Question

A. Given Kw for water is 2.4x10-14 at 37 deg C. What is the pH of a natural aqueous solution at 37 deg C?

B. If Ka of a weak acid is 5.0x10-6. What is the pH of a 0.43 M solution of this acid?

C. What concentration of SO3 2- is in equilibrium with Ag2SO3(s) and 7.60x10-3 M Ag+?

Answer 1

Ans A:

Expression for Kw for water can be written as follows:

Kw = [H⁺][OH^-]

In natural water [H⁺]= [OH^-]

Therefore, Kw = [H⁺][H⁺] = [H⁺]²

Substitute Kw = 2.4 x 10^-14

2.4 x 10^-14 = [H⁺]²

[H+] = (2.4 x 10^-14)^1/2

[H+] = 1.549 x 10^-7

Expression for pH can be written as follows:

pH = -log[H⁺]

pH = -log (1.549 x 10^-7 )

pH = 6.810

Therefore, pH of natural aqueous solution is 6.180

Ans B:

Expression for Ka of weak acid HA can be written as follows:

Ka = [H+][A-] / [HA]

Since, [H+] = [A-]

Ka = [H+][H+] / [HA]

Ka = [H+]² / [HA]

Substitute Ka = 5.0x10^-6 and [HA] = 0.43 M

5.0 x 10^-6 = [H+]² / 0.43

[H+]² = 5.0 x 10^-6 x 0.43

[H+]² =2.15 x 10^-6

[H+] =1.47 x 10^-3

Expression for pH can be written as follows:

pH = -log[H+]

pH = -log (1.47 x 10^-3 )

pH = 2.83

Therefore pH of a 0.43 M solution of given acid is 2.83

Ans C:

Expression for solubility product of Ag₂SO₃ can be written as follows:

Ksp = [Ag+]² [SO₃]

Ksp for Ag₂SO₃ is 1.50 x 10^-14

Let s be the moles of Ag₂SO₃ dissociated to give 2s moles of Ag+ and s moles of SO₃^2-then, Ksp can be written as follows:

Ksp = (2s)² x s

Since, Ag+ already present is 7.60 x 10^-3 M

Therefore, total [Ag+] is 2s + 7.60 x 10^-3 M

Hence, Ksp can be written as follows:

Ksp = (2s + 7.60x10^-3)² x s

Substitute value of Ksp = 1.50 x 10^-14

1.50 x 10^-14 = (2s + 7.60x10^-3)² x s

Assuming s to be very very small, 2s can be neglected.

1.50 x 10^-14 = (7.60x10^-3)² x s

s = (1.50 x 10^-14) / (7.60x10^-3)²

s = (1.50 x 10^-14) / (5.776 x 10^-5)

s = 0.260 x 10^-9 M

Therefore, 0.260 x 10^-9 M SO₃^2- is in equilibrium with Ag₂SO₃(s)