Question

PbCl_2 is 2.4x10^-4. Its solutibility in g/L is (Pb= 207.2, Cl=35.5)? PbCl_2 (s) +H_2O (l) ⇌ Pb^+2 (ac) + 2Cl^- (ac)

List item

Answer 1

10,9 g/L

Explanation:

if s are the mol of `PbCl_2` that solubilize, you from the reaction have s
the mol of `Pb^(2+)` and 2s the mol of `Cl^-`
Kps is
`Kps = [Pb^(2+)] xx [Cl^-]^2 = s xx (2s)^2 = 4 s^3 = 2.4 xx 10^(-4)`
hence
`s = root(3)((2.4 xx 10^(-4))/4)=root(3)( 60 xx 10^-6) = 3,9 xx 10^-2 (mol)/L`
Since a mol of `PbCl_2` is 278.2 g, solubility in g/L is:
`S= 3,9 xx 10^-2 (mol)/L xx 278.2 g/(mol) = 10,9 g/L`