Question

Suppose you add 0.2489 g of P b C l 2 ( s ) to 50.0 mL of water. In the resulting saturated solution, you find that the concentration of P b 2 + ( a q ) is 0.0159 M and the concentration of C l − ( a q ) is 0.0318 M. What is the value of the equilibrium constant, Ksp, for the dissolution of P b C l 2 ?

Answer 1

Answer:

$K_{sp}=1.61\times 10^{-5}$

Explanation:

$PbCl_2$ will form its respective ions in the solution as:

Consider the ICE take for $PbCl_2$ as:

                                     PbCl₂         ⇄             Pb²⁺ +       2Cl⁻

At t =equilibrium                                              x               2x            

The expression for dissociation constant of $PbCl_2$ is:

Solubility product = $[Pb^{2+}][Cl^-]^2$

Given that:

$[Pb^{2+}]=0.0159\ M$

$[Cl^-]=0.0318\ M$

So,  

Solubility product = $[Pb^{2+}][Cl^-]^2$  =  $0.0159\times {0.0318}^2$

$K_{sp}=1.61\times 10^{-5}$