Normally, Mn^2+ is in group III of the cation qualitative scheme and the hydroxide does NOT ppt in a solution of NH3 and NH4Cl. The NH4Cl is there to keep the OH^- DOWN so Ksp for Mn(OH)2 is not exceeded. However, this problem has no NH4Cl and that will let the OH^- be higher.
First calculate OH^- from 6M NH3.
...........NH3 + HOH ==> NH4 + OH^-
I..........6...............0.....0
C..........-x.............x.....x
E.........6-x.............x.....x
Kb = 1.8E-5 = (NH4^+)(OH^-)/(NH3)
Substitute and solve for x = OH^-
My answer is approximately 0.008 but you need to do that more accurately. Check that Kb value also and use the one in your text or notes. Texts differ on that.
Then Mn(OH)2 ==> Mn^2+ + 2OH^-
Qsp = (Mn^2+)(OH^-)^2
Qsp = (0.01M)(0.08)^2 = about 6E-5
Compare that with Ksp.
If Qsp > Ksp a ppt forms.
If Qsp < Ksp no ppt forms.
If a .01M Mn(NO3)2 solution has an ammonia concentration of 6M, would a precipitate form? Calculate the ion product and compare to Ksp.
I'm way too confused. I'm pretty sure a precipitate would form so I set up an ice table but don't know where to go from there? Please help!!
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