Question

The solubility of pb(oh) 2 in water is 6.7 × 10 –6 m. Calculate the solubility of pb(oh)2 in a buffer solution of ph = 8.

Answer 1

pOH of the solution : 6(14-pH)

hence OH- concentration: 10^-6

Now in water its solubility is 6.7*10^-6

Pb(OH)2—→ Pb2+(x) + 2(OH)-(2x)

Ksp=x*(2x)^2

Ksp=4x^3=4*(6.7*10^-6)^3

Also for the given buffer Ksp=(Pb2+)*(OH-)^2

(Pb2+)(10^-6)^2=4*(6.7*10^-6)^3

Solving we get [Pb2+]=1.203×10-3

P.S > Units not mentioned in the question so am also leaving it as it is, it doesn’t mean solubility don’t have units, it’s mostly in mol/l or g/l.