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Solubility of Al(OH)3 in presence of 0.1 M Ca(OH)2 is [ksp (Al(OH)3)=4*10-34?
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Solubility of Al(OH)3 in presence of 0.1 M Ca(OH)2 is [ksp (Al(OH)3)=4...
Solubility of Al(OH)3 in the presence of 0.1 M Ca(OH)2
Introduction:
Solubility refers to the maximum amount of a substance that can dissolve in a given solvent at a specific temperature and pressure. In this case, we are interested in determining the solubility of Al(OH)3 in the presence of Ca(OH)2.
Explanation:
To determine the solubility of Al(OH)3 in the presence of Ca(OH)2, we need to consider the common ion effect. The presence of the common ion (OH-) from Ca(OH)2 will affect the solubility of Al(OH)3.
Common ion effect:
The common ion effect states that the solubility of a salt is decreased when a common ion is present in the solution. This is because the common ion reduces the effective concentration of the solute in the solution.
Equilibrium equation:
The equilibrium equation for the dissolution of Al(OH)3 can be written as follows:
Al(OH)3 (s) ⇌ Al3+ (aq) + 3OH- (aq)
Ksp expression:
The solubility product constant (Ksp) expression for Al(OH)3 can be given as:
Ksp = [Al3+][OH-]^3
Effect of Ca(OH)2:
When Ca(OH)2 is added to the solution, it dissociates into Ca2+ and 2OH-. The OH- ion is the common ion with Al(OH)3.
Common ion effect on solubility:
The presence of the common ion OH- from Ca(OH)2 will shift the equilibrium of Al(OH)3 to the left, reducing the concentration of Al3+ and OH- ions in the solution. This ultimately decreases the solubility of Al(OH)3.
Calculating solubility:
To calculate the solubility of Al(OH)3, we need to consider the concentration of OH- ions from Ca(OH)2. Since Ca(OH)2 is a strong electrolyte and dissociates completely, the concentration of OH- ions will be 2 times the concentration of Ca(OH)2.
Using the Ksp expression:
We can write the Ksp expression for Al(OH)3 as:
Ksp = [Al3+][OH-]^3
Since the concentration of OH- ions is 2 times the concentration of Ca(OH)2, we can substitute the value in the Ksp expression:
Ksp = [Al3+](2[Ca(OH)2])^3
Conclusion:
In conclusion, the solubility of Al(OH)3 in the presence of 0.1 M Ca(OH)2 can be determined by considering the common ion effect. The presence of the common ion OH- from Ca(OH)2 will decrease the solubility of Al(OH)3. To calculate the solubility, we can use the Ksp expression and substitute the concentration of OH- ions from Ca(OH)2.
Introduction:
Solubility refers to the maximum amount of a substance that can dissolve in a given solvent at a specific temperature and pressure. In this case, we are interested in determining the solubility of Al(OH)3 in the presence of Ca(OH)2.
Explanation:
To determine the solubility of Al(OH)3 in the presence of Ca(OH)2, we need to consider the common ion effect. The presence of the common ion (OH-) from Ca(OH)2 will affect the solubility of Al(OH)3.
Common ion effect:
The common ion effect states that the solubility of a salt is decreased when a common ion is present in the solution. This is because the common ion reduces the effective concentration of the solute in the solution.
Equilibrium equation:
The equilibrium equation for the dissolution of Al(OH)3 can be written as follows:
Al(OH)3 (s) ⇌ Al3+ (aq) + 3OH- (aq)
Ksp expression:
The solubility product constant (Ksp) expression for Al(OH)3 can be given as:
Ksp = [Al3+][OH-]^3
Effect of Ca(OH)2:
When Ca(OH)2 is added to the solution, it dissociates into Ca2+ and 2OH-. The OH- ion is the common ion with Al(OH)3.
Common ion effect on solubility:
The presence of the common ion OH- from Ca(OH)2 will shift the equilibrium of Al(OH)3 to the left, reducing the concentration of Al3+ and OH- ions in the solution. This ultimately decreases the solubility of Al(OH)3.
Calculating solubility:
To calculate the solubility of Al(OH)3, we need to consider the concentration of OH- ions from Ca(OH)2. Since Ca(OH)2 is a strong electrolyte and dissociates completely, the concentration of OH- ions will be 2 times the concentration of Ca(OH)2.
Using the Ksp expression:
We can write the Ksp expression for Al(OH)3 as:
Ksp = [Al3+][OH-]^3
Since the concentration of OH- ions is 2 times the concentration of Ca(OH)2, we can substitute the value in the Ksp expression:
Ksp = [Al3+](2[Ca(OH)2])^3
Conclusion:
In conclusion, the solubility of Al(OH)3 in the presence of 0.1 M Ca(OH)2 can be determined by considering the common ion effect. The presence of the common ion OH- from Ca(OH)2 will decrease the solubility of Al(OH)3. To calculate the solubility, we can use the Ksp expression and substitute the concentration of OH- ions from Ca(OH)2.
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Solubility of Al(OH)3 in presence of 0.1 M Ca(OH)2 is [ksp (Al(OH)3)=4...
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Solubility of Al(OH)3 in presence of 0.1 M Ca(OH)2 is [ksp (Al(OH)3)=4*10-34?
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Solubility of Al(OH)3 in presence of 0.1 M Ca(OH)2 is [ksp (Al(OH)3)=4*10-34? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Solubility of Al(OH)3 in presence of 0.1 M Ca(OH)2 is [ksp (Al(OH)3)=4*10-34? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solubility of Al(OH)3 in presence of 0.1 M Ca(OH)2 is [ksp (Al(OH)3)=4*10-34?.
Solubility of Al(OH)3 in presence of 0.1 M Ca(OH)2 is [ksp (Al(OH)3)=4*10-34? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Solubility of Al(OH)3 in presence of 0.1 M Ca(OH)2 is [ksp (Al(OH)3)=4*10-34? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solubility of Al(OH)3 in presence of 0.1 M Ca(OH)2 is [ksp (Al(OH)3)=4*10-34?.
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