[高中] 醋酸銀Ksp
已知CH3COOAg之Ksp=3.6×10-3,CH3COOH之Ka=1.8×10-5,則CH3COOAg在一最後pH=5
的溶液中之溶解度為
(A)4.5×10-2M (B)6.0×10-2M (C)7.5×10-2M (D)1.5×10-2M
我的想法是
CH3COOAg==>CH3COO- + Ag+ Ksp
CH3COO- +H+ ==> CH3COOH (1/Ka)
CH3COOAg+ H+ ==>CH3COOH + Ag+ (Ksp/Ka)
[Ag+]^2/[H+] =(Ksp/Ka)
最後算出來 [Ag+] =4.5×10-2M
可是答案是7.5×10-2M
想請問各位高手我在哪個環節出錯了
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