Question

The expression for the ion product of water is:

K_w = γ_H+ [H⁺] x γ_OH-[OH^-] = 1.00 x 10^-14.

In 0.10 M Na₂SO₄: γ_H+ = 0.86 and γ_OH- = 0.81.

Given this information, please calculate the pH of an aqueous solution of 0.10 M Na₂SO₄, using activities.

Enter your answer below, to a precision of two digits after the decimal point. Do not enter units or use scientific notation.

Hint: pH = -log (γ_H+ [H⁺])

Answer 1

Ans.

The expression for K_b is

(Ka of H₂SO₄ is 1.00 x 10³)

The ICE table for the below equilibrium is

	[SO42-]	[H2O]	[H2SO4]	[OH-]
I	0.10M	-	0	0
C	-x	-	+x	+2x
E	0.10-x	-	x	2x

Make the assumption that since the value of K_b is so small, the value of x is even smaller and x in the denominator can be neglected.

Therefore,

(since the question requires pH calculation involving activities)

Therefore, pH of the aqueous 0.10M Na₂SO₄ is 7.96