Question

Ka1 = 1.5 ✕ 10-2 Ka2 = 2.6 ✕ 10-7

A 50.0 mL sample of 0.15 M of maleic acid is titrated with 0.13 M NaOH. Calculate the pH at the following points in the titration curve. (a) the first half-equivalence point (b) the first equivalence point (c) the second half-equivalence point (d) the second equivalence point

Answer 1

pKa1 = -log Ka1

pKa1 = -log (1.5 x 10^-2)

pKa1 = 1.82

pKa2 = 6.585

(a) the first half-equivalence point

here : pH = pKa1

pH = 1.82

(b) the first equivalence point

pH = 1/2 [pKa1 + pKa2]

pH = 1/2 [1.82 + 6.58]

pH = 4.20

(c) the second half-equivalence point

pH = pKa2

pH = 6.58

(d) the second equivalence point

H2A + 2NaOH ------------------> Na2A + 2H2O

50x 0.15 / 1 = 0.13 x V / 2

V = 115 mL

volume of NaOH needed = 115 mL

total volume = 115 + 50 = 165 mL

here only salt remains = 50 x 0.15 / 165

                                    = 0.0454 M

A-2 + H2O --------------------> HA- + OH-

0.0454 -x                                 x            x

Kb1 = x^2 / 0.0454-x

3.85 x 10^-8 = x^2 / 0.0454-x

x^2 + 3.85 x 10^-8 x - 1.75 x 10^-9 = 0

x = 4.18 x 10^-5

[OH-]= 4.18 x 10^-5 M

pOH = 4.38

pH + pOH = 14

pH = 9.62