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6-2.Equilibrium-II (Ionic Equilibrium)
medium
Calculate the solubility of $A_{2} X_{3}$ in pure water, assuming that neither kind of ion reacts with water. The solubility product of $A _{2} X _{3}, K_{ sp }=1.1 \times 10^{-23}$
Option A
Option B
Option C
Option D
Solution
$A _{2} X _{3} \rightarrow 2 A ^{3+}+3 X ^{2-}$
$K_{ sp }=\left[ A ^{3+}\right]^{2}\left[ X ^{2-}\right]^{3}=1.1 \times 10^{-23}$
If $S =$ solubility of $A _{2} X _{3},$ then
$\left[A^{3+}\right]=2 S ;\left[X^{2-}\right]=3 \,S$
therefore, $K_{ sp }=(2 S )^{2}(3 S )^{3}=108 \,S ^{5}$
$=1.1 \times 10^{-23}$
thus, $S^{5}=1 \times 10^{-25}$
$S=1.0 \times 10^{-5} \,mol / L$
Standard 11
Chemistry
