Im confused regarding a question;Ksp that Mg(OH)2 is 1.2 x 10 ^-11. In which systems would Mg(OH)2 be many soluble?They give the answer together 0.4 M HBr, but i dont understand why it i do not know be 0.6M Ba(OH)2 due to the fact that it would reason the reaction to transition towards the right..

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If you select the Ba(OH)2 solution, the reaction should change toward the left. Think about Qsp vs Kspgiven Mg(OH)2 => Mg++ + 2OH- & Ksp = ^2 = 1.2*10^-11lets very first find the concentration of each & . Allows assume X = . Ksp = X*(2X)^2 = 4X^3 = 1.2*10^-11 then, X = 1.4*10^-4. If you pick the 0.6 M of Ba(OH)2 solution, the Qsp = (1.4*10^-4)(0.3)^2 = 1.26*10^-5 > provided Ksp = 1.2*10^-11. It means the reaction shift to the left. (precipitation occurred) Thus, we cannot speak the selection of the Ba(OH)2 is an excellent for make Mg(OH)2 soluble. Acidic solution such together HBr need to be an excellent to react with OH- from the Mg(OH)2. Mg(OH)2 consists of 2*1.4*10^4 M that OH- therefore it can be reacted with 0.4M that H+ indigenous the HBr. That is my thought...
simply, any kind of solution that includes or will certainly decrease the solubility as result of the usual ion effect
If you choose the Ba(OH)2 solution, the reaction should transition toward the left. Think about Qsp vs Kspgiven Mg(OH)2 => Mg++ + 2OH- & Ksp = ^2 = 1.2*10^-11lets very first find the concentration of each & . Lets assume X = . Ksp = X*(2X)^2 = 4X^3 = 1.2*10^-11 then, X = 1.4*10^-4. If you pick the 0.6 M of Ba(OH)2 solution, the Qsp = (1.4*10^-4)(0.3)^2 = 1.26*10^-5 > provided Ksp = 1.2*10^-11. It way the reaction shift to the left. (precipitation occurred) Thus, us cannot to speak the selection of the Ba(OH)2 is good for do Mg(OH)2 soluble. Acidic solution such together HBr need to be good to react v OH- from the Mg(OH)2. Mg(OH)2 includes 2*1.4*10^4 M the OH- thus it might be reacted through 0.4M the H+ indigenous the HBr. It is mine thought...
great explanation, but why didnt you use, .6 as the concetration that (OH)so Ksp= 1.2x10-11Qsp= (X)(2x + .6)2 = (x)(~.6)2 = .36x1.4x10-4 x .36 = ~ 5 x 10-5 a preciptate tho forms, moving leftits late so ns am probably wrong, why am i up? however yes merely put anything through the commone ion will shift the rxn left. Additionally note the anything that is basic will be much more soluble in one acidid solution, so the a good way come think about it. The more acidic the more soluble.and...im out
Thanks because that correction, recyrb!! i remember friend for great explanation about nasty polyprotic Ka difficulty last time... GLuck for her exam and also am expecting your nice breakdown.
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+1 to common ion explanation. Also, IIRC native Chad (it provides sense logically anyway) a compound that provides a basic solution by dissociating, below will always be more soluble in an acidic solution than basic. And also vice versa. Questioning the French guy Le Chatelier.
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