Question

The solubility product of pb(oh)2 in water is 1.2 × 10–15. determine the solubility of pb(oh)2 i

Answer 1

We have,             Ksp of Pb(OH)2       =         4s3

=             4×(6.7×10-6)3

=             1.203×10-15

Now,     pH of buffer       =             8 

Therefore,             pOH       =             6

Or,                        [OH- ]   =             10-6

Let solubility of Pb(OH)2  in buffer solution be ‘s’ mol litre-1

Therefore,   [Pb2+][OH- ]2 =  Ksp

Or,    [Pb2+](10-6)2    =   1.203×10-15

Therefore,     [Pb2+]  =   (1.203×10-15)/10-12

=   1.203×10-3 mol litre-1