Question #ed5f3

1 Answer
May 19, 2017

Answer:

#PbCl_2(s)rightleftharpoonsPb^(2+) + 2Cl^(-)#

Explanation:

The solubility expression for lead chloride can be written using the #"solubility product"#, another equilibrium constant, that must be measured........

At #298*K#,

#K_"sp"=[Pb^(2+)][Cl^-]^2=1.7xx10^-4#.

In a saturated solution: #K_"sp"=(S)(2S)^2=4S^3#

#S=""^(3)sqrt((1.7xx10^-4)/4)=3.49xx10^-2*mol*L^-1#; which gives a gram solubility of under #10*g*L^-1#.