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A 50.00 ml aliquot of 0.1000 M benzoic acid (Ka= 6.28x10^-5) was titrated with 0.2500 M...

A 50.00 ml aliquot of 0.1000 M benzoic acid (Ka= 6.28x10^-5) was titrated with 0.2500 M KOH. Calculate the pH of the solution after the addition of 5.00 ml of base.

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Answer #1

Millimoles of Benzoic acid added = Molarity x Volume (in mL)

=> Millimoles of Benzoic acid added = 0.1 x 50 = 5

Millimoles of NaOH added = 0.25 x 5 = 1.25

The reaction is,

C6H5COOH + NaOH ----> C6H5COONa + H2O

After complete reaction,

Millimoles of CH3COONa (Salt) formed = 1.25

Millimoles of Benzoic Acid remaining = 5 - 1.25 = 3.75

This solution will now act as an acidic buffer,

pH of an acidic buffer = pKa + log (Salt / Acid)

pKa = - log Ka = - log (6.28 x 10^-5) = 4.202

=> pH = 4.202 + log (1.25 / 3.75) = 3.725

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