What is the PH of a buffer solution containing 0.2moldm^-3 ethanoic acid and 0.5moldm^-3 sodium ethanoate?Ka=1.75×10^-5 moldm^-3
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PH=−log1.75×10−5+log[0.2×10−30.5×10−3]PH=-log1.75×10^{-5}+log[\frac{0.2×10^{-3}}{0.5×10^{-3}}]PH=−log1.75×10−5+log[0.5×10−30.2×10−3]
PH=4.757−0.398=4.359PH=4.757-0.398=4.359PH=4.757−0.398=4.359
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