Question

Even though most metal sulfides are sparingly soluble in water, their solubilities differ by several orders of magnitude. This difference is sometimes used to separate the metals by controlling the pH. Use the following data to find the pH at which you can separate \(0.10 M \mathrm{Cu}^{2+}\) and \(0.10 \mathrm{M} \mathrm{Ni}^{2+}\) Saturated \(\mathrm{H}_{2} \mathrm{~S}=0.10 \mathrm{M}\) \(K_{a t}\) of \(\mathrm{H}_{2} \mathrm{~S}=9 \times 10^{-8} \quad K_{a 2}\) of \(\mathrm{H}_{2} \mathrm{~S}=1 \times 10^{-17}\) \(K_{v}\) of NiS \(=1.1 \times 10^{-18} \quad K_{\mathrm{sp}}\) of \(\mathrm{CuS}=8 \times 10^{-34}\)

Short answer

The metals Cu(II) and Ni(II) can be separated at pH 4.97.

Step-by-step solution

- Write the solubility product expressions

For the metal sulfides CuS and NiS, write the solubility product expressions from their given Ksp values. For CuS: \[ K_{sp}(\text{CuS}) = [\text{Cu}^{2+}][\text{S}^{2-}] = 8 \times 10^{-34} \]For NiS:\[ K_{sp}(\text{NiS}) = [\text{Ni}^{2+}][\text{S}^{2-}] = 1.1 \times 10^{-18} \]

- Determine concentration of sulfide ions

Using the solubility product expressions, rearrange to solve for the sulfide ion concentration [S\(^{2-}\)]:For CuS: \[ [\text{S}^{2-}] = \frac{K_{sp}(\text{CuS})}{[\text{Cu}^{2+}]} = \frac{8 \times 10^{-34}}{0.10} = 8 \times 10^{-33} \]For NiS: \[ [\text{S}^{2-}] = \frac{K_{sp}(\text{NiS})}{[\text{Ni}^{2+}]} = \frac{1.1 \times 10^{-18}}{0.10} = 1.1 \times 10^{-17} \]

- Write the expression for sulfide ion concentration using Ka values

The sulfide ion concentration [S\(^{2-}\)] can be derived from the dissociation of H\( _2 \)S. H\( _2 \)S dissociates in two steps:\[ \text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^- \quad K_{a1} = 9 \times 10^{-8} \] and \[ \text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-} \quad K_{a2} = 1 \times 10^{-17} \].Using the initial concentration of H\( _2 \)S (0.10 M), the concentration of S\(^{2-}\) is given by:\[ [\text{S}^{2-}] = \frac{K_{a1} K_{a2} [\text{H}_2\text{S}]}{[\text{H}^+]^2} = \frac{(9 \times 10^{-8})(1 \times 10^{-17})(0.10)}{[\text{H}^+]^2} = 9 \times 10^{-26} [\text{H}^+]^{-2} \]

- Equate the sulfide ion concentrations

Set the expressions for [S\(^{2-}\)] equal to the values calculated in step 2 to solve for [H\( ^+ \)].For CuS: \[ 9 \times 10^{-26} [\text{H}^+]^{-2} = 8 \times 10^{-33} \]Solving for [H\( ^+ \)]: \[ [\text{H}^+]^2 = \frac{9 \times 10^{-26}}{8 \times 10^{-33}} = 1.125 \times 10^8 \]\[ [\text{H}^+] = \frac{1}{\text{pH}}^2 = 1.06 \times 10^4 \]\[ \text{pH} = -\text{log} (\text{H}+) = 4.97 \]

- Calculate pH for NiS

Set the other expression for [S\(^{2-}\)] equal to the value found in step 2 for NiS. \[ 9 \times 10^{-26} [\text{H}^+]^{-2} = 1.1 \times 10^{-17} \]Solving for [H\( ^+ \)] \[ [\text{H}^+]^2 = 9 \times 10^{-26}/1.1 \times 10^{-26}= 8.18*10^{-3} \]\[ \text{pH} = -\text{log} (\text{H}+) = 3.15 \]

- Determining Final pH

Compare the pH values calculated above. The separation of Cu\( ^{2+} \) and Ni\( ^{2+} \) occurs at the pH value where Cu\( ^{2+} \) will precipitate, but Ni\( ^{2+} \) won't:The separation pH is higher value of the pH values calculated ,\[ \text{pH} =4.97 \]

Key concepts

solubility product constant (Ksp)

The solubility product constant, or Ksp, is a measure of the solubility of a compound. It represents the equilibrium between a solid and its ions in a saturated solution. For the metal sulfides CuS and NiS, we can write their Ksp expressions as follows:
For CuS: \[ K_{sp}(\text{CuS}) = [\text{Cu}^{2+}][\text{S}^{2-}] = 8 \times 10^{-34} \] And for NiS: \[ K_{sp}(\text{NiS}) = [\text{Ni}^{2+}][\text{S}^{2-}] = 1.1 \times 10^{-18} \] These expressions help us determine the concentration of metal ions in a solution.

pH and precipitation

The pH of a solution affects the solubility of metal sulfides. By controlling the pH, we can selectively precipitate metal ions. For example, to separate Cu²⁺ and Ni²⁺, we find the pH at which Cu²⁺ will precipitate as CuS, but Ni²⁺ will not.
In our problem, the separation occurs at a pH where Cu²⁺ starts to form a solid, while Ni²⁺ remains dissolved. Using the equilibrium calculations from the dissociation constants, we can identify that the Cu²⁺ will precipitate at a higher pH than Ni²⁺.

equilibrium constants

Equilibrium constants describe the balance point of a chemical reaction in a closed system. For metal sulfides, the relevant equilibrium constants are the Ksp values. These values provide the ratio of ions to solid at equilibrium.
For example, in the dissociation of H₂S, there are two equilibrium constants: \[ \text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^- \ (K_{a1} = 9 \times 10^{-8}) \] and \[ \text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-} \ (K_{a2} = 1 \times 10^{-17}) \] These constants help us derive the sulfide ion concentration and relate it to the pH of the solution.

sulfide ion concentration

To solve problems involving metal sulfides, we need to understand sulfide ion concentration. This concentration is influenced by the dissociation of H₂S. By rearranging the Ksp expressions for CuS and NiS, we can solve for [S²⁻].
For CuS: \[ [\text{S}^{2-}] = \frac{K_{sp}(\text{CuS})}{[\text{Cu}^{2+}]} = \frac{8 \times 10^{-34}}{0.10} = 8 \times 10^{-33} \] For NiS: \[ [\text{S}^{2-}] = \frac{K_{sp}(\text{NiS})}{[\text{Ni}^{2+}]} = 1.1 \times 10^{-17} \] These calculations give us values that we equate to find the pH at which these metals precipitate.

dissociation of H2S

The dissociation of H2S in water is described by two dissociation constants, Kₐ1 and Kₐ2. This dissociation affects the sulfide ion concentration in solution. The relevant reactions are:
\[ \text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^- \ (K_{a1} = 9 \times 10^{-8}) \] and \[ \text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-} \ (K_{a2} = 1 \times 10^{-17}) \] Using these constants, for a given concentration of H₂S, we can calculate [S²⁻]: \[ [\text{S}^{2-}] = \frac{K_{a1} K_{a2} [\text{H}_2\text{S}]}{[\text{H}^+]^2} \] This relation helps us understand how the pH influences the sulfide ion concentration, determining whether a metal sulfide will precipitate.