What is the molar solubility of #PbCl_2# with a Ksp of #1.6 xx 10^(-5)#?
1 Answer
Explanation:
Lead(II) chloride,
Instead of dissociating completely, an equilibrium rection governed by the solubility product constant,
#"PbCl"_text(2(s]) rightleftharpoons "Pb"_text((aq])^(2+) + color(red)(2)"Cl"_text((aq])^(-)#
Now, the molar solubility of the compound,
Notice that every mole of lead(II) chloride will produce
#" " "PbCl"_text(2(s]) " "rightleftharpoons" " "Pb"_text((aq])^(2+) " "+" " color(red)(2)"Cl"_text((aq])^(-)#
By definition, the solubility product constant will be equal to
#K_"sp" = ["Pb"^(2+)] * ["Cl"^(-)]^color(red)(2)#
#K_"sp" = s * (color(red)(2)s)^color(red)(2) = 4s^3#
This means that the molar solubility of lead(II) chloride will be
#4s^3 = 1.6 * 10^(-5) implies s = root(3)(1.6/4 * 10^(-5)) = color(green)("0.0159 M")#