For the equilibrium Al(OH)3(s) Al³+ (aq) + 3 OH(aq) Ksp = 1.3 x 10-33 a. What is the minimum pH at which Al(OH)3(s) will precipitate from a solution that is 0.075 M in Al³+? b. A solution has [A13+] = 0.075 M and [CH3COOH] = 1.00 M. What is the maximum quantity of NaCH3COO, in grams, that can be added to 250.0 mL of this solution before precipitation of Al(OH)3(s) begins? (Ka of CH3COOH = 1.8 x 10-5)
For the equilibrium Al(OH)3(s) Al³+ (aq) + 3 OH(aq) Ksp = 1.3 x 10-33 a. What is the minimum pH at which Al(OH)3(s) will precipitate from a solution that is 0.075 M in Al³+? b. A solution has [A13+] = 0.075 M and [CH3COOH] = 1.00 M. What is the maximum quantity of NaCH3COO, in grams, that can be added to 250.0 mL of this solution before precipitation of Al(OH)3(s) begins? (Ka of CH3COOH = 1.8 x 10-5)
Introductory Chemistry: A Foundation
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Chapter17: Equilibrium
Section: Chapter Questions
Problem 70QAP: . The solubility product of iron(III) hydroxide is very small: Ksp=41038at 25 °C. A classical method...
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Ksp and Precipitation
![For the equilibrium Al(OH)3(s) Al³+ (aq) + 3 OH(aq) Ksp = 1.3 x
10-33
a. What is the minimum pH at which Al(OH)3(s) will precipitate from
a solution that is 0.075 M in Al³+?
b. A solution has [A1³+] = 0.075 M and [CH3COOH] = 1.00 M.
What is the maximum quantity of NaCH3COO, in grams, that can
be added to 250.0 mL of this solution before precipitation of
Al(OH)3(s) begins? (Ka of CH3COOH = 1.8 x 10-5)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F32869abc-9e7b-4303-a456-c0d169ed7c77%2Ff77d189d-1499-4c8c-b3d4-434a785d21c6%2Fsbuns_processed.jpeg&w=3840&q=75)
Transcribed Image Text:For the equilibrium Al(OH)3(s) Al³+ (aq) + 3 OH(aq) Ksp = 1.3 x
10-33
a. What is the minimum pH at which Al(OH)3(s) will precipitate from
a solution that is 0.075 M in Al³+?
b. A solution has [A1³+] = 0.075 M and [CH3COOH] = 1.00 M.
What is the maximum quantity of NaCH3COO, in grams, that can
be added to 250.0 mL of this solution before precipitation of
Al(OH)3(s) begins? (Ka of CH3COOH = 1.8 x 10-5)
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