For a sparingly soluble salt {Aₚ}{Bq}, relationship of its solubility product ({L_S}) with its sol

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6-2.Equilibrium-II (Ionic Equilibrium)

hard

For a sparingly soluble salt ${A_p}{B_q},$ the relationship of its solubility product $({L_S})$ with its solubility $(S)$ is

A

${L_s} = {S^{p + q}}.{p^p}.{q^q}$

B

${L_s} = {S^{p + q}}.{p^q}.{q^p}$

C

${L_s} = {S^{pq}}.{p^p}.{q^q}$

D

${L_s} = {S^{pq}}.{(p.q)^{p + q}}$

(IIT-2001)

Solution

(a) ${A_p}{B_q}$ $ \rightleftharpoons $ $p{A^{1 + }} + q{B^{p – }}$

${L_s} = {[{A^{q + }}]^p}{[{B^{p – }}]^q} = {(p \times S)^p}{(q \times S)^q} = {S^{p + q}}.{p^p}.{q^q}$.

Standard 11

Chemistry

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