Question

Calculate the Ve₁, Ve₂, and pH at each point listed for the titration 30mL of 0.030 M diprotic acid (pKa₁ = 2.97 and pKa₂ = 9.61) with 0.050 M NaOH

V_b = 0.0 mL

V_b = 5.0 mL

V_b = 9.0 mL

V_b = 18.0 mL

V_b = 20.0 mL

V_b = 36.0 mL

V_b = 50.0 mL

V_b = 70.0 mL

*Please show all work!! Thank you!*

Answer 1

Titration of diprotic acid H2A with NaOH

Vb = 0.0 ml

H2A <==> H+ + HA-

Ka1 = [H+][HA-]/[H2A]

1.07 x 10^-3 = x^2/0.03

x = [H+] = 5.67 x 10^-3 M

pH = -log[H+] = 2.25

Vb = 5 ml

[H2A] remained = (0.03 M x 30 ml - 0.05 M x 5 ml)/35 ml = 0.0186 M

[HA-] formed = 0.05 M x 5 ml/35 ml = 0.007 M

pH = pKa1 + log(base/acid)

      = 2.97 + log(0.007/0.0186)

      = 2.55

Vb = 9 ml

[H2A] remained = (0.03 M x 30 ml - 0.05 M x 9 ml)/39 ml = 0.001154 M

[HA-] formed = 0.05 M x 9 ml/39 ml = 0.01154 M

First 1/2 equivalence point

pH = pKa1 + log(base/acid)

      = 2.97

Vb = 18 ml

1st equivalence point

pH = 1/2(pKa1 + pKa2)

      = 1/2(2.97 + 9.61)

      = 6.29

Vb = 20 ml

moles H2A = 0.03 M x 30 ml = 0.9 mmol

moles NaOH = 0.05 M x 20 ml = 1 mmol

[HA-] remained = 0.8 mmol/50 ml = 0.016 M

[A^2-] formed = 0.1 mmol/50 ml = 0.002 M

pH = pKa2 + log(base/acid)

      = 9.61 + log(0.002/0.016)

      = 8.71

Vb = 36 ml

moles H2A = 0.03 M x 30 ml = 0.9 mmol

moles NaOH = 0.05 M x 36 ml = 1.8 mmol

IInd equivalence point

[A^2-] formed = 0.9 mmol/66 ml = 0.0136 M

A^2- + H2O <==> HA- + OH-

Kw/Ka2 = [HA-][OH-]/[A^2-]

1 x 10^-14/2.45 x 10^-10 = x^2/0.0136

x = [OH-] = 7.45 x 10^-4 M

pOH = -log[OH-] = 3.13

pH = 14 - pOH = 10.87

Vb = 50 ml

moles H2A = 0.03 M x 30 ml = 0.9 mmol

moles NaOH = 0.05 M x 50 ml = 2.5 mmol

excess [OH-] = 0.7/80 ml = 0.00875 M

pOh = -log[OH-] = 2.06

pH = 14 - pOH = 11.94

Vb = 70 ml

excess [OH-] = (0.05 M x 70 ml - 2 x 0.03 M x 30 ml)/100 ml = 0.017 M

pOH = -log[OH-] = 1.77

pH = 14 - pOH = 12.23