[讨厌化学]

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 K_sp[Mg(OH)2] = 1.2 x 10^-11; K_b[NH3] = 1.8 x 10^-5; M_NH4Cl = 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  Mg²⁺ + 2OH^-

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

K_sp[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 + H₂0  NH₄⁺ + OH^-

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

K_b[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?

[savy2020]

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

K_sp[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 K_spexpression

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