Question

1) If 25.2 mL of 0.109 M acid with a pKa of 5.55 is titrated with 0.102 M NaOH solution, what is the pH of the titration mixture after 13.7 mL of base solution is added?

2) If 28.8 mL of 0.108 M acid with a pKa 4.15 is titrated with 0.108 M NaOH solution, what is the pH of the acid solution before any base solution is added?

3)If 27.9 mL of 0.107 M acid with a pKa of 5.48 is titrated with 0.106 M NaOH solution, what is the pH of the titration mixture at halfway to the equivalence point?

Answer 1

Q1)

pKa of acid = 5.55

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

25.2x0.109=2.7468 13.7x0.102=1.3974 0 0 initial mmoles

1.3494 0 1.3974 - after reaction

So the solution forms a buffer of weak acid and its conjugate base.The pH of buffer is given by Hendersen equation as

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

= 5.55 + log 1.3974/1.3494

= 5.565

Q2) molarity of acid = 0.108M

pKa of acid = 4.15

pH of the acid before any base is added is given by

pH = 1/2[pKa -logC]

= 1/2[ 4.15 -log 0.108]

= 5.1165

Q3) At the equivalence

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

27.9x 0.107= 2.985 0 0 - initial mmoles

1.492 0 1.492 - at half equivalence

thus the solution is a buffer

pH = Pka + log[conjugatebase]/[acid]

= 5.48 + log 1.492/1.492

= 5.48