] {[I O_{3}^{-}]}^{2}$ $[Pb^(2+)][IO_3^-]^2$

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

$K_{s p}$ $K_(sp)$ $=$ $=$ $(S) {(2 S)}^{2}$ $(S)(2S)^2$ $=$ $=$ $4 S^{3}$ $4S^3$ .

So now we work out the solubility of $P b {(I O_{3})}_{2}$ $Pb(IO_3)_2$ .

We are given $S_{mass} = 0.76 \cdot g \cdot L^{- 1}$ $S_("mass")=0.76*g*L^-1$

$S_{molar} = \frac{0.76 \cdot g}{557.00 \cdot g \cdot m o l^{- 1}} \times \frac{1}{1 \cdot L}$ $S_("molar")=(0.76*g)/(557.00*g*mol^-1)xx1/(1*L)$

$=$ $=$ $1.37 \times 10^{- 3} \cdot m o l \cdot L^{- 1}$ $1.37xx10^-3*mol*L^-1$

And thus $K_{s p} = 4 \times {(1.37 \times 10^{- 3})}^{3}$ $K_(sp)=4xx(1.37xx10^-3)^3$ $=$ $=$ $? ?$ $??$

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(IO3)2Pb(IO_3)_2, is 0.76 g/L at 25*C. How do you calculate the Value of Ksp at this same temperature?