Question

Question #b333c

Answer 1

The pH of the solution is 9

We know that
`pH+pOH=14`

`=>9+pOH=14`

`=>-log_(10)[OH^-]=5`

`=>[OH^-]=10^-5`

`CdCl_2` ionises in water as follows

`CdCl_2rightleftharpoonsCd^(2+)+2Cl^-`

So 0.01M `CdCl_2` solution will have 0.01M `Ca^(2+)` and 0.02M `Cl^-` concentration.

The ionic product of `Cd^(2+) and OH^-`

`=[Cd^(2+)]xx[OH^-]`

`=10^-2*10^-5=10^-7`

This ionic product is greater than sulubility product`" "K_(sp)=2.5xx10^-14`

So `Cd(OH)_2` will be precipitated