Question

Calculate the pH of 25.0 mL of 0.110 M aqueous lactic acid being titrated with 0.150 M NaOH(aq) [See text for Ka value.]
(a) initially

(b) after the addition of 5.0 mL of base

(c) after the addition of a 5.0 mL more of base, Vt=10.0mL

(d) at the equivalence point

(e) after the addition of 5.0 mL of base beyond the equivalence point

(f) after the addition of 10.0 mL of base beyond the equivalence point

(g) Pick a suitable indicator from your text, or your Laboratory Manual. (I can do this part too)

Answer 1

lactic acid Ka = 1.38 x 10^-4

pKa = -log Ka = -log (1.38 x 10^-4)

pKa = 3.86

a)

pH = 1/2 (pKa - log C)

     = 1/2 (3.86 - log 0.110)

pH = 2.41

b)

millimoles of lactic acid = 25 x 0.110 = 2.75

millimoles of NaOH = 5 x 0.150 = 0.75

lactic acid + NaOH   ----------------> sodium lactate + H2O

2.75              0.75                                      0                   0

2                   0                                       0.75

pH = pKa + log [salt / acid]

     = 3.86 + log [0.75 / 2]

pH = 3.43

c)

millimoles of NaOH =10 x 0.150 = 1.50

lactic acid + NaOH   ----------------> sodium lactate + H2O

2.75              1.50                                     0                   0

1.25                   0                                      1.50

pH = 3.86 + log [1.50 / 1.25]

pH = 3.94

d)

At equivalence point :

millimoles of acid = millimoles of base

2.75 = 0.150 x V

V = 18.33 mL

here salt only remains.

salt concentration = 2.75 / 25 + 18.33 = 0.0635 M

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

    = 7 + 1/2 (3.86 + log 0.0635)

pH = 8.33