May 07, Friday

I wasn't late today, so I volunteered to do the blog. :) :)


First, we did the questions 3 & 4 in the Solubility Equilibria Assignment #1. Obtain the answers from Ms. K. Note: in letters b and d you have to change the unit to moles/liter before you can solve for the Ksp.


We also went over the pages 31-35 on the Chemical Equilibrium booklet. In that, we learned a new lesson, determining ion concentration from Ksp. For all of you who hates ICE, you'll life will be easier. It's no different from the ICE method we've been doing, except that:

• the initial concentration of the ions (products), before dissolving, is zero.

• reaction is consists of a saturated solution as reactants and ions as products so you get to IGNORE the saturated solution, the solid one, in your ice table.

Example: The Ksp of magnesium hydroxide is 8.9 x 10 ⁻12. Determine the solubility of Mg(OH)₂.

• Mg(OH)₂(s) <=> Mg2+ (aq) + 2OH⁻(aq)

I...............___.................0................0
C.............___................+x................+2x
E..............___...............+x................+2x


To solve for x, use the formula, Ksp = [products] and substitute the values. We'll find that the solubility of Mg(OH)₂ is 1.3 x 10-4.


That's determining ion concentration using Ksp in PURE WATER. Another thing we learned is determining ion concentration with an ION IN COMMON WITH THE COMPOUND.

Example: The Ksp of magnesium hydroxide is 8.9 x 10 ⁻12. What is the solubility of magnesium hydroxide in a 0.10 M solution of NaOH.

Step 1: Determine the concentration of the common ion, OH ⁻.

NaOH(s) <=> Na+(aq) + OH -(aq)
[OH⁻] = [NaOH] = 0.10 M

Step 2: Set up an ICE table for Mg(OH )₂.

• Mg(OH)₂(s) <=> Mg2+ (aq) + 2OH⁻(aq)

I...............___.................0................0
C.............___................+x................+2x
E..............___...............+x................0.10 +2x

***You have to include the concentration of OH - from NaOH(s) in the table because when we dissolve Mg(OH)₂ in the solution, OH ions are present. According to Le Chatelier's Principle, adding more OH ions to a saturated solution would increase the overall concentration of the products shifting the equilibrium to the left. This would result in more solid formed and a decrease solubility.***


Step 3: Substitute values into the solubility product expression.

Ksp = [Mg2+][OH-]2

8.9 x 10⁻12= (x)(0.10 + 2x)2

**at this point, you can see that you'll have a quadratic equation if you solve further, so, ignore the x of the ion in common, OH. Ms. K said that it's too tiny to matter.**

8.9 x 10⁻12= (x)(0.10)2

8.9 x 10⁻12= 0.01x

8.9 x 10⁻10M = x

Let's check if the value makes sense. The solubility of Mg(OH)₂ in pure water is 1.3 x 10-4 M and in a 0.10 solution of NaOH is 8.9 x 10⁻10M. The solubility decreased, just as we predicted.



I dont know if I have to pick for someone to do the next scribe... if so, i'll pick Ericka! :) :)