Question

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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