Question

1- Calculate the concentrations of all species in a 0.690 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10^–2 and Ka2 = 6.3× 10^–8.

2-For the diprotic weak acid H2A, Ka1 = 2.1 × 10^-6 and Ka2 = 8.7 × 10^-9.

What is the pH of a 0.0800 M solution of H2A?

What are the equilibrium concentrations of H2A and A2– in this solution?

3- NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.070 M in NH4Cl at 25 °C?

Answer 1

0.690 M

Na2SO3 --------------------> 2 Na+ + SO3-2

[Na+] = 2 x 0.690 = 1.38 M

Ka1 = 1.4 x 10^-2 ,

Ka2 = 6.3 x 10^-8

Kw = 1.0 x 10^-14

Kb1 = Kw / Ka2 = 1.0 x 10^-14 / 6.3 x 10^-8 = 1.59 x10^-7

Kb2 = Kw / Ka1 = 1.0 x 10^-14 / 1.4 x 10^-2 = 7.14 x 10^-13

SO3^-2 + H2O -------------------> HSO3-   + OH-

0.69                                               0               0    ------------------> initial

0.69-x                                            x                x ---------------------> equilibrium

Kb1 = [HSO3-][OH-]/[SO3-2]

Kb1 = x^2 / 0.69 -x

1.59 x 10^-7   = x^2 / 0.69 -x

x^2 + 1.59 x 10^-7 x - 1.097 x 10^-7 = 0

x = 3.31 x 10^-4

x = [OH-] = [HSO3-] = 3.31 x 10^-4 M

[OH-] = [HSO3-] = 3.31 x 10^-4 M

[SO3^-2] = 1.10 - x

[SO3^-2]      = 0.690 M

[H+] = Kw / [OH-] = 1.0 x 10^-14 / 3.31 x 10^-4

[H+] = 3.02 x 10^-11 M

H2SO3 = Kb2

[H2SO3 ] = 7.14 x10^-13 M

2)

H2A -----------------> HA-   +    H+

0.08                           0            0

0.08 - x                      x             x

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

2.1 x 10^-6 = x^2 / 0.08 - x

x = 4.09 x 10^-4

[H+] = 4.09 x 10^-4 M

pH = -log[H+] = -log (4.09 x 10^-4)

pH = 3.39

[H2A] = 0.08 - 4.09 x 10^-4 = 0.0446

[H2A] = 0.0796 M

[A2-] = Ka2 = 8.7 × 10^-9 M

3)

pH = 5.21