Question #37594

May 27, 2017

None of these options are correct.

Explanation:

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

And ${K}_{\text{sp}} = \left[P {b}^{2 +}\right] {\left[{I}^{-}\right]}^{2}$ (Note that $P b {I}_{2}$ does not appear in the equation in that as a solid it does not participate in the equilibrium, as it CANNOT express a concentration.)

And given the prior equation, we can represent the solubility of $P b {I}_{2} \left(s\right)$ as $S$, and given the stoichiometry.........

${K}_{\text{sp}} = \left(S\right) \times {\left(2 S\right)}^{2} = 4 {S}^{3}$

So $S = {\text{^3sqrt(K_"sp"/4)=}}^{3} \sqrt{\frac{8.0 \times {10}^{-} 9}{4}} = 1.26 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

See here for another example of a solubility equilibrium.

May 27, 2017

None of the options are correct

$n \left(P b {I}_{2}\right) = 1.3 \cdot {10}^{-} 3 \text{ mol}$

Explanation:

The equilibrium reaction is $P b {I}_{2} \text{ <--> } P {b}^{2 +} + 2 {I}^{-}$

Let $x = \left[P {b}^{2 +}\right]$

Then due to the stoichiometry

$\left[{I}^{-}\right] = 2 x$

For a saturated solution

${K}_{s p} = 8.0 \cdot {10}^{-} 9 = \left[P {b}^{2 +}\right] \cdot {\left[{I}^{-}\right]}^{2} = x {\left(2 x\right)}^{2}$

$8.0 \cdot {10}^{-} 9 = 4 {x}^{3}$

$x = {\left(\frac{8.0 \cdot {10}^{-} 9}{4}\right)}^{\frac{1}{3}}$

$x = \left[P {b}^{2 +}\right] = 1.3 \cdot {10}^{-} 3 \text{ M}$

$n \left(P {b}^{2 +}\right) = C \cdot V = 1 \cdot 1.3 \cdot {10}^{-} 3 \text{ mol}$

As one mole of $P b {I}_{2}$ releases one mole of $P {b}^{2 +}$, the number of moles of $P b {I}_{2}$ dissolved is the same as $n \left(P {b}^{2 +}\right)$

$n \left(P b {I}_{2}\right) = 1.3 \cdot {10}^{-} 3 \text{ mol}$

This is saying that you can dissolve up to $1.3 \cdot {10}^{-} 3 \text{ mol}$ of $P b {I}_{2}$ in 1 L of water, which is the solubility. Any more $P b {I}_{2}$ added will remain a solid.