其他討論
消閒生活手機討論遊戲地帶影視娛樂校園生活數碼科技寵物樂園學術文化體育世界購物廣場時事投資貼圖影片上班一族美容纖體戀愛婚姻汽車討論成人資訊博彩娛樂資源交流站務管理
發帖
註冊 登入/註冊 微博



收藏 訂閱 打印

[化學 Chem] Ksp of Ag2CrO4 & AgCl



Ksp of Ag2CrO4 & AgCl

[隱藏]
Why Ag2CrO4 is more soluble than agcl although ksp of agcl is larger than k2cro4?
ksp of AgCl = 1*10^ -10
ksp of AgCrO4 = 1*10^ -12

thx

[ 本帖最後由 ScareGhostAr? 於 2011-7-20 04:59 PM 編輯 ]

回覆 引用 TOP

因為K2CrO4 的Ksp係[K]^2[CrO4]
而AgCl 的Ksp是[Ag][Cl]






回覆 引用 TOP

Are you asking about Ksp of Ag2CrO4 or that of K2CrO4?
If you are talking about K2CrO4, then it is not worthwhile to give Ksp of Ag2CrO4 or even that of K2CrO4 itself(if you can find it).
This is because essentially all ionic compounds of K are totally soluble in water if it is still not saturated.
If you are talking about solubilities of Ag2CrO4 and AgCl in the same solvent, then refer to the equations:
Ksp[AgCl] = [Ag+][Cl-], Ksp[Ag2CrO4] = [Ag+]^2[CrO4 2-]
And solubilities refer to the no. of moles of the compound which 1dm^3 of solvent(usually pure water) can dissolve.




回覆 引用 TOP

引用:
原帖由 Hokkon1 於 2011-7-20 02:55 PM 發表
Are you asking about Ksp of Ag2CrO4 or that of K2CrO4?
If you are talking about K2CrO4, then it is not worthwhile to give Ksp of Ag2CrO4 or even that of K2CrO4 itself(if you can find it).
This is because essentially all ionic compounds of K are totally soluble in water if it is still not saturated.
If you are talking about solubilities of Ag2CrO4 and AgCl in the same solvent, then refer to the equations:
Ksp[AgCl] = [Ag+][Cl-], Ksp[Ag2CrO4] = [Ag+]^2[CrO4 2-]
And solubilities refer to the no. of moles of the compound which 1dm^3 of solvent(usually pure water) can dissolve.


係wo, 打錯左

In  Ag+ and Cl- titration,
用K2CrO4 做indicator 的原因
係因為Ag2CrO4比 AgCl  soluble
E個arm 唔 arm


但係ksp of AgCl 又比ksp of Ag2CrO4 大
AgCl(s)<--> Ag+  十 Cl-     .........(1)        , Ag2CrO4(s) <---> 2Ag+ 十 CrO4 2-   ..........(2)

(1) 會更lie on RHS
即係 AgCl 更 soluble

咁到底which is more soluble
明唔明我想問咩?

[ 本帖最後由 ScareGhostAr? 於 2011-7-20 04:55 PM 編輯 ]






回覆 引用 TOP

[隱藏]
Obviously whether I understand or not is not the core of our discussion as I have given excess information as an overflow. Anyways I will fine-tune my previous answer to make it easier to understand.
As I have mentioned, the solubility of a barely soluble ionic compound in a particular solvent is determined by the no. of moles of ionized compound per dm^3 of solvent according to its empirical formula such that the solubility of AgCl in pure water, according to the ionization equilibrium equation is given by
Solubility[AgCl] = [Ag+] = [Cl-] = (Ksp[AgCl])^(1/2)
Similarly, the solubility of Ag2CrO4 is givent by
Solubility[Ag2CrO4] = (1/2)[Ag+] = [CrO4 2-] = {(1/4)Ksp[Ag2CrO4]}^(1/3)
And the one not with the greater Ksp but the greater solubility value calculated from its Ksp is the more soluble ionic compound.
Also note that it is incorrect to use K2CrO4 as an indicator in the titration between Ag+ and Cl- because at the end point white ppt. AgCl(s) which is observable is produced.



熱門搜尋: 加拿大升學 bba 升學 學日文 日文 課程

回覆 引用 TOP



 提示:支持鍵盤翻頁 ←左 右→ 發新話題發佈投票
請先登入
小貼士:
依家可以用“@”tag會員啦!
預覽帖子  恢復數據  清空內容