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)

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1 Answer
Andrea B.
Apr 27, 2018

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#

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