Question

A student performs the following experiment: 25.0 mL of 0.1 M acetic acid is titrated to the equivalence point with 25.0 mL of 0.1 M NaOH. Calculate the pH at the equivalence point. The equilibrium constant Ka for acetic acid is 1.8*10^-5. Compare this value with the experimental value.

Experimental value for acetic acid is pH = 8.8 and volume of base = 24 mL at equivalence point.

Answer 1

millimoles of acetic acid = 25 x 0.1 = 2.5

At equivalence point :

millimoles of acid = millimoles of base

here salt remains.

salt concentration = millimoles / total volume

                             = 2.5 / 25 + 25

                             = 0.05 M

Ka = 1.8 x 10^-5

pKa = -log Ka = -log (1.8 x 10^-5) = 4.74

pH = 7 + 1/2 (pKa + log C)

     = 7 + 1/2 (4.74 + log 0.05)

pH = 8.72

Theortical value 8.72 is less than the Experimental value.