If acetic acid is the only acid that vinegar contains (Ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.
the particular sample of vinegar has a PH of 3.15
CH3COOH ---> CH3COO-
+ H+
X
X
pH = -log [H+]
3.15 = -log [H+]
[H+] = 7.08*10^-4 M
so,
x = 7.08*10^-4 M
Ka = [CH3COO-] [H+] / [CH3COOH]
1.8*10^-5 = x*x / [CH3COOH]
1.8*10^-5 = (7.08*10^-4)*(7.08*10^-4) / [CH3COOH]
[CH3COOH] = 0.028 M
Answer: 0.028 M
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