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This question was asked a few years back on sdn, but I don't get the solution.
The question asked: 1M NaOH is added to the following solutions (each is 1M) which will precipitate first?
AgOH (Ksp = 1.5 x 10 ^-8)
Al(OH)3 (ksp = 3.7 x 10^-15)
Mg(OH)2 (ksp = 1.2 x 10 ^-11)
Mn(OH)2 (ksp = 2 x 10^ -13)
The answer is AgOH.
I solved for x(where x=solubility) and I understand that the higher the x, the lower the precipitation. But with my calculations, I get Mn(OH)2 as the answer with the lowest solubility (x=3.6X10^-5)
For AgOH, I get x=1.22X10^-4
For Al(OH)3, I get x=1.08X10^-4
For Mg(OH)2, I get x=1.44X10^-4
In these calculations, I accounted for the stochiometric coefficients, but obviously must have messed up somewhere. So could someone please explain with detailed calculations what I messed up on.
Thanks
The question asked: 1M NaOH is added to the following solutions (each is 1M) which will precipitate first?
AgOH (Ksp = 1.5 x 10 ^-8)
Al(OH)3 (ksp = 3.7 x 10^-15)
Mg(OH)2 (ksp = 1.2 x 10 ^-11)
Mn(OH)2 (ksp = 2 x 10^ -13)
The answer is AgOH.
I solved for x(where x=solubility) and I understand that the higher the x, the lower the precipitation. But with my calculations, I get Mn(OH)2 as the answer with the lowest solubility (x=3.6X10^-5)
For AgOH, I get x=1.22X10^-4
For Al(OH)3, I get x=1.08X10^-4
For Mg(OH)2, I get x=1.44X10^-4
In these calculations, I accounted for the stochiometric coefficients, but obviously must have messed up somewhere. So could someone please explain with detailed calculations what I messed up on.
Thanks