Here following reactions takes place :
a) 2KI(s) + Pb(NO3) 2(s) → 2KNO3(aq) + PbI2(s)
When you add lead nitrate to potassium iodide, their particles combine and form a white solid of potassium nitrate & yellow solid of lead iodide and a.
b) Na2CO3 + Pb(NO3)2 → NaNO3 + PbCO3
This is a precipitation reaction: PbCO 3 is the formed precipitate.
Point 1 :
Where,
8.91x 10-3 = [KI]
1.37x 10-3 = [Na2CO3]
Means [KI] > [Na2CO3]
Point 2 :
According to Appendix J i.e. list of compound and their respective Ksp = Solubility product
Ksp at 25 °C for PbI2 = 1.4 × 10−8 & PbCO3 = 1.5 × 10−15
we can say that KPbI2 < KPbCO3
it means solubility of PbCO3 ppt is more than solubility of ppt PbI2
So, first ppt will form of PbI2
Q. A , Answer : Formula for first ppt formed = PbI2
Q. B , Answer : [Pb2+] = 0.00000157126 = 1.5710-6 M
Substance |
Ksp at 25 °C |
PbI2 |
1.4 × 10−8 |
PbCO3 |
1.5 × 10−15 |
In reaction,
a) 2KI + Pb(NO3) 2 → 2KNO3 + PbI2
Here Ksp = KspKI = [K+][I-]=8.91x 10-3 M
Let solubility of each[ K+] & [I-] at equilibrium is ‘S’
KKI = S x S = S2
[I-] = square root of 8.91x 10-3 M = 0.094M = 9.4x 10-2 M
According to Appendix J i.e. list of compound and their respective Ksp = Solubility product
Ksp at 25 °C for PbI2 = 1.4 × 10−8 M/L = KPbI2
KPbI2 = [Pb2+][I-]2
Put solubility of each [ I-] at equilibrium is = 9.4x 10-2 M
KPbI2 = [Pb2+] x[ 9.4x 10-2 ]2
1.4 × 10−8 =[Pb2+] x 8.91x 10-3
Concentration of [Pb2+] = 0.00000157126 = 1.5710-6 M
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