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The pH of an aqueous solution of 1.0 M solution of a weak monoprotic acid which is 1% ionised is:
- a)1
- b)2
- c)3
- d)11
Correct answer is option 'B'. Can you explain this answer?
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Explanation:
The ionization constant of a weak monoprotic acid is given by:
Ka = [H+][A-]/[HA]
where [H+] is the concentration of hydrogen ions, [A-] is the concentration of the conjugate base of the acid, and [HA] is the concentration of the acid.
If the acid is 1% ionized, then [H+] = 0.01 M and [A-]/[HA] = 0.01.
The pH of the solution can be calculated using the equation:
pH = -log[H+]
Substituting [H+] = 0.01 M, we get:
pH = -log(0.01) = 2
Therefore, the pH of the solution is 2.
Answer: b) 2
The ionization constant of a weak monoprotic acid is given by:
Ka = [H+][A-]/[HA]
where [H+] is the concentration of hydrogen ions, [A-] is the concentration of the conjugate base of the acid, and [HA] is the concentration of the acid.
If the acid is 1% ionized, then [H+] = 0.01 M and [A-]/[HA] = 0.01.
The pH of the solution can be calculated using the equation:
pH = -log[H+]
Substituting [H+] = 0.01 M, we get:
pH = -log(0.01) = 2
Therefore, the pH of the solution is 2.
Answer: b) 2
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The pH of an aqueous solution of 1.0 M solution of a weak monoprotic acid which is 1% ionised is:a)1b)2c)3d)11Correct answer is option 'B'. Can you explain this answer?
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