91Ó°ÊÓ

Write the balanced chemical equations for the dissolution of Ag2S and CuS in water: \[ \mathrm{Ag}_{2}\mathrm{S}(s) \rightleftharpoons 2\mathrm{Ag}^{+}(aq) + \mathrm{S}^{2-}(aq) \] \[ \mathrm{CuS}(s) \rightleftharpoons \mathrm{Cu}^{2+}(aq) + \mathrm{S}^{2-}(aq) \]

Write the solubility product constant (Ksp) expressions for Ag2S and CuS: For Ag2S: \[ K_{\mathrm{sp}} = [\mathrm{Ag}^{+}]^{2} [\mathrm{S}^{2-}] \] For CuS: \[ K_{\mathrm{sp}} = [\mathrm{Cu}^{2+}] [\mathrm{S}^{2-}] \]

Define molar solubility as the number of moles of solute that can dissolve per liter of solution to form a saturated solution. Let the molar solubility of Ag2S be x and the molar solubility of CuS be y. For Ag2S, the dissolution of 1 mole of Ag2S will produce 2 moles of Ag+ and 1 mole of S2-. Therefore, \[ [\mathrm{Ag}^{+}] = 2x \] \[ [\mathrm{S}^{2-}] = x \] For CuS, the dissolution of 1 mole of CuS will produce 1 mole of Cu2+ and 1 mole of S2-. Therefore, \[ [\mathrm{Cu}^{2+}] = y \] \[ [\mathrm{S}^{2-}] = y \]

Substitute the expressions of molar solubility into the Ksp expressions for Ag2S and CuS. For Ag2S: \[ K_{\mathrm{sp}} = (2x)^{2} (x) = 4x^{3} \] For CuS: \[ K_{\mathrm{sp}} = (y)(y) = y^{2} \]

From step 4, it's apparent that the relationship between Ksp and molar solubility is different for Ag2S and CuS. The molar solubility of Ag2S is proportional to the cube root of Ksp (\(x\propto K_{\mathrm{sp}}^{\frac{1}{3}}\)), while the molar solubility of CuS is proportional to the square root of Ksp (\(y\propto K_{\mathrm{sp}}^{\frac{1}{2}}\)). Even though Ag2S has a smaller Ksp value, the cube root function for Ag2S (Ksp)^(1/3) might give a larger molar solubility value (x) compared to the square root function for CuS (Ksp)^(1/2) if the Ksp of Ag2S is sufficiently small. This is because the cube root function will increase more rapidly than the square root function when dealing with small input values. Consequently, it is possible for Ag2S to have a larger molar solubility despite having a smaller Ksp value.