Question

Solid Mg(OH)₂ has a K_sp value of 1.8 x 10^-11 at room temperature. You are given a 0.150L solution of a 0.100M MgCl₂ solution to mix with 0.150L of 0.100M NH₃.

How many grams of (NH₄)₂(SO₄) would you need to add to this solution to keep Mg(OH)₂ from precipitating?

The K_b (NH₃) = 1.8x10^-5

Answer 1

The reaction can be represented as follows:

2 NH₄OH + MgCl₂ Mg(OH)_2(s) + 2 NH₄⁺ + 2Cl^-

Number of moles of NH₃ added = 0.100 0.150 = 0.015 NH3 or NH₄OH.

Number of moles of MgCl₂ = 0.100 0.150 = 0.015 MgCl₂

Molarity of Mg⁺⁺ = 0.015 0.300 = 0.05

Molarity OH^- = 0.015 0.300 = 0.05

Molarity of OH^- is the as the molarity of the Mg⁺⁺, we will get the same Molarity of Mg(OH)₂ as we had in Mg⁺⁺.

The next step is to drive the OH^- concentration down so low that Mg(OH)₂ will not precipitate out.

Ksp = l.8 x l0^-11 = [Mg] [OH^-]²

We know the Mg++ concentration to be 0.05 molar ; so let X = [OH^-]

l.8 x l0^-11 = [0.05] [X]²

[X]² = l.8 x l0^-11 0.05 = 3.6 x l0^-10

Therefore, X = 1.9 x l0^-5 moles per liter [OH^-] that would be the [OH^-] where a solid precipitate would begin to form.

Addition of NH₄⁺ ions (act as an acid by donating H⁺) can lower the [OH^-] to this level.

The available [OH^-] is 0.05 down to 1.9 x l0^-5 molar OH^- by adding NH₄⁺ ions.

Hence the desired concentration of OH- = 0.05 - 1.9 x l0^-5 = 0.0499 moles per liter

This is the desired concentration of OH^- which is to be neutralized by adding NH₄⁺.

So we require 0.0499 moles per liter NH₄⁺ concentration to neutralize the 0.0499 moles per liter of OH^- in solution.

We have already 0.015 moles/liter of [NH4⁺], so subtracting the 0.015 moles of NH₄⁺ from 0.0499 moles of OH^-, we will get the remaining OH^-.

0.0499 - 0.015 = 0.0349 molar NH₄⁺

Therefore, the extra moles of NH₄⁺ needed is:

0.0349 molar NH₄⁺ 0.300 L = 0.0105 moles NH₄⁺

For (NH₄)₂SO₄ we have 2 moles of NH₄⁺ in each mole of this substance, so we require one half times. Therefore,

0.0105 moles of (NH₄)₂SO₄ or 0.0053 moles of this substance.

Therefore, 0.0053 moles of (NH₄)₂SO₄ times 132 grams per mole will give 0.7 grams of (NH₄)₂SO₄.

Hence 0.7 grams of (NH₄)₂SO₄ is added to this solution inorder to prevent precipitation.