Question #b333c

Question

1899 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-28T17:12:31

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