Question

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

Answer 1

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