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+])
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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