What is the solubility product of lead(II) chloride? I was thinking it is [Pb²⁺] [Cl^-]² , but my book says [Pb²⁺] [2Cl^-]².  I am pretty sure the book is wrong. Can anyone plz confirm?

 

 

I think you are right. But the concentration of [Cl^- ] Should be double than [Pb²⁺].

You are right and if the book says so, it is wrong.

PbCl₂(s) Pb²⁺ + 2 Cl^-
K_sp = [Pb²⁺][Cl^-]²

The concentration of Cl^- is squared because the balanced equation for the reaction shows a 2 as the coefficient in front of Cl^-.

[Pb²⁺] [Cl^-]²  is the correct answer as u stated.

PbCl₂(s) Pb²⁺ + 2 Cl^-

^{X                 x          2x}
K_sp = [Pb²⁺][Cl^-]²

K_sp = [x][2x]²

what is Solubility Product?

For sparingly soluble salts(solubility<0.01mol/L), it is experimental fact that the mass action product of the concentrations of the ions is a constant at constant temperature. This product Ks is called the solubility product. For a binary electrolyte

AB=A⁺ + B^-  then Ks_(AB)=[A⁺][B^-]

In general

For electrolyte ApBq, which ionizes into pA^q+ and qB^p- ions, then

ApBq = pA^q+ + qB^p-

and solubility product K_s(ApBq) = [A^q+]^p[B^p-]^q

Lets discuss your question: PbCl₂ ionises to give Pb⁺ and 2Cl^- ions

PbCl₂ = Pb⁺ + 2Cl^-

Ks = [Pb⁺][Cl^-]²

correction to lead ion it is Pb⁺² not Pb⁺

i am sure you are wrong and the book is right.
and santa is right too.i am sure you must have confused.
however, the book is right.

 You are right Raacal

Solubility product = [Pb²⁺] [Cl^-]² 

 But you should have to multiply the concentration of Cl^- by 2. 

 

 You are right Raacal

Solubility product = [Pb²⁺] [Cl^-]² 

 But you should have to multiply the concentration of Cl^- by 2. 

 

Thanks everyone. You guys rock!!!