Many Metal ions are precipitated from sulfides. As an example, consider treating a solution of copper(II) sulfate with sodium sulfide solution: CuSO(aq) + NaS(aq) CuS(s) + Na;SO4 What volume of 0.10S M Na:S solution would be required to precipitate all the copper(II) ion form 27.5 mL of 0.121 M CuSO, solution?

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### Precipitation of Metal Ions from Sulfides

Many metal ions can be precipitated from solutions using sulfides. Here is an example where a solution of copper(II) sulfate reacts with sodium sulfide to form a precipitate of copper(II) sulfide and sodium sulfate:

\[ \text{CuSO}_4 (\text{aq}) + \text{Na}_2\text{S} (\text{aq}) \rightarrow \text{CuS} (\text{s}) + \text{Na}_2\text{SO}_4 (\text{aq}) \]

**Question:**
What volume of a 0.105 M Na₂S solution would be required to precipitate all the copper(II) ions from 27.5 mL of a 0.121 M CuSO₄ solution?

### Detailed Explanation:

To determine the required volume of Na₂S solution, use the stoichiometry of the balanced chemical equation:
\[ \text{CuSO}_4 (\text{aq}) + \text{Na}_2\text{S} (\text{aq}) \rightarrow \text{CuS} (\text{s}) + \text{Na}_2\text{SO}_4 (\text{aq}) \]

1. **Calculate Moles of CuSO₄:**
   - Molarity (M) = moles/volume (L)
   - Moles of CuSO₄ = 0.121 M * 0.0275 L = 0.0033275 moles

2. **Determine the Stoichiometric Ratio:**
   - 1 mole of CuSO₄ reacts with 1 mole of Na₂S.

3. **Calculate Moles of Na₂S Required:**
   - Moles of Na₂S = moles of CuSO₄ = 0.0033275 moles

4. **Calculate Volume of Na₂S Solution:**
   - Volume (L) = moles/Molarity
   - Volume of Na₂S = 0.0033275 moles / 0.105 M = 0.03169 L
   - Convert to mL = 0.03169 L * 1000 mL/L = 31.69 mL

Therefore, **31.69 mL** of 0.105 M Na₂S solution is required to precipitate all the
Transcribed Image Text:### Precipitation of Metal Ions from Sulfides Many metal ions can be precipitated from solutions using sulfides. Here is an example where a solution of copper(II) sulfate reacts with sodium sulfide to form a precipitate of copper(II) sulfide and sodium sulfate: \[ \text{CuSO}_4 (\text{aq}) + \text{Na}_2\text{S} (\text{aq}) \rightarrow \text{CuS} (\text{s}) + \text{Na}_2\text{SO}_4 (\text{aq}) \] **Question:** What volume of a 0.105 M Na₂S solution would be required to precipitate all the copper(II) ions from 27.5 mL of a 0.121 M CuSO₄ solution? ### Detailed Explanation: To determine the required volume of Na₂S solution, use the stoichiometry of the balanced chemical equation: \[ \text{CuSO}_4 (\text{aq}) + \text{Na}_2\text{S} (\text{aq}) \rightarrow \text{CuS} (\text{s}) + \text{Na}_2\text{SO}_4 (\text{aq}) \] 1. **Calculate Moles of CuSO₄:** - Molarity (M) = moles/volume (L) - Moles of CuSO₄ = 0.121 M * 0.0275 L = 0.0033275 moles 2. **Determine the Stoichiometric Ratio:** - 1 mole of CuSO₄ reacts with 1 mole of Na₂S. 3. **Calculate Moles of Na₂S Required:** - Moles of Na₂S = moles of CuSO₄ = 0.0033275 moles 4. **Calculate Volume of Na₂S Solution:** - Volume (L) = moles/Molarity - Volume of Na₂S = 0.0033275 moles / 0.105 M = 0.03169 L - Convert to mL = 0.03169 L * 1000 mL/L = 31.69 mL Therefore, **31.69 mL** of 0.105 M Na₂S solution is required to precipitate all the
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