\(\alpha=\dfrac{C}{C_0}\Rightarrow C=0.002\cdot1.2\%=2.4\cdot10^{-5}\left(M\right)\)
\(CH_3COOH⇌H^++CH_3COO^-\)
\(0.002..................0..................0\left(BĐ\right)\)
\(2.4\cdot10^{-5}.....2.4\cdot10^{-5}...2.4\cdot10^{-5}\left(PL\right)\)
\(1.976\cdot10^{-3}....2.4\cdot10^{-5}...2.4\cdot10^{-5}\left(CB\right)\)
\(pH=-log\left[H^+\right]=-log\left(2.4\cdot10^{-5}\right)=4.6\)
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