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Author Topic: Add how much ammonium chloride to prevent precipitation of magnesium hydroxide?  (Read 2861 times)

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讨厌化学

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Here is the question:

You mix 50ml of 0.2 mol.L-1 Mg2+ with 50ml of 2.0 mol.L-1 NH3 aqueous solution. How much NH4Cl needs to be added to prevent precipitation of Mg(OH)2? You are given Ksp[Mg(OH)2] = 1.2 x 10-11; Kb[NH3] = 1.8 x 10-5; MNH4Cl = 53.5 g.

I have an answer but no explanation of the method used. Here is how I would proceed, but I think the method is wrong somehow:

Mg(OH)2  ::equil:: Mg2+ + 2OH-

Find the concentration of OH- needed for precipitation of Mg(OH)2 by using

Ksp[Mg(OH)2] = 1.2 x 10-11 = [Mg2+] [OH-]2

giving [OH-] = 0.0346 mol.L-1.

Now OH- is produced in the reaction NH3 + H2::equil:: NH4+ + OH-

So we can find what concentration of NH4+ would give a corresponding concentration of OH- using

Kb[NH3] = 1.8 x 10-5 = [NH4+] [OH-] / [NH3] =  [NH4+] 0.0346 / (2x0.05/0.1)

giving [NH4+] = 5.20 x 10-4 mol.L-1.

So, to give this concentration of NH4+, we would have to add mass m of NH4Cl such that (m/53.5)/0.1 = 5.20 x 10-4, which gives m = 2.78 x 10-3 g.

OK, I know this is wrong, and the answer I have is 8.8 g, but how do I calculate it?


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savy2020

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Find the concentration of OH- needed for precipitation of Mg(OH)2 by using

Ksp[Mg(OH)2] = 1.2 x 10-11 = [Mg2+] [OH-]2

giving [OH-] = 0.0346 mol.L-1.
What did you do here?.. How did you get [OH-]=0.0346 mol.L-1? I couldn't follw that
OK..anyway you can find [Mg2+] from the data of the question after you mix the 2 solutions.
From that you get [OH-].. using Kspexpression

Now NH3 - NH4Cl combination is a buffer. You could be able to find the pOH of the buffer as pOH=pKb+ log([salt]/[base]). With this [NH4Cl] i.e, salt can be obtained

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:-) SKS

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