Lead(II) chloride:

PbCl_{2}(s) 
Pb^{2+}(aq) 
+ 
2 Cl^{–}(aq) 



Initial 


0 

0 
Change 


+ x 

+ 2x 
Equilibrium 


x 

2x 
1.6×10^{–5} = [x][2x]^{2} = 4x^{3}
x = 1.6×10^{–2}
The solubility of lead(II) chloride is 1.6×10^{–2} M