The acid dissociation constant, K<sub>a</sub>, for ethanoic acid is given by the expression
$$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$$
The value of K<sub>a</sub> for ethanoic acid is 1.74 × 10<sup>-5</sup> mol dm<sup>-3</sup> at 25 °C
A buffer solution with a pH of 3.87 was prepared using ethanoic acid and sodium ethanoate - AQA - A-Level Chemistry - Question 2 - 2017 - Paper 1
Question 2
The acid dissociation constant, K<sub>a</sub>, for ethanoic acid is given by the expression
$$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$$
The value of K<sub>a</sub... show full transcript
Worked Solution & Example Answer:The acid dissociation constant, K<sub>a</sub>, for ethanoic acid is given by the expression
$$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$$
The value of K<sub>a</sub> for ethanoic acid is 1.74 × 10<sup>-5</sup> mol dm<sup>-3</sup> at 25 °C
A buffer solution with a pH of 3.87 was prepared using ethanoic acid and sodium ethanoate - AQA - A-Level Chemistry - Question 2 - 2017 - Paper 1
Step 1
Calculate the concentration of the acid
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Answer
Start with the expression for the acid dissociation constant:
Ka=[CH3COOH][CH3COO−][H+]
Rearranging gives:
[CH3COOH]=Ka[CH3COO−][H+]
Use the provided values:
Ka=1.74×10−5mol dm−3
[CH3COO−]=0.136mol dm−3
Use the formula for [H+] derived from pH:
[H+]=10−pH=10−3.87=1.3498×10−4mol dm−3
Substitute into the rearranged equation:
[CH3COOH]=1.74×10−5(0.136)(1.3498×10−4)
Calculate the concentration:
[CH3COOH]=1.74×10−51.834×10−8≈1.06mol dm−3
Thus, the concentration of the ethanoic acid in the buffer solution is approximately 1.06 mol dm<sup>-3</sup>.
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