What is the molar solubility of PbCl_2 with a Ksp of 1.6 xx 10^(-5)?

Nov 1, 2015

$\text{0.0159 M}$

Explanation:

Lead(II) chloride, ${\text{PbCl}}_{2}$, is an insoluble ionic compound, which means that it does not dissociate completely in lead(II) cations and chloride anions when placed in aqueous solution.

Instead of dissociating completely, an equilibrium rection governed by the solubility product constant, ${K}_{\text{sp}}$, will be established between the solid lead(II) chloride and the dissolved ions.

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

Now, the molar solubility of the compound, $s$, represents the number of moles of lead(II) chloride that will dissolve in aqueous solution at a particular temperature.

Notice that every mole of lead(II) chloride will produce $1$ mole of lead(II) cations and $\textcolor{red}{2}$ moles of chloride anions. Use an ICE table to find the molar solubility of the solid

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

color(purple)("I")" " " " " "-" " " " " " " " " " "0" " " " " " " " " " "0
color(purple)("C")" " " "color(white)(x)-" " " " " " " " "(+s)" " " " " " " " "(+color(red)(2)s)
color(purple)("E")" " " "color(white)(x)-" " " " " " " " " " "s" " " " " " " " " " "color(red)(2)s

By definition, the solubility product constant will be equal to

K_"sp" = ["Pb"^(2+)] * ["Cl"^(-)]^color(red)(2)

${K}_{\text{sp}} = s \cdot {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}} = 4 {s}^{3}$

This means that the molar solubility of lead(II) chloride will be

$4 {s}^{3} = 1.6 \cdot {10}^{- 5} \implies s = \sqrt[3]{\frac{1.6}{4} \cdot {10}^{- 5}} = \textcolor{g r e e n}{\text{0.0159 M}}$