Problem 11 - Chemistry

Question

\(\mathrm{Ag}_{2} \mathrm{S}(s)\) has a larger molar solubility than CuS even though \(\mathrm{Ag}_{2} \mathrm{S}\) has the smaller \(K_{\mathrm{sp}}\) value. Explain how this is possible.

Step-by-Step Solution

Answer

The molar solubility of Ag2S is proportional to the cube root of its Ksp value (\(x\propto K_{\mathrm{sp}}^{\frac{1}{3}}\)), while the molar solubility of CuS is proportional to the square root of its Ksp value (\(y\propto K_{\mathrm{sp}}^{\frac{1}{2}}\)). The cube root function increases more rapidly than the square root function for small input values. Therefore, even though Ag2S has a smaller Ksp, its molar solubility can be larger if its Ksp value is sufficiently small.

Step 1: Write the balanced chemical equations for Ag2S and CuS dissolution

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

Step 2: Write the Ksp expressions for Ag2S and CuS

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

Step 3: Define molar solubility

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, \[ [\mathrm{Cu}^{2+}] = y \] \[ [\mathrm{S}^{2-}] = y \]

Step 4: Compare the Ksp expressions for Ag2S and CuS with respect to molar solubility

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: \[ K_{\mathrm{sp}} = (y)(y) = y^{2} \]

Step 5: Compare the relationship between Ksp and molar solubility for Ag2S and CuS

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.