- Home
- Standard 11
- Chemistry
6-2.Equilibrium-II (Ionic Equilibrium)
medium
The molar solubility of $PbI_2$ in $0.2\,M\,\, Pb(NO_3)_2$ solution in terms of solubility product of $PbI_2$ is
A
$\sqrt {({K_{sp}}/\,0.2)} $
B
$\sqrt {{K_{sp}}/\,0.4} $
C
$\sqrt[3]{{{K_{sp}}/\,0.8}}$
D
$\sqrt {{K_{sp}}/\,0.8} $
Solution
The solution contains:
$\quad\quad\quad\quad\quad PbI _2 \quad \Longrightarrow \quad Pb ^{+2}+2 I^{-}$
solubitity $S\quad\quad\quad\quad\quad\quad\quad S\quad\quad2S$
$Pb \left( NO _3\right)_2 \longrightarrow Pb ^{+2}+2 NO _3^{-}$
$0.2\quad\quad\quad\quad\quad\quad0.2 \quad\quad 2 \times 0.2$
$K_{s p}=\left[p b^{2+}\right]\left[I^{-}\right]^2$
$K_{s p}=(S+0.2)(2S)^2$
$S\,<\,<\,2$
$K_{s p}=0.2 \times 4 S^2$
$K_{s p}=0.8 S^2$
$S =\sqrt{\frac{K s p}{0.8}}$
$=(K_{s p}/ 0.8)^{1 / 2}$
Standard 11
Chemistry
