Chemistry 112

Lead(II) chloride:

 

PbCl2(s)

Pb2+(aq)

+

2 Cl(aq)

 

Initial

 

 

0

 

0

Change

 

 

+ x

 

+ 2x

Equilibrium

 

 

x

 

2x


1.6×10–5 = [x][2x]2 = 4x3

x = 1.6×10–2

The solubility of lead(II) chloride is 1.6×10–2 M



Return