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Calculation of K_a for ethanoic acid

Mass of ethanoic acid used = 0.017 g

pH measured = 3.4

CH₃COOH = 60

Original number of moles of ethanoic acid = 0.017/60

\ Original concentration of ethanoic acid = 0.017/60 × 10 = 0.00283 mol dm^-3

pH = - log₁₀[H⁺]

Rearranging this pH equation, log₁₀[H⁺] = - pH

To remove the log term we antilog both sides of the equation - this will appear on your calculator as INV LOG or 10^x:

[H⁺] = 10^-pH = 10^-3.4 = 0.000398 mol dm^-3

K_a = [H⁺]_eqm × [CH₃COO^-]_eqm ÷ [CH₃COOH]_eqm

We make the assumption that there is no H⁺ from the water so that [H⁺]_eqm = [CH₃COO^-]_eqm = 0.000398 M

The amount of ethanoic acid present at equilibrium is the amount we started with less any that has dissociated into hydrogen ions. As we know how many hydrogen ions have been formed from the acid, we can work out how many ethanoic acid molecules are left at equilibrium:

[CH₃COOH]_eqm = [CH₃COOH]_initial - [H⁺]_eqm

[CH₃COOH]_eqm = 0.00283 - 0.000398 = 0.00243 mol dm^-3

K_a = 0.000398 × 0.000398 ÷ 0.00243 = 6.5 × 10^-5 mol dm^-3

This compares only roughly with the accepted value of 1.7 × 10^-5 mol dm^-3. The result of this experiment is highly dependent on the accuracy of the pH meter. The measured answer may appear to be well out from the accepted value, but it is worth calculating the effect of a small difference in pH reading.