Question

The expression for the ion product of water is: Kw = γH+ [H+] x γOH-[OH-] =...

The expression for the ion product of water is:

Kw = γH+ [H+] x γOH-[OH-] = 1.00 x 10-14.

In 0.10 M Na2SO4: γH+ = 0.86 and γOH- = 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 (γH+ [H+])

Homework Answers

Answer #1

Ans.

The expression for Kb 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 Kb 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

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