Kep=|A|^3| B|^2 for the salt where A and B are the cation and anion as the case may be stands tr

step 1 Dear students, in problem 138 of chapter 7, we have asked that KSP is equal to A to the power 3

Kep=|A|^3| B|^2 for the salt where A and B are the cation...

$K_{\mathrm{ep}}=|\mathrm{A}|^{3}|\mathrm{~B}|^{2}$ for the salt where $\mathrm{A}$ and $\mathrm{B}$ are the cation and anion as the case may be stands true for (1) $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ (2) $\mathrm{As}_{2} \mathrm{~S}_{3}$ (3) $\mathrm{Bi}_{2} \mathrm{~S}_{3}$ (4) All are correct

$K_{\mathrm{ep}}=|\mathrm{A}|^{3}|\mathrm{~B}|^{2}$ for the salt where $\mathrm{A}$ and $\mathrm{B}$ are the
cation and anion as the case may be stands true for
(1) $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$
(2) $\mathrm{As}_{2} \mathrm{~S}_{3}$
(3) $\mathrm{Bi}_{2} \mathrm{~S}_{3}$
(4) All are correct

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Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant is an equilibrium constant that applies to the dissolution of a sparingly soluble salt. It is expressed as the product of the ion concentrations raised to the power of their stoichiometric coefficients. This means that for a salt dissociating into its constituent ions, the equilibrium determined by the salt's dissociation is governed by the number of each type of ion produced, which in turn determines the exponents in the Ksp expression.

Stoichiometry of Salt Dissociation

When an ionic compound dissociates in solution, it produces ions in fixed ratios defined by its chemical formula. For example, if a salt dissociates to yield three cations and two anions, the equilibrium constant expression will reflect these stoichiometric coefficients (the ion concentrations are raised to the respective powers of 3 for the cation and 2 for the anion). This concept is crucial to correctly formulating and understanding equilibrium expressions for various salts.

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