# Using the appropriate values of K_(sp) and K _f, what is the equilibrium constant for the following reaction: PbCl_2(s) + 3OH^(-)(aq) rightleftharpoons Pb (OH)_3^(-)(aq) + 2Cl^(-)(aq) ?

Aug 11, 2016

${K}_{\text{eq}} = 6.5 \cdot {10}^{9}$

#### Explanation:

The first thing to do here is look up the values of the solubility product constant, ${K}_{s p}$, for lead(II) chloride, ${\text{PbCl}}_{2}$, and the formation constant, ${K}_{f}$, for the trihydroxoplumbate(II) complex ion, "Pb"("OH")_3^(-)

${K}_{s p} = 1.70 \cdot {10}^{- 5}$

${K}_{f} = 3.8 \cdot {10}^{14}$

http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf

http://bilbo.chm.uri.edu/CHM112/tables/Kftable.htm

Now, the idea here is that lead(II) chloride is considered Insoluble in aqueous solution, which means that a dissociation equilibrium is established when you place this salt in water

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

As shown by the value of ${K}_{s p}$, this equilibrium lies to the left, meaning that you will get very small concentrations of lead(II) cations and chloride anions in solution.

Now, if you add a strong base, which can be symbolized by ${\text{OH}}^{-}$, to this solution, a complexation reaction will take place

"Pb"_ ((aq))^(2+) + 3"OH"_ ((aq))^(-) rightleftharpoons "Pb"("OH")_ (3(aq))^(-)

Since ${K}_{f}$ has such a large value, this equilibrium will lie to the right, meaning that the reaction will proceed in the forward direction

"Pb"_ ((aq))^(2+) + 3"OH"_ ((aq))^(-) -> "Pb"("OH")_ (3(aq))^(-)

The thing to notice here is that adding hydroxide anions, to the aqueous solution of lead(II) chloride will consume the lead(II) cations dissolved in solution.

As a result, the dissociation equilibrium for lead(II) chloride will shift to the right to compensate for the fact that the concentration of lead(II) cations is decreasing $\to$ think Le Chatelier's Principle here.

This means that you'll have

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

If you were to add these two reactions, you would get

{(color(white)(aaaaaaa)"PbCl"_ (2(s)) rightleftharpoons "Pb"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)), ("Pb"_ ((aq))^(2+) + 3"OH"_ ((aq))^(-) rightleftharpoons "Pb"("OH")_ (3(aq))^(-)) :}
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

${\text{PbCl"_ (2(s)) + color(red)(cancel(color(black)("Pb"_ ((aq))^(2+)))) + 3"OH"_ ((aq))^(-) rightleftharpoons "Pb"("OH")_ (3(aq))^(-) + color(red)(cancel(color(black)("Pb"_ ((aq))^(2+)))) + 2"Cl}}_{\left(a q\right)}^{-}$

This is equivalent to

${\text{PbCl"_ (2(s)) + 3"OH"_ ((aq))^(-) rightleftharpoons "Pb"("OH")_ (3(aq))^(-) + 2"Cl}}_{\left(a q\right)}^{-}$

Now, because you're added two equilibrium reactions to get your overall equilibrium, you must multiply the value of their equilibrium constants to get the equilibrium constant, let's say ${K}_{\text{eq}}$, for this overall equilibrium

${K}_{\text{eq}} = {K}_{s p} \cdot {K}_{f}$

Plug in your values to find

${K}_{\text{eq}} = 1.70 \cdot {10}^{- 5} \cdot 3.8 \cdot {10}^{14} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{6.5 \cdot {10}^{9}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Once again, the fact the ${K}_{\text{eq " ">> }} 1$ tells you that this equilibrium will lie mostly to the right, which essentially tells you that you can increase the solubility of lead(II) chloride by increasing the pH of the solution.