The Ka for acetic acid, CH3COOH, is 1.8 × 10-5. A buffer, made from 0.10 M CH3COOH and 0.10 M CH3COO- has a pH of ________.
a. 4.74
b. 14.00
c. 1.00
d. 9.26
e. 7.00
Ka = 1.8*10^-5
PKa = -logKa
= -log1.8*10^-5
= 4.74
[CH3COOH] = 0.1M
[CH3COO^-] = 0.1M
PH = PKa + log[CH3COO^-]/[CH3COOH]
= 4.74 + log0.1/0.1
= 4.74 + log1
= 4.74 +0
= 4.74
>>>>>answer
a. 4.74 >>>>answer
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