Chemistry

Chemistry

Problem

A saturated solution of the salt XY has a molar solubility of 1.5×10−4 M1.5\times10^{-4}\,\text{M} at 25 °C, and Ksp(XY)=2.3×10−8K_{sp}(XY)=2.3\times10^{-8}. Another salt, X₃Y₂, has the same molar solubility in pure water at the same temperature. Calculate KspK_{sp} for X₃Y₂.
Intuitive explanation

The key is recognizing that molar solubility (how many moles dissolve) differs from ion concentration depending on the salt's formula. For XY, dissolving 1.5×10−4 M1.5\times10^{-4}\,\text{M} produces equal concentrations of X⁺ and Y⁻, which you can use to verify the given KspK_{sp}. For X₃Y₂, the same molar solubility produces different ion concentrations—three times as many X⁺ ions and twice as many Y⁻ ions—so you must account for these stoichiometric coefficients when calculating KspK_{sp}.

Intuitive explanation

The key is recognizing that molar solubility (how many moles dissolve) differs from ion concentration depending on the salt's formula. For XY, dissolving 1.5×10−4 M1.5\times10^{-4}\,\text{M} produces equal concentrations of X⁺ and Y⁻, which you can use to verify the given KspK_{sp}. For X₃Y₂, the same molar solubility produces different ion concentrations—three times as many X⁺ ions and twice as many Y⁻ ions—so you must account for these stoichiometric coefficients when calculating KspK_{sp}.

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