# For the dissolution of lead(II) chloride in "0.2 M NaCl", find the molar solubility at SATP? How do I do this without a K_(sp)?

Jun 9, 2017

About $4.25 \times {10}^{- 4}$ $\text{M}$ via the small $x$ approximation.

Well, many if not all textbook appendices that list ${K}_{s p}$ list them for ${25}^{\circ} \text{C}$ and $\text{1 atm}$, but SATP has $\text{1 bar}$. We approximate that ${K}_{s p}$ is the same at $\text{1 bar}$ ($\text{0.987 atm}$) as at $\text{1 atm}$ since molar solubility of solids is negligibly pressure-dependent.

We have

${K}_{s p} = 1.7 \times {10}^{- 5}$

for

${\text{PbCl"_2(s) rightleftharpoons "Pb"^(2+)(aq) + 2"Cl}}^{-} \left(a q\right)$.

In a $\text{0.2 M NaCl}$ soln, the conc. of ${\text{Cl}}^{-}$ is also $\text{0.2 M}$, which induces a common-ion effect on the ${\text{Cl}}^{-}$ that dissociates from ${\text{PbCl}}_{2}$, thereby forcing a shift in the equilibrium leftwards from Le Chatelier's Principle.

The ICE table has:

${\text{PbCl"_2(s) rightleftharpoons "Pb"^(2+)(aq) + 2"Cl}}^{-} \left(a q\right)$

$\text{I"" "-" "" "" "" ""0 M"" "" "" ""0.2 M}$
$\text{C"" "-" "" "" "+s" "" "" } + 2 s$
$\text{E"" "-" "" "" "" "s" "" "" "" "(0.2 + 2s) "M}$

And the ${K}_{s p}$ expression therefore is:

$1.7 \times {10}^{- 5} = s {\left(0.2 + 2 s\right)}^{2}$

$= s \left(0.04 + 0.8 s + 4 {s}^{2}\right) = 4 {s}^{3} + 0.8 {s}^{2} + 0.04 s$

where $s$ is the molar solubility of ${\text{PbCl}}_{2}$ (or ${\text{Pb}}^{2 +}$ I suppose, by a $1 : 1$ mol ratio).

We then get:

$4 {s}^{3} + 0.8 {s}^{2} + 0.04 s - 1.7 \times {10}^{- 5} = 0$

which would be cumbersome to solve. However, since ${K}_{s p}$ is small (around ${10}^{- 5}$ or less), we can say that $2 s$ is small compared to $0.2$.

$\implies 0.2 + 2 s \approx 0.2$

$\implies \textcolor{b l u e}{s} \approx \frac{1.7 \times {10}^{- 5}}{0.2} ^ 2 = \textcolor{b l u e}{4.25 \times {10}^{- 4}}$ $\textcolor{b l u e}{\text{M}}$

(compared to the true answer of $4.21 \times {10}^{- 4}$ $\text{M}$.)

And $s$ is, again, the molar solubility of ${\text{PbCl}}_{2}$ when placed in this solution.