hóa lớp 11 nhờ các bạn tính dùng

H

hiepchau96

[TEX]CH_3COONa \to CH_3COO^{-} + Na^{+}[/TEX]
0,2M.........................0,2M

[TEX]CH_3COOH \to CH_3COO^{-} + H^{+}[/TEX]
0,1M......................0,2M
x............................x...........................x
0,1-x....................0,2+x....................x

ta có: 1,75.10^-6 ( 0,1-x ) = x.(0,2+x)
giải tìm ra x

pH = -log[x]
 
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