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[化學 Chem] Ksp of Ag2CrO4 & AgCl

Ksp of Ag2CrO4 & AgCl E-mail 此主題給朋友

[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 ]

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

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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.




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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 ]



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[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.

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