Question #b333c

1 Answer
Aug 28, 2016

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