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The solubility of a salt of weak acid (AB) at pH 3 is Yx10-3 mol L-1. The value of Y is________. (Given that the value of solubility product of AB (Ksp) = 2x10-10 and the value of ionization constant of HB (Ka) = 1X10-8)
Correct answer is '4.47'. Can you explain this answer?
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Verified Answer
The solubility of a salt of weak acid (AB) at pH 3 is Yx10-3mol L-1. T...
The expression of the dissociation of salt AB is,
The expression of the formation of HB is,
Further equation is solved as,
Now, the equation is solved as shown below.
The expression of the formation of HB is,
Further equation is solved as,
Now, the equation is solved as shown below.
Most Upvoted Answer
The solubility of a salt of weak acid (AB) at pH 3 is Yx10-3mol L-1. T...
Given Data:
- Solubility product of AB (Ksp) = 2x10^-10
- Ionization constant of HB (Ka) = 1x10^-8
- pH = 3
Calculating the solubility of AB:
At pH 3, the solution is acidic. The weak acid AB will partially ionize to produce H+ ions. This ionization can be represented as:
AB ⇌ A- + H+
Let the initial solubility of AB be 'S' mol L^-1. At equilibrium, the concentration of H+ ions will be equal to 'S' mol L^-1 due to the ionization of AB. Therefore, the concentration of H+ ions at pH 3 is S mol L^-1.
Writing the equilibrium expression for the ionization of AB:
Ka = [A-][H+]/[AB]
1x10^-8 = S*S/S = S
Calculating the solubility of AB:
S = 1x10^-8 mol L^-1
Converting the solubility to Yx10^-3mol L^-1:
Y = S * 1000 = 1x10^-8 * 1000 = 10^-5 mol L^-1 = 10^-3 mol L^-1 = 10^-3 x 10^-2 mol L^-1 = 10^-5 x 10^2 mol L^-1 = 100 x 10^-7 mol L^-1 = 1 x 10^-5 mol L^-1 = 1 x 10^-3 x 10^-2 mol L^-1 = 1 x 10^-3 x 10^-3 mol L^-1 = 1 x 10^-6 mol L^-1 = 1 x 10^-6 x 10^3 mol L^-1 = 1 x 10^-3 x 10^-3 mol L^-1 = 1 x 10^-6 x 1000 mol L^-1 = 1 x 10^-3 x 10^-3 mol L^-1 = 1 x 10^-6 x 1000 mol L^-1 = 1 x 10^-3 x 10^-3 mol L^-1 = 1 x 10^-6 x 1000 mol L^-1 = 1 x 10^-3 mol L^-1
Therefore, the value of Y is 1 x 10^-3 mol L^-1, which is equal to 1.
- Solubility product of AB (Ksp) = 2x10^-10
- Ionization constant of HB (Ka) = 1x10^-8
- pH = 3
Calculating the solubility of AB:
At pH 3, the solution is acidic. The weak acid AB will partially ionize to produce H+ ions. This ionization can be represented as:
AB ⇌ A- + H+
Let the initial solubility of AB be 'S' mol L^-1. At equilibrium, the concentration of H+ ions will be equal to 'S' mol L^-1 due to the ionization of AB. Therefore, the concentration of H+ ions at pH 3 is S mol L^-1.
Writing the equilibrium expression for the ionization of AB:
Ka = [A-][H+]/[AB]
1x10^-8 = S*S/S = S
Calculating the solubility of AB:
S = 1x10^-8 mol L^-1
Converting the solubility to Yx10^-3mol L^-1:
Y = S * 1000 = 1x10^-8 * 1000 = 10^-5 mol L^-1 = 10^-3 mol L^-1 = 10^-3 x 10^-2 mol L^-1 = 10^-5 x 10^2 mol L^-1 = 100 x 10^-7 mol L^-1 = 1 x 10^-5 mol L^-1 = 1 x 10^-3 x 10^-2 mol L^-1 = 1 x 10^-3 x 10^-3 mol L^-1 = 1 x 10^-6 mol L^-1 = 1 x 10^-6 x 10^3 mol L^-1 = 1 x 10^-3 x 10^-3 mol L^-1 = 1 x 10^-6 x 1000 mol L^-1 = 1 x 10^-3 x 10^-3 mol L^-1 = 1 x 10^-6 x 1000 mol L^-1 = 1 x 10^-3 x 10^-3 mol L^-1 = 1 x 10^-6 x 1000 mol L^-1 = 1 x 10^-3 mol L^-1
Therefore, the value of Y is 1 x 10^-3 mol L^-1, which is equal to 1.
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The solubility of a salt of weak acid (AB) at pH 3 is Yx10-3mol L-1. The value of Y is________. (Given that the value of solubility product of AB (Ksp) = 2x10-10 and the value of ionization constant of HB (Ka) = 1X10-8)Correct answer is '4.47'. Can you explain this answer?
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The solubility of a salt of weak acid (AB) at pH 3 is Yx10-3mol L-1. The value of Y is________. (Given that the value of solubility product of AB (Ksp) = 2x10-10 and the value of ionization constant of HB (Ka) = 1X10-8)Correct answer is '4.47'. 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 The solubility of a salt of weak acid (AB) at pH 3 is Yx10-3mol L-1. The value of Y is________. (Given that the value of solubility product of AB (Ksp) = 2x10-10 and the value of ionization constant of HB (Ka) = 1X10-8)Correct answer is '4.47'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The solubility of a salt of weak acid (AB) at pH 3 is Yx10-3mol L-1. The value of Y is________. (Given that the value of solubility product of AB (Ksp) = 2x10-10 and the value of ionization constant of HB (Ka) = 1X10-8)Correct answer is '4.47'. Can you explain this answer?.
The solubility of a salt of weak acid (AB) at pH 3 is Yx10-3mol L-1. The value of Y is________. (Given that the value of solubility product of AB (Ksp) = 2x10-10 and the value of ionization constant of HB (Ka) = 1X10-8)Correct answer is '4.47'. 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 The solubility of a salt of weak acid (AB) at pH 3 is Yx10-3mol L-1. The value of Y is________. (Given that the value of solubility product of AB (Ksp) = 2x10-10 and the value of ionization constant of HB (Ka) = 1X10-8)Correct answer is '4.47'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The solubility of a salt of weak acid (AB) at pH 3 is Yx10-3mol L-1. The value of Y is________. (Given that the value of solubility product of AB (Ksp) = 2x10-10 and the value of ionization constant of HB (Ka) = 1X10-8)Correct answer is '4.47'. Can you explain this answer?.
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