Let's say CH3COOH = HAc
...........HAc ==> H^+ + Ac^-
I........0.2853........0.........0
C........-x...............x..........x
E......0.2853-x.......x..........x
Substitute the E line into the Ka expression and solve for x = (H^+) and convert that to pH. pH= -log(H^+)
Note that it's a little ridiculous to have a concn of 0.2853 when the Ka value is not know to that accuracy.
Post your work if you get stuck.
Acetic acid is a weak acid with the formula , CH3COOH, the Ka for acetic acid is 1.76 x 10-5.
In aqueous solution, acetic acid partially dissociates according to the following reaction:
CH3COOH ⇔ CH3COO- + H+
Use the Ka equation to calculate the pH of the acetic acid solution described below:
Volume: 350 mL
Concentration: 0.2853 M
Since this is a weak acid, you can assume the amount of acid dissociated is << 5% of the total amount of acid present.
Report your answer with the correct number of significant digits.
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