Calculate the saturation indices of the solutions describe below and classify each as undersaturated, supersaturated, or in equilibrium. Saturation state with respect to: Anhydrite (CaSO4) [Ca²+] = 0.00002 mM [SO42-] = 25 mM "Amakinite" (Fe(OH)2) [Fe²+] = 0.1 mM; [OH-] = 0.0001 mM Siderite (FECO;) [Fe2*] = 3 mM; [CO;²-] = 10 mM
Calculate the saturation indices of the solutions describe below and classify each as undersaturated, supersaturated, or in equilibrium. Saturation state with respect to: Anhydrite (CaSO4) [Ca²+] = 0.00002 mM [SO42-] = 25 mM "Amakinite" (Fe(OH)2) [Fe²+] = 0.1 mM; [OH-] = 0.0001 mM Siderite (FECO;) [Fe2*] = 3 mM; [CO;²-] = 10 mM
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Calculate the saturation indices of the solutions describe below and classify each as undersaturated, supersaturated, or in equilibrium.
Saturation state with respect to:
Anhydrite (CaSO4) [Ca2+] = 0.00002 mM [SO42-] = 25 mM “Amakinite” (Fe(OH)2) [Fe2+] = 0.1 mM; [OH-] = 0.0001 mM Siderite (FeCO3) [Fe2+] = 3 mM; [CO32-] = 10 mM
![**Title: Understanding Saturation Indices in Chemical Solutions**
**Introduction:**
Calculating the saturation indices of solutions is a vital process in determining whether a solution is undersaturated, supersaturated, or at equilibrium. Let’s explore how to calculate these indices with a few examples.
**Saturation State with Respect to Various Compounds:**
1. **Anhydrite (CaSO₄):**
- \([ \text{Ca}^{2+} ] = 0.00002 \, \text{mM}\)
- \([ \text{SO₄}^{2-} ] = 25 \, \text{mM}\)
2. **"Amakinite" (Fe(OH)₂):**
- \([ \text{Fe}^{2+} ] = 0.1 \, \text{mM}\)
- \([ \text{OH}^- ] = 0.0001 \, \text{mM}\)
3. **Siderite (FeCO₃):**
- \([ \text{Fe}^{2+} ] = 3 \, \text{mM}\)
- \([ \text{CO₃}^{2-} ] = 10 \, \text{mM}\)
**Explanation:**
For each compound, the concentration of ions in the solution is specified. By calculating the ion product and comparing it to the solubility product (Ksp) of each compound, you can determine if the solution is:
- **Undersaturated:** Ion product < Ksp
- **Supersaturated:** Ion product > Ksp
- **In Equilibrium:** Ion product = Ksp
These calculations help in predicting whether a precipitate will form or dissolve in a given solution. Understanding such processes is crucial for applications in geochemistry, environmental science, and industrial chemistry.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ed8e3bd-e447-4950-851f-6c9d8eeaf9bd%2F6090efd1-88e1-46dd-92e6-22647ae8c061%2Ferkxp4f_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Understanding Saturation Indices in Chemical Solutions**
**Introduction:**
Calculating the saturation indices of solutions is a vital process in determining whether a solution is undersaturated, supersaturated, or at equilibrium. Let’s explore how to calculate these indices with a few examples.
**Saturation State with Respect to Various Compounds:**
1. **Anhydrite (CaSO₄):**
- \([ \text{Ca}^{2+} ] = 0.00002 \, \text{mM}\)
- \([ \text{SO₄}^{2-} ] = 25 \, \text{mM}\)
2. **"Amakinite" (Fe(OH)₂):**
- \([ \text{Fe}^{2+} ] = 0.1 \, \text{mM}\)
- \([ \text{OH}^- ] = 0.0001 \, \text{mM}\)
3. **Siderite (FeCO₃):**
- \([ \text{Fe}^{2+} ] = 3 \, \text{mM}\)
- \([ \text{CO₃}^{2-} ] = 10 \, \text{mM}\)
**Explanation:**
For each compound, the concentration of ions in the solution is specified. By calculating the ion product and comparing it to the solubility product (Ksp) of each compound, you can determine if the solution is:
- **Undersaturated:** Ion product < Ksp
- **Supersaturated:** Ion product > Ksp
- **In Equilibrium:** Ion product = Ksp
These calculations help in predicting whether a precipitate will form or dissolve in a given solution. Understanding such processes is crucial for applications in geochemistry, environmental science, and industrial chemistry.
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