Question

1. From the Kb values for arsenate in the following equations, calculate the three Ka values of arsenic acid. (Assume that Kw is 1.01???10?14.)

a) Kb1 = 3.19???10?3

b) Kb3 = 1.76???10?12

2. A solution contains 0.0490 M Ca²⁺ and 0.0316 M Ag⁺ (The K_sp of CaSO₄ is 2.4???10^?5, and theK_sp of Ag₂SO₄ is 1.5???10^?5.)

a) What will be the concentration of Ca²⁺ when Ag₂SO₄ begins to precipitate?

I would greatly appreciate assistance with the both of these problems.

Answer 1

1. Given,

Kb1 Arsenate = 3.19 x 10^-3

pKb = -log Kb = -log (3.19 x 10^-3 ) = 2.50

Now, Ka x Kb = Kw [Kw = Ionic product of water = 10^-14 at 25^OC]

Putting -log on both sides,

-log Ka x -log Kb = -log Kw

pKa + pKb = 14

Hence pKa = 14 - pKb

= 14 - 2.50

Hence, pKa1 = 11.50

pKa = -log Ka

Ka = 10^-pKa

Ka1 = 3.1 x 10^-12

Kb3 Arsenate = 1.76 x 10^-12

pKb = -log Kb = -log (1.76 x 10^-12 ) = 11.76

pKa = 14 - pKb

= 14 - 11.76

Hence, pKa3 = 2.24

pKa = -log Ka

Ka = 10^-pKa

Ka3 = 5.7 x 10^-3

pKa1 = 11.50, pKa3 = 2.24, pKa2 = ?

According to Pauling's rule of succesive pKa values, pKa = 8 - 5n, where n= number of ketonic oxygens present in the compound.

Hence to get succesive pKa values, we have to add approximately 5 units with the previous pKa values.

Therefore, pKa2 = 2.24 + 5 = 7.24

Ka2 = 10^-7.24

Ka2 = 6 x 10^-8

2. Ag₂SO₄ 2Ag⁺ + SO₄^2-

Let Ksp be the solubility product of Ag₂SO₄

Ksp = [Ag⁺]² [SO₄^2-] --------------------------(i)

Ksp = 1.5 x 10^-5

[Ag⁺] = 0.0316 M

Putting the values in eq (i)

1.5 x 10^-5 = (0.0316)²[SO₄^2-]

[SO₄^2-] = 1.5 x 10^-5 / (0.0316)²

[SO₄^2-] = 15.02 x 10^-3 M

CaSO₄ Ca²⁺ + SO₄^2-

Let Ksp be the solubility product of CaSO₄

Ksp = [Ca²⁺][SO₄^2-] --------------------------(ii)

Ksp = 2.4 x 10^-5

Putting the values in eq (ii)

2.4 x 10^-5 = [Ca²⁺] (15.02 x 10^-3)

[Ca²⁺] = 2.4 x 10^-5 / 15.02 x 10^-3

[Ca²⁺] = 1.6 x 10^-3 M

Hence the concentration of Ca²⁺ when Ag₂SO₄ begins to precipitate is 1.6 x 10^-3 M