JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Equilibrium

The solubility product of

PbI_{2}

8.0 \times 1 0^{- 9}

. The solubility of lead iodide in 0.1 molar solution of lead nitrate is

x \times 1 0^{- 6} mol / L

. The value of x is

____

(Rounded off to the nearest integer) [Given

2 = 1.41

]

Answer: 141

Solution

Lead nitrate is fully dissociated, so the solution already contains

0.1

mol/L of

P b^{2 +}

and the lead contributed by

Pb I_{2}

itself is negligible beside it. Taking the solubility of

Pb I_{2}

s

, the iodide concentration is

2 s

and

K_{s p} = [P b^{2 +}] [I^{-}]^{2} = (0.1) (2 s)^{2} = 0.4 s^{2}

0.4 s^{2} = 8.0 \times 1 0^{- 9} \Rightarrow s^{2} = 2.0 \times 1 0^{- 8}

s = 2 \times 1 0^{- 4} = 1.41 \times 1 0^{- 4} mol/L

Written in the required form

x \times 1 0^{- 6}

mol/L, this solubility gives

x = 141

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Integer answer