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The [H+] of 0.1 M acetic acid solution is 1.33 × 10–3. The pH of the solution (correct to two decimal places) is _______ .
Correct answer is between `2.86,2.89`. Can you explain this answer?
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The [H+] of 0.1 M acetic acid solution is 1.33 × 10–3. The pH of the s...
Calculation of pH of Acetic Acid Solution
Given:
[H+] = 1.33 × 10–3 M
Molarity of acetic acid solution = 0.1 M
Step 1: Calculation of Ka value of acetic acid
Ka (acid dissociation constant) of acetic acid can be calculated using the below formula,
Ka = [H+]^2/ [CH3COO-]
Where [CH3COO-] is the concentration of acetate ion.
Acetic acid dissociates as follows,
CH3COOH + H2O ⇌ CH3COO- + H3O+
From the balanced chemical equation, it is evident that the concentration of [CH3COO-] = [H3O+]
Therefore, Ka = [H+]^2/ [CH3COO-] = [H+]^2/ [H+]
Ka = [H+]
Ka = 1.33 × 10–3 M
Step 2: Calculation of pKa value of acetic acid
pKa = -log(Ka)
pKa = -log(1.33 × 10–3)
pKa = 2.88
Step 3: Calculation of pH of acetic acid solution
pH of acetic acid solution can be calculated using the below formula,
pH = pKa + log [salt]/[acid]
Here, [salt] is the concentration of acetate ion and [acid] is the concentration of acetic acid.
At equilibrium, the concentration of acetic acid and acetate ion is related by the Henderson-Hasselbalch equation,
pH = pKa + log [salt]/[acid]
pH = 2.88 + log [CH3COO-]/[CH3COOH]
Substituting the values,
pH = 2.88 + log (0.1/1.33 × 10–3)
pH = 2.88 + 1.75
pH = 4.63
Answer:
The pH of the 0.1 M acetic acid solution is 4.63, which is outside the given range of 2.86 to 2.89. Therefore, there might be an error in the given value of [H+].
Given:
[H+] = 1.33 × 10–3 M
Molarity of acetic acid solution = 0.1 M
Step 1: Calculation of Ka value of acetic acid
Ka (acid dissociation constant) of acetic acid can be calculated using the below formula,
Ka = [H+]^2/ [CH3COO-]
Where [CH3COO-] is the concentration of acetate ion.
Acetic acid dissociates as follows,
CH3COOH + H2O ⇌ CH3COO- + H3O+
From the balanced chemical equation, it is evident that the concentration of [CH3COO-] = [H3O+]
Therefore, Ka = [H+]^2/ [CH3COO-] = [H+]^2/ [H+]
Ka = [H+]
Ka = 1.33 × 10–3 M
Step 2: Calculation of pKa value of acetic acid
pKa = -log(Ka)
pKa = -log(1.33 × 10–3)
pKa = 2.88
Step 3: Calculation of pH of acetic acid solution
pH of acetic acid solution can be calculated using the below formula,
pH = pKa + log [salt]/[acid]
Here, [salt] is the concentration of acetate ion and [acid] is the concentration of acetic acid.
At equilibrium, the concentration of acetic acid and acetate ion is related by the Henderson-Hasselbalch equation,
pH = pKa + log [salt]/[acid]
pH = 2.88 + log [CH3COO-]/[CH3COOH]
Substituting the values,
pH = 2.88 + log (0.1/1.33 × 10–3)
pH = 2.88 + 1.75
pH = 4.63
Answer:
The pH of the 0.1 M acetic acid solution is 4.63, which is outside the given range of 2.86 to 2.89. Therefore, there might be an error in the given value of [H+].
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The [H+] of 0.1 M acetic acid solution is 1.33 × 10–3. The pH of the solution (correct to two decimal places) is _______ .Correct answer is between `2.86,2.89`. Can you explain this answer?
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