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 Na_{2}SO_{4}: γ_{H+} = 0.86 and γ_{OH-} = 0.81.
Given this information, please calculate the pH of an aqueous solution of 0.10 M Na_{2}SO_{4}, 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^{+}])
Ans.
The expression for K_{b} is
(Ka of H_{2}SO_{4} is 1.00 x 10^{3})
The ICE table for the below equilibrium is
[SO_{4}^{2-}] | [H_{2}O] | [H_{2}SO_{4}] | [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 Na_{2}SO_{4} is 7.96
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