Question

Consider the titration of 50.00 mL of 0.200 M thioglycolic acid, H2SC2H2O2 (Ka1 = 2.3 x 10 -4 and Ka2 = 2.5 x 10 -11) with 0.500 M NaOH and calculate the pH after the following volumes of NaOH have been added. a) 0.00 mL b) 7.00 mL c) 20.00 mL d) 32.00 mL e) 40.00 mL f) 43.00 mL

Answer 1

pKa1 = -log Ka1 = 3.64

pKa2 = 10.6

suppose acid is H2A

millimoles of H2A = 50 x 0.2 = 10

a) 0.00 mL

H2A ----------------------> H+   + HA-

0.2                              0               0---------------------> initial

0.2-x                           x                x -----------------> equilibrium

Ka1 = x^2 /0.2-x

2.3 x 10^-4 =x^2 /0.2-x

x^2 + 2.3 x 10^-4 x - 4.6 x 10^-5 = 0

x = 0.0067

x = [H+] = 0.0067M

pH= -log[H+] = -log(0.0067)

pH = 2.17

b) 7.00 mL

millimoles of NaOH = 7 x 0.5 = 3.5

H2A + NaOH ------------------> NaHA + H2O

10          3.5                              0              0 --------------------> initial

6.5            0                             3.5             3.5----------------> equilibrium

pH = pKa 1 + log [3.5/6.5]

pH = 3.64 +log [3.5/6.5]

pH = 3.37

c) 20.00 mL

millimoles of NaOH = 20 x 0.5 = 10

H2A + NaOH ------------------> NaHA + H2O

10          10                              0              0

0           0                             10            10

it is equivaelce point . at this main species NaHA

so

pH = 1/2 [pKa1 + pKa2]

pH = 1/2 [3.64 + 10.6]

pH = 7.12