Question

The solubility of lead (ii) chloride (pbcl2) is 1.6 ⋅ 10-2 m. what is the ksp of pbcl2? the solubility of lead (ii) chloride (pbcl2) is 1.6 10-2 m. what is the ksp of pbcl2? 3.1 ⋅ 10-7 5.0 ⋅ 10-4 1.6 ⋅ 10-5 1.6 ⋅ 10-2 4.1 ⋅ 10-6

Answer 1

According to the equilibrium equation for this reaction:

PbCl2(s)  ↔ Pb2+(aq)   +  2Cl-(aq)

So when Ksp is the solubility product constant for a solid substance when it

dissolved in the solution. and measure how a solute dissolves in the solution 

So, Ksp expression = [Pb2+] [Cl-]^2

and when the solubility is the maximum quantity of solute which can dissolve in a certain solute.
 
So, we assume the solubility = X 

∴[Pb2+] = X = 1.6 x 10^-2 M

[Cl-] = 2X = 2 * 1.6 x 10^-2 = 0.032 M

by substitution:

∴ Ksp = (1.6 x 10^-2) * (0.032)^2

           = 1.64 x 10^-5  

Answer 2

Answer:The Solubility product of $PbCl_2$ is $1.63\times 10^{-5}$.

Explanation;

Solubility of $PbCl_2,S=1.6\times 10^{-2} M$

                         $PbCl_2\rightleftharpoons Pb^{2+}+2Cl^-$

                                                 S        2S

The expression of $K_{sp}$ is given as:

$K_{sp}=S\times (2S)^2=4S^3$

$K_{sp}=4\times (1.6\times 10^{-2})^3=1.63\times 10^{-5}$

The Solubility product of $PbCl_2$ is $1.63\times 10^{-5}$.