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A 100ml solution of 0.50M acetic acid (Ka=1.78x10^-5) is titrated to equivalence point with 100ml of...

A 100ml solution of 0.50M acetic acid (Ka=1.78x10^-5) is titrated to equivalence point with 100ml of 0.50M NaOH what is the Ph of the resulting solution

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Answer #1

This is a reaction between weak acid and a strong base.

So corresponding conjugate base would be strong and conjugate acid would be weak. Thus, it would form a buffer because strong conjugate base would react with weak conjugate acid leading to the the solution to be slightly higher in pH

pKa = -Log(Ka)

Ka value is 1.78 x 10^-5

Therefore pKa = 4.75

Using hendersson haselbalch equation, we can find the pH of the solution

concentration of CH3COONa and H2O would be equal to the concentration of NaOH and CH3COOH

pH = 4.75 + log(0.5/0.5)

pH = 4.75 + 0

Hence, pH = 4.75

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