Bài tập pH hay

N

nhoc_maruko9x

[tex]CH_3COOH[/tex]
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[tex]CH_3COO^- + H^+[/tex]

[tex]CH_3COONa \rightarrow CH_3COO^- + Na^+[/tex]

[tex]K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]

[tex]\Rightarrow [H^+]=\frac{K_a[CH_3COOH]}{[CH_3COO^-]}[/tex]

Do [tex]CH_3COOH[/tex] phân ly rất yếu nên [tex]CH_3COO^-[/tex] coi như từ [tex]CH_3COONa[/tex] phân ly ra [tex]\Rightarrow [CH_3COO^-]=0.2M[/tex]

[tex]\Rightarrow [H^+]=\frac{1.8*10^{-5}*0.3}{0.2}=2.7*10^{-5}[/tex]

[tex]\Rightarrow pH=4.57[/tex]
 
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