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) ?
1 Answer
Explanation:
The first thing to do here is look up the values of the solubility product constant,
K_(sp) = 1.70 * 10^(-5)
K_f = 3.8 * 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
"PbCl"_ (2(s)) rightleftharpoons "Pb"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)
As shown by the value of
Now, if you add a strong base, which can be symbolized by
"Pb"_ ((aq))^(2+) + 3"OH"_ ((aq))^(-) rightleftharpoons "Pb"("OH")_ (3(aq))^(-)
Since
"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
This means that you'll have
"PbCl"_ (2(s)) -> "Pb"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)
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))^(-)) :}
color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)
"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"_ ((aq))^(-)
This is equivalent to
"PbCl"_ (2(s)) + 3"OH"_ ((aq))^(-) rightleftharpoons "Pb"("OH")_ (3(aq))^(-) + 2"Cl"_ ((aq))^(-)
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_"eq" = K_(sp) * K_f
Plug in your values to find
K_"eq" = 1.70 * 10^(-5) * 3.8 * 10^(14) = color(green)(|bar(ul(color(white)(a/a)color(black)(6.5 * 10^9)color(white)(a/a)|)))
Once again, the fact the