Posted by Taylor on Sunday, February 26, 2012 at 9:51pm.
pH = -log(H^+)
2.7 = -log(H^+).
(H^+) = 0.002
.........CH3COOH ==> H^+ + CH3COO^-
...........x.........0.002..0.002
Ka = (H^+)(CH3COO^-)/(CH3COOH)
Substitute 0.002 for H^+ and CH3COO^-, and x0.002 for CH3COOH, look up Ka, solve for x which = (CH3COOH)
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