A solution which is 0.1 M in Na I and also 0.1 M in Na 2 SO 4 is treated with solid Pb(NO 3 ) 2 . Which compound, Pb I 2 or PbSO 4 , will precipitate first ? What is the concentration of anion of the least soluble compound when the more soluble one starts precipitating ? K sp (Pb I 2 ) = 9 × 10 –9 , K sp (PbSO 4 ) = 1.8 × 10 –8 . Assume no hydrolysis of Pb 2+ ion.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(PbSO 4 , 0.02 M)
Sol. [NaI] = 0.1 M
NaI 
Na + + I –
0.1 M 0.1 M 0.1 M
Na 2 SO 4 
2Na + + SO 4 –2
0.1 M 0.2 M 0.1 M
For precipitation of PbI 2 ,
K sp (PbI 2 ) = [Pb +2 ] [I – ] 2 = 9 × 10 –9 
[Pb +2 ] × (0.1) 2 = 9 × 10 –9
[Pb +2 ] req = 9 × 10 –7 M
For precipitation of PbSO 4 ,
K sp (PbSO 4 ) = [Pb +2 ] [SO 4 –2 ] = 1.8 × 10 –8 
[Pb +2 ] × 0.1 = 1.8 × 10 –8
[Pb +2 ] req = 1.8 × 10 –7 M
For precipitation of PbSO 4 , required conc. of Pb +2 is less. So, PbSO 4 precipitates first.
When PbI 2 (more soluble compound) starts precipitating, [Pb 2+ ] = 9 × 10 –7 M.
Then, conc. of anion of less soluble salt can be obtained as follows :
K sp (PbSO 4 ) = [Pb +2 ] [SO 4 –2 ] = 1.8 × 10 –8 ⇒ (9 × 10 –7 ) [SO 4 –2 ] = 1.8 × 10 –8
[SO 4 –2 ] = 0.02 M
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