Question

Follow up questions 01. Find the accepted values for the pKa and Ka of acetic acid. How well do the accepted values compare with your calculated values? Explain.

Answer 1

Hi there, here we have to consider Henderson-Hasselbalch equation for buffers for weak acid (CH3COOH) and its corresponding conjugate base (CH3COO-)

HH equation pH = pKa + log[conjugate base / weak acid] = pKa + log[A-/HA]

we know that at the half-equivalence point, the concentrations of the weak acid (CH3COOH) and its conjugate base (CH3COO-) are the same. Therefore log ([A-]/[HA]) = log 1 = 0, and pH = pKa.

from table we find pH at the half-equivalence point = 5.10 & 5.18 for trial 1 & 2 respectively, just average them

(5.10 + 5.18) / 2 = 5.14 --------------------------> this is pKa of acetic acid

pKa = 5.14 (calculated value)

Ka = 10^-pKa = 10^-5.14 = 0.00000724435 = 7.24 x 10^-6

Ka = 7.24 x 10^-6 (calculated value)

The accepted (literature value) value of Ka for acetic acid is 1.8×10^–5 , so there is an error in experiment.

percent error = (|calculated - accepted| / accepted) x 100% = [(7.24 x 10^-6 - 1.8×10^–5) / (1.8×10^–5)] x 100%

percent error = 59.7 %

The result of this experiment showed there is more than 50% error for calculated vs accepted values of Ka, hence its better to repeat whole experiment or try to find where we did mistake like calibrations, readings, calculations etc.

Hope this helped you!

Thank You So Much!

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