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The value of the dissociation constant for ethanoic acid is 1.8 x 10^-5 mol dm^-3 - Leaving Cert Chemistry - Question d - 2002

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The value of the dissociation constant for ethanoic acid is 1.8 x 10^-5 mol dm^-3. Calculate the pH of a 0.01 M solution of ethanoic acid.

Worked Solution & Example Answer:The value of the dissociation constant for ethanoic acid is 1.8 x 10^-5 mol dm^-3 - Leaving Cert Chemistry - Question d - 2002

Step 1

Calculate the concentration of hydrogen ions [H⁺]

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Answer

Using the dissociation constant expression:

Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}

For the weak acid dissociation, we can assume that [HA] is approximately the initial concentration of the acid. Therefore:

[H+]=Ka×M=1.8×10−5×0.01[H^+] = \sqrt{K_a \times M} = \sqrt{1.8 \times 10^{-5} \times 0.01}

Calculating this gives:

[H+]=1.8×10−7=1.34×10−4 M[H^+] = \sqrt{1.8 \times 10^{-7}} = 1.34 \times 10^{-4} \text{ M}

Step 2

Calculate the pH

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Answer

The pH is calculated using the formula:

pH=−log⁡[H+]pH = -\log[H^+]

Substituting in the value of [H⁺]:

pH=−log⁡(1.34×10−4)pH = -\log(1.34 \times 10^{-4})

This results in:

pH≈3.87pH \approx 3.87

For practical purposes, the pH can be rounded to:

pH≈3.8pH \approx 3.8

Step 3

Alternative calculation method

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Answer

Alternatively, we can express [H⁺] directly in terms of Ka and C:

[H+]2=Ka⋅0.01[H^+]^2 = K_a \cdot 0.01

Calculating:

[H+]=1.8×10−5×0.01[H^+] = \sqrt{1.8 \times 10^{-5} \times 0.01} gives the same result as before, confirming that:

pH≈3.8pH \approx 3.8.

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