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?
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?
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2f4e1179-0845-4cce-a522-94afec860b33%2F37e29b5a-a849-40a9-a338-c1340d97cdcd%2Fgeghtys_processed.jpeg&w=3840&q=75)
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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