Re: [高中] 醋酸銀Ksp
※ 引述《pingshung (孤獨之狼)》之銘言:
: 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
找錯方向了 最後pH=5 所以[A-]/[HA] = 1.8 [A-] = 1.8[HA]
[Ag+] = 2.8 [HA] 所有Ag+都是從CH3CO2Ag來的
所以 2.8[HA]*1.8[HA] = 3.6*10-3
[HA]=2.7*10-2
溶解度就是Ag+的濃度 所以 [Ag+] = 2.7*10-2 * 2.8 = 7.48* 10-2 =7.5*10-2
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