What is the concentration of lead ions and sulfide ions in a saturated solution of lead sulfide (#PbS#) solution at 25°C?

1 Answer
Manish Bhardwaj
Apr 11, 2016

1.73 x #10^(-14)# Mol/liter

Explanation:

PbS (aq) #<=># # Pb^(2+)# (aq) + # S^(2-)# (aq)

#K_(sp)# = [ # Pb^(2+)# (aq) ] . [ # S^(2-)# (aq) ]

On dissociation the number of lead ions and sulfide ion in solution are in equal concentration we can reduce the above equation to ,
[ # Pb^(2+)# (aq) ] = [ # S^(2-)# (aq) ]

#K_(sp)# = [ # Pb^(2+)# (aq) ] . [ # S^(2-)# (aq) ]

#K_(sp)# = # [ # Pb^(2+)# (aq) ] ^2#

[ # Pb^(2+)# (aq) ] = #sqrt (#K(sp)# #

[ # Pb^(2+)# (aq) ] = #sqrt # 3 x #10^(-28)# 

                               =  1.73 x  #10^(-14)# Mol/ liter
Related questions
See all questions in Molarity
Impact of this question
7684 views around the world
You can reuse this answer
Creative Commons License