Ion Concentration

Disssociation:

-Ionic Particles are made up of 2 parst:

• Cation: + charged particles
• Anion: - charged particles

-When ionic compounds are dissolved in Water the cation and anion separate from each other
-This process is called dissociation.
-When writing dissociation equ'ns the atoms and charges must be ballanced
-The dissociation equ'n for NaCl is Na (+) + Cl (-)

EX// 

What are the following compound's Dissociation equ'ns?

-BaSO4 --> Ba (2+) + SO4 (2-)

-Al2(SO4)3 --> 2Al (3+) + 3SO4 (2-)

-If the volume doesn't change then the concentration of the individual ions depends on the balanced coefficients in the dissociation equ'n.

EX//

Determine the Ion Concentrations...

- 0.250 M of the sol'n KOH:

KOH --> K (+) + OH (-) 

0.250 mol/L x 1/1 + 0.250 M of [K+] 
*because there is a 1:1 ratio for this compound, [OH-] has the same ion concentration as [K+]. 

- A sol'n made from dissolving 2.5g of Ba3(PO4)2 in 50mL of water:

Ba3(PO4)2 --> 3Ba (2+) + 2PO4 (3+)

2.5g x 1 mol/601.9g= 0.00415mol / 0.050L = 0.831 mol/L x 3/1 = 0.249 M of [Ba 2+]
   "                     "                       "                           "              x 2/1 = 0.166 M of [PO4 3+]

A helpful video... :)

Solution Stiochiometry

What is Solution Stiochiometry? : Solution stoichiometry deals with reactions in solutions.

Types of Solution Reactions

   - Precipitation

   - Acid-base

   - Oxidation-reduction

   - Titrations

Example:

100 mL of 0.250 M Iron (II)chloride reacts with excess copper. How many grams of Iron are produced?

FeCl2 + Cu----> Fe + CuCl2

0.250 mol    x 0.100 L x 1     x 55.8 g  = 0.0250g 

           L                          1         mol

Example:

A beaker contains 100 mL of 1.5 M Hcl. Excess Zinc added to the beaker. Determine how many litres of hydrogen gas should be produced.

2HCl + Zn ---> H2 + ZnCl2

1.5 mol    x 0.100L x 1   x    22.4 L  = 1.68 L

        L                        2         mol     = 1.7 L

Here is a step-by- step video that shows you how to solve solution stoichiometry problems

Just in case if you didnt understand the previous video...

Solutions and Molarity

Solutions are homogeneous mixtures composed of a solute & a solvent
Solute: is the chemical present in lesser amount
(whatever is dissolved)
Solvent: the chemical present is the greater amount


Molarity


Concentration can be expressed in many ways. g/L, g/mL,mol/L, etc..

• But the most common is Molarity mol/L= M
• [HCL] concentration of HCl
• [NaCl] concentration of NaCl

Heres a video that teaches you how to calculate molarity problems :)

Molarity= moles    

                Volume

Example:

What is the concentration of a Sodium chloride solution made from 0.75 mol of NaCl dissolved in 250 mL of water?

0.75 mol x     1      =   30 mol/L

Example:

What is the concentration of the solution when 2.75g of Sodium Carbonate are dissolved in 50 mL of water?

2.75g x     1   mol    = 0.0259

                 106 g      = 0.0260

              

Titrations

A Titration is a common laboratory experimental technique used to determine the concentration of an unknown solution. Because volume measurements play a key role in titration, it is also known as volumetric analysis.


Needed for a titration lab:


1. Buret
2. Stopcock
3. Pipet (Glass tube)

4. Erlenmeyer Flask
5. Indicators- Used to identify the end point of filtration
6. Stock solution - Known Solution


Mike the explorer completed a titration of 0.330 M NaOH with 15.00 mL samples of HI of unknown concentration. The data he gathered is below. Determine the concentration of HI.




Subtract the final reading from the initial reading to get the Volume used.


Next we add up 11.2 mL + 10.9 mL + 11.4 mL = divide by 3 = 11.16666667 mL
Doing this gets us the average, but notice how we don't count the volume that was far off.


Next we change 11.16 mL into L

11.16 x 1L/1000 mL
= .0111L

Then, we take our concentration and change that into Moles of HI

0.330 M x .011 L = .00363 x 1 = 0.00363 mol
L 1


Finally, we take 0.00363 mol
and divide it by the 15.00 mL, but remember to convert into L!!



15.00 mL x 1 L/1000 mL
= .015 L


0.00363 mol/ 0.015 L
= .246 M

TItration Quiz!!

Dilutions

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 does change!

Because concentration is mol/L we can write:

C= n/C, n= CV, and V= n/C
n=  number of moles, v= volume in litres


So we can use:

C1V1= C2V2
C1V1 is the intial concentration and volume
C2,V2 is the final concentration and volume


Here are some examples to follow by:


Example 1) 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?


C1V1= C2V2


(11.9mol/L)(.03L)= C2 (0.3L)
.357 mol = C2 (0.3L)
C2= 1.19 M


Example 2) 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/L) = 0.033 mol/L

0.60 L x (0.125 mol/L) = 0.075 mol/L

0.033 mol + 0.075 mol = 0.108 mol

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


0.108 mol/ 0.7L
= 0.154 M




Example 3) 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

Try some Dilution questions!!