ION CONCENTRATION (March 10th,2011)

Today we learned about Ion concentration!


Involved with Ion Concentration is Dissociation.
Dissociation:

• Ionic compounds that are made up of 2 parts 
• An ionic compound is a chemical compound in which ions are held together in structure by ionic bonds. Usually, the positively charged portion consists of metal cations (positively charged particles) and the negatively charged portion is an anion (negatively charged particle.)
• When ionic compounds are dissolved in water, the cation and the anion separate from eachother.

Becareful because...

Dissociation is not Decomposition! When dissociation happens the ionic compounds separate from eachother, while decomposition in which organic material is broken down into simpler forms of matter.

• When writing dissociation equations, the atoms and charges MUST BALANCE

Here are a few examples:

• If the volume does not change then the concentration of individual ions depends on the balanced coefficient in the dissociation equation

Determine each of the following ion concentrations when the following solutions dissociate:

a) 0.250 M solution of KOH

b) 0.56 M solution of Fe(OH)3

c) 50.0 mL of 1.50 M NH4Cl is mixed with 75.0 mL of 1.25 M Ca(No3)2

Here's how to do Letter A:

Here's how to do letter B:

Here's how to do Letter C:

And that is all!

Post by Ren Ren Flores

DILUTIONS (March 4th, 2011)

Today we learned about Dilutions!
When two solutions are mixed the concentration changes

Dilution is the process of decreasing the concentration by adding a solvent (usually water)
the amount of solute does not change (n1=n2), but the concentration changes!


- Because concentration is mol/L we can write:

• C= n/C...... and n= CV..... and V= n/C

n= # of moles, v= volume in litres


So.... we use:

• C1V1= C2V2

C1V1 is the intial concentration and volume
C2,V2 is the final concentration and volume


Time for some questions:


A 30.0 mL of 11.9 M of Hydrochloric Acid is diluted to a final volume of 30 mL. What is the concentration of the new solution?
*if you look at the volume, it is in mL so don't forget to change that to L!


C1V1= C2V2


(11.9mol/L)(.03L)= C2 (0.3L)
.357 mol = C2 (0.3L)
.357mol= C2
0.3


C2= 1.19 M


100 mL of 0.330 M Li2CO3 is added to 600 mL 0.125 M Li2Co3. Determine the [Li2Co3]


0.10 L x 0.330 mol = 0.033 mol
                        L


0.60 x 0.125 mol = 0.075 mol
                     L


0.033 mol + 0.075 mol = 0.108 mol


Take your volumes (100 mL + 600 mL) Add them together and convert it into units of Liters.


0.108 mol
0.7 L


= 0.154 M


One more?


If 250 mL of water are added to 50.0 mL of 1.00 M BaCo3, what will the final concentration be?


(1.00 mol/L) (.05 L) = C2 (.25L)


0.05 mol = C2
0.25 L


C2 = 0.2 M


Post by Ren

Titration of Vinegar LAB

Today we did a lab on the Titration of Household vinegar.

The problem: What is the Molar Concentration of Acetic Acid in household vinegar?

Materials used:

• Lab Stand
• Buret
• Buret Clamp
• Funnel
• Erlenmeyer Flask
• Vinegar
• 1.00 M naOH solution
• Phenolphthalein Indicator 
• Pipet
• Lab coats and goggles for protection

A picture depicting our first ever titration lab experiment.... o0o so exciting:

Procedure:

1. Set up the buret and clamp, place the funnel on top of the buret, and put on lab coats and goggles
2. Add 1.00 M of NaOH to the buret being careful not to overfill the buret
3. Pipet 10.00 mL of Vinegar solution into a dry erlenmeyer flask
4. Add 2-3 drops of Phenolphthalein indicator to erlenmeyer flask, this will tell you when the reactants fully react
5. Record the initial volume of NaOH on the buret using the meniscus
6. Carefully add NaOH through the buret into the Erlenmeyer Flask until solution changes from colourless to light pink
7. Record the final volume of NaOH in the buret
8. Repeat steps 1-3 times until consistent results are obtained

So the accepted value for the concentration of household vinegar is 0.85 M

We got 0.92 M through our trials

Our percent error is 8.2 %, not bad but not good either..... haha

-Post by Ren