ksp and solubiluty

[spoog74]

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Im confused regarding a question;

Ksp of Mg(OH)2 is 1.2 x 10 ^-11. In which solution would Mg(OH)2 be most soluble?

They give the answer as 0.4 M HBr, but i dont understand why it cant be 0.6M Ba(OH)2 since it would cause the reaction to shift towards the right..

 

[ScratchMan]

If you choose the Ba(OH)2 solution, the reaction should shift toward the left. Think about Qsp vs Ksp

given Mg(OH)2 => Mg++ + 2OH- & Ksp = [Mg++][OH-]^2 = 1.2*10^-11
lets first find the concentration of each [Mg++] & [OH-]. Lets assume X = [Mg++].
Ksp = X*(2X)^2 = 4X^3 = 1.2*10^-11 then, X = 1.4*10^-4.

If you choose the 0.6 M of Ba(OH)2 solution, the Qsp = (1.4*10^-4)(0.3)^2 = 1.26*10^-5 > given Ksp = 1.2*10^-11. It means the reaction shift to the left. (precipitation occurred) Thus, we cannot say the choice of the Ba(OH)2 is good for making Mg(OH)2 soluble.

Acidic solution such as HBr should be good to react with OH- from the Mg(OH)2. Mg(OH)2 includes 2*1.4*10^4 M of OH- thus it could be reacted with 0.4M of H+ from the HBr.

It is my thought...

 

[seunglim87]

simply, any solution that contains [Mg+2] or [OH-] will decrease the solubility due to the common ion effect

 

[recyrb]

If you choose the Ba(OH)2 solution, the reaction should shift toward the left. Think about Qsp vs Ksp

given Mg(OH)2 => Mg++ + 2OH- & Ksp = [Mg++][OH-]^2 = 1.2*10^-11
lets first find the concentration of each [Mg++] & [OH-]. Lets assume X = [Mg++].
Ksp = X*(2X)^2 = 4X^3 = 1.2*10^-11 then, X = 1.4*10^-4.

If you choose the 0.6 M of Ba(OH)2 solution, the Qsp = (1.4*10^-4)(0.3)^2 = 1.26*10^-5 > given Ksp = 1.2*10^-11. It means the reaction shift to the left. (precipitation occurred) Thus, we cannot say the choice of the Ba(OH)2 is good for making Mg(OH)2 soluble.

Acidic solution such as HBr should be good to react with OH- from the Mg(OH)2. Mg(OH)2 includes 2*1.4*10^4 M of OH- thus it could be reacted with 0.4M of H+ from the HBr.

It is my thought...

Great explanation, but why didnt you use, .6 as the concetration of (OH)
so Ksp= 1.2x10-11

Qsp= (X)(2x + .6)2 = (x)(~.6)2 = .36x

1.4x10-4 x .36 = ~ 5 x 10-5

a preciptate still forms, shifting left



its late so i am probably wrong, why am i up? but yes simply put anything with the commone ion will shift the rxn left. also note that anything that is basic will be more soluble in an acidid solution, so that a good way to think about it. the more acidic the more soluble.
and...im out

 

[ScratchMan]

Thanks for correction, recyrb!! I remember you for great explanation about nasty polyprotic Ka problem last time... GLuck for your exam and am expecting your nice breakdown. 🙂

 

[2th Doc]

+1 to common ion explanation. Also, IIRC from Chad (it makes sense logically anyway) a compound that makes a basic solution by dissociating, here [OH-] will always be more soluble in an acidic solution than basic. And vice versa. Ask the French guy Le Chatelier.