For a sparingly soluble salt A_ p B_ q , the relationship of its...

For a sparingly soluble salt

A_{p} B_{q}

, the relationship of its solubility product

L s \rightarrow K_{s p}

with its solubility (S) is

Options

A $\quad \mathrm{Ls} \rightarrow \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{\mathrm{pq}}(\mathrm{pq})^{\mathrm{P}+\mathrm{q}}$
B $\mathrm{Ls}=\mathrm{S}^{\mathrm{p}+\mathrm{q}} \cdot \mathrmp^{\mathrm{p}} \mathrmq^{\mathrm{q}}$
C $\quad \mathrm{Ls} \rightarrow \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{\mathrm{p}+\mathrm{q}} \cdot \mathrmp^{\mathrm{q}} \mathrmq^{\mathrm{p}}$
D $\quad \mathrm{Ls} \rightarrow \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{\mathrm{pq}} \mathrmp^{\mathrm{P}} \mathrmq^{\mathrm{q}}$

B. $\mathrm{Ls}=\mathrm{S}^{\mathrm{p}+\mathrm{q}} \cdot \mathrmp^{\mathrm{p}} \mathrmq^{\mathrm{q}}$

\mathrm{ApBq} \rightleftharpoons \mathrm{pA}_{\mathrm{Sp}}^{+}+\mathrm{qB}_{\mathrm{Sq}}

Let the solubility be

\mathrm{S} \mathrm{mol} /

liter Thus,

\begin{aligned}\mathrm{K}_{\mathrm{sp}}&=\left[\mathrm{A}^{+}\right]^{\mathrm{p}}\left[\mathrm{B}^{-}\right]^{\mathrm{q}}\\ &=[\mathrm{Sp}]^{\mathrm{p}}[\mathrm{Sq}]^{\mathrm{q}}=\mathrmp^{\mathrm{p} \cdot} \mathrmq^{\mathrm{q}}(\mathrm{S})^{\mathrm{p}+\mathrm{q}} \end{aligned}