[高中] 難溶鹽

看板Chemistry作者 (浪子回頭)時間13年前 (2011/06/20 02:10), 編輯推噓1(1推 0噓 3→)
留言4則, 1人參與, 最新討論串1/1
因為脫離高中化學太久了 突然忘了怎麼算 希望大家幫個忙囉~~ The solubility of Mg(OH)2 (Ksp = 8.9* 10^-12) in 1.0 L of a solution buffered (with large capacity) at pH 9.85 is: a) 4.5 10^8 moles b) 1.8 10^-3 moles c) 1.3 10^-7 moles d) 7.1 10^-5 moles e) none of these 我知道M= 1.45* 10^-10 了 後面就不知道了 謝謝囉~~~ -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 120.105.159.89
推
06/20 17:55, , 1F
pOH = 14 - 9.85 = 4.15 = 5-0.85 = 5-log7
06/20 17:55, 1F
→
06/20 17:55, , 2F
所以水溶液中的[OH-] = 7x10^-5
06/20 17:55, 2F
→
06/20 17:56, , 3F
Ksp = 8.9*10^-12 = [Mg2+]*[OH-]^2 帶入[OH-]的值
06/20 17:56, 3F
→
06/20 17:58, , 4F
[Mg2+]=1.8*10^-3 M
06/20 17:58, 4F
‣ 更多 ru18284 的文章
文章代碼(AID): #1D_ZkX7K (Chemistry)