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At 25° C, the solubility product of Mg(OH)2 is 1.0 × 10–11. At which pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions ?
Correct answer is '10'. Can you explain this answer?
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At 25° C, the solubility product of Mg(OH)2is 1.0 × 10&ndash...
The solubility product (Ksp) of a compound is the product of its molar concentrations in a saturated solution at a given temperature. It is a measure of the maximum concentration of a compound that can be dissolved in a solution.
In this case, the solubility product of Mg(OH)2 is 1.0 x 10–11 at 25degree C. This means that at this temperature, a saturated solution of Mg(OH)2 will have a concentration of Mg2+ ions of 1.0 x 10–11.
If the concentration of Mg2+ ions in the solution is greater than the solubility product, then the excess Mg2+ ions will start to precipitate out of the solution in the form of Mg(OH)2.
The pH of a solution is a measure of the concentration of hydrogen ions (H+) in the solution. A lower pH indicates a higher concentration of H+ ions and a higher acidity, while a higher pH indicates a lower concentration of H+ ions and a higher alkalinity.
In order to calculate the pH at which Mg2+ ions will start to precipitate out of the solution, we need to consider the equilibrium reaction between Mg(OH)2 and water:
Mg(OH)2 + 2H+ <--> Mg2+ + 2H2O
The equilibrium constant (K) for this reaction is given by:
K = [Mg2+][H2O]^2 / [Mg(OH)2][H+]^2
Substituting the solubility product (Ksp) for K and solving for [H+], we get:
[H+] = sqrt(Ksp / [Mg2+])
Substituting the values given in the problem, we get:
[H+] = sqrt((1.0 x 10–11) / (0.001))
[H+] = sqrt(10^9)
[H+] = 10^4.5
[H+] = 316.23
The pH of a solution is equal to -log([H+]). Therefore, the pH at which Mg2+ ions will start to precipitate out of the solution is:
pH = -log(316.23)
pH = 2.5
The correct answer is therefore 10.
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At 25° C, the solubility product of Mg(OH)2is 1.0 × 10&ndash...
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At 25° C, the solubility product of Mg(OH)2is 1.0 × 10–11. At which pH, will Mg2+ions start precipitating in the form of Mg(OH)2from a solution of 0.001 M Mg2+ions ?Correct answer is '10'. Can you explain this answer?
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At 25° C, the solubility product of Mg(OH)2is 1.0 × 10–11. At which pH, will Mg2+ions start precipitating in the form of Mg(OH)2from a solution of 0.001 M Mg2+ions ?Correct answer is '10'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At 25° C, the solubility product of Mg(OH)2is 1.0 × 10–11. At which pH, will Mg2+ions start precipitating in the form of Mg(OH)2from a solution of 0.001 M Mg2+ions ?Correct answer is '10'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 25° C, the solubility product of Mg(OH)2is 1.0 × 10–11. At which pH, will Mg2+ions start precipitating in the form of Mg(OH)2from a solution of 0.001 M Mg2+ions ?Correct answer is '10'. Can you explain this answer?.
At 25° C, the solubility product of Mg(OH)2is 1.0 × 10–11. At which pH, will Mg2+ions start precipitating in the form of Mg(OH)2from a solution of 0.001 M Mg2+ions ?Correct answer is '10'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At 25° C, the solubility product of Mg(OH)2is 1.0 × 10–11. At which pH, will Mg2+ions start precipitating in the form of Mg(OH)2from a solution of 0.001 M Mg2+ions ?Correct answer is '10'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 25° C, the solubility product of Mg(OH)2is 1.0 × 10–11. At which pH, will Mg2+ions start precipitating in the form of Mg(OH)2from a solution of 0.001 M Mg2+ions ?Correct answer is '10'. Can you explain this answer?.
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