Question

25.000 mL of 0.1114 M propanoic acid solution is titrated with 0.1308 M NaOH. Calculate the pH of titrant mixture at the following volumes of NaOH added

a) 12.72 mL

b) 21.29 mL (Equivalence point)

Answer 1

HA + NaOH -------> NaA + H2O

25 x 0.114=2.85 0 0 0 Initial mmoles

2.85 12.72x0.1308 0 0 after adding NaoH

1.19 0 1.663 1.66 after reaction

The solution behaves as a buffer whose pH is given by

pH = pKa + log [(conjugate base)/(acid)]

Ka of propanoic acid is 1.3x 10^-5

Thus pH = -log (1.3x10^-5) + log 1.66/1.19 =5.0306

At equivalence point

[conjugate base ] = 2.85/46.29 M

The salt being a salt of strong base and weak acid , will be basic in nature .

pH of such a salt is given by pH = 1/2 pkW + 1/2pKa + 1/2logC

= 7.0 + 1/2(4.886) + 1/2 log {2.85/46.29}

= 8.840