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6-2.Equilibrium-II (Ionic Equilibrium)
medium
Solubility of $Pb{I_2}$ is $0.005\,\, M$. Then, the solubility product of $Pb{I_2}$ is
A
$6.8 \times {10^{ - 6}}$
B
$6.8 \times {10^6}$
C
$2.2 \times {10^{ - 9}}$
D
None of these
Solution
(d) $Pb{I_2}\; \to \mathop {Pb}\limits_x + \mathop {{I_2}}\limits_{2x} $
${K_{sp}} = 4{x^3} = 4{(.005)^3}$$ = 4 \times .005 \times .005 = .4 \times {10^{ – 6}}$.
Standard 11
Chemistry
