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The solubility of a salt of weak acid (AB) at pH 3 is Y × 10–3 mol L−1. The value of Y is ____.
(Given that the value of solubility product of AB (Ksp) = 2 × 10–10 and the value of ionization constant of HB (Ka) = 1 × 10–8)
(Given that the value of solubility product of AB (Ksp) = 2 × 10–10 and the value of ionization constant of HB (Ka) = 1 × 10–8)
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 Y × 10&nda...
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The solubility of a salt of weak acid (AB) at pH 3 is Y × 10&nda...
The solubility of a salt of weak acid (AB) at pH 3 is Y. This means that at pH 3, the salt AB is partially dissociated into its ions, with a concentration of Y. The equilibrium for this dissociation reaction can be represented as follows:
AB ⇌ A- + BH+
The equilibrium constant (Ka) for this reaction can be calculated using the expression:
Ka = [A-][BH+]/[AB]
At pH 3, the concentration of H+ ions is relatively high, which will push the equilibrium towards the formation of more BH+. This means that the concentration of A- ions will be relatively low. Therefore, we can assume that the concentration of BH+ is equal to the concentration of AB, which is equal to Y. Substituting these values into the expression for Ka, we get:
Ka = Y^2/[AB]
Solving for [AB], we get:
[AB] = Y^2/Ka
Therefore, the solubility of the salt AB at pH 3 is given by Y^2/Ka.
AB ⇌ A- + BH+
The equilibrium constant (Ka) for this reaction can be calculated using the expression:
Ka = [A-][BH+]/[AB]
At pH 3, the concentration of H+ ions is relatively high, which will push the equilibrium towards the formation of more BH+. This means that the concentration of A- ions will be relatively low. Therefore, we can assume that the concentration of BH+ is equal to the concentration of AB, which is equal to Y. Substituting these values into the expression for Ka, we get:
Ka = Y^2/[AB]
Solving for [AB], we get:
[AB] = Y^2/Ka
Therefore, the solubility of the salt AB at pH 3 is given by Y^2/Ka.
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The solubility of a salt of weak acid (AB) at pH 3 is Y × 10–3 mol L−1. The value of Y is ____.(Given that the value of solubility product of AB (Ksp) = 2 × 10–10 and the value of ionization constant of HB (Ka) = 1 × 10–8)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 Y × 10–3 mol L−1. The value of Y is ____.(Given that the value of solubility product of AB (Ksp) = 2 × 10–10 and the value of ionization constant of HB (Ka) = 1 × 10–8)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 Y × 10–3 mol L−1. The value of Y is ____.(Given that the value of solubility product of AB (Ksp) = 2 × 10–10 and the value of ionization constant of HB (Ka) = 1 × 10–8)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 Y × 10–3 mol L−1. The value of Y is ____.(Given that the value of solubility product of AB (Ksp) = 2 × 10–10 and the value of ionization constant of HB (Ka) = 1 × 10–8)Correct answer is '4.47'. Can you explain this answer?.
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