Question ed5f3

1 Answer
May 19, 2017

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

Explanation:

The solubility expression for lead chloride can be written using the $\text{solubility product}$, another equilibrium constant, that must be measured........

At $298 \cdot K$,

${K}_{\text{sp}} = \left[P {b}^{2 +}\right] {\left[C {l}^{-}\right]}^{2} = 1.7 \times {10}^{-} 4$.

In a saturated solution: ${K}_{\text{sp}} = \left(S\right) {\left(2 S\right)}^{2} = 4 {S}^{3}$

S=""^(3)sqrt((1.7xx10^-4)/4)=3.49xx10^-2*mol*L^-1#; which gives a gram solubility of under $10 \cdot g \cdot {L}^{-} 1$.