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This question is about acidic solutions - AQA - A-Level Chemistry - Question 2 - 2017 - Paper 1
Question 2
This question is about acidic solutions. 1. The acid dissociation constant, K_a, for ethanoic acid is given by the expression $$K_a = \frac{[CH_3COO^-][H^+]}... show full transcript
Worked Solution & Example Answer:This question is about acidic solutions - AQA - A-Level Chemistry - Question 2 - 2017 - Paper 1
Step 1
Calculate the concentration of the ethanoic acid in the buffer solution
Answer
To find the concentration of ethanoic acid ([CH_3COOH]), we can rearrange the initial expression for the acid dissociation constant:
From the problem, we have:
- (K_a = 1.74 \times 10^{-5}) mol dm⁻³
- ([CH_3COO^-] = 0.136) mol dm⁻³
- pH = 3.87, thus ([H^+] = 10^{-3.87} \approx 1.3489 \times 10^{-4}) mol dm⁻³.
Now substitute these values into the equation:
Rearranging and solving for ([CH_3COOH]):
Thus, the concentration of the ethanoic acid is (1.06 \text{ mol dm}^{-3}).
Step 2
Calculate the pH of the buffer solution after the sodium hydroxide was added
Answer
In the second part, we have:
- Initial ([CH_3COOH] = 0.260) mol dm⁻³
- ([CH_3COO^-] = 0.121) mol dm⁻³
- Amount of NaOH added = 7.00 \times 10⁻³ mol to 500 cm³ of buffer solution.
First, calculate how the addition of NaOH affects the concentrations: NaOH will react with (CH_3COOH), creating more (CH_3COO^-):
After reaction:
- ([CH_3COOH]_{final} = 0.260 - 0.007 = 0.253) mol dm⁻³,
- ([CH_3COO^-]_{final} = 0.121 + 0.007 = 0.128) mol dm⁻³.
Now we apply the Henderson-Hasselbalch equation:
Since (pK_a = -\log(1.74 \times 10^{-5}) \approx 4.76):
Thus, the pH of the buffer solution after adding the sodium hydroxide is approximately (4.56) (to two decimal places).