POLAR MOLECULES

Today in Chemistry class, we continues our discussion on Polar Molecules from last class.    

                     

        o       Polar Molecules have an overall charge separation.

        o       Unsymmetrical molecules are usually polar. 

              

o       Molecular dipoles are the result of unequal sharing of electrons in a molecule. A perfect example of this would be water bending, which Mr. Doktor demonstrated in class. Here’s a video showing it : 

o       If a molecule is symmetrical the pull of e- is usually balanced.

o       Molecules can be un-symmetrical in two ways :

    Different Atoms

    Different number of Atoms

-         Ex. Find the Polarity

                 

          

Bonding and Electronegativity:

In today's class we learned about different types of bonds and also electronegativity. Electronegativity is just measure of an atom's attraction for electrons in a bond.

Types of bonds:
1. Ionic (metal-non metal)
    -electrons are transfered from metal to non-metal
2. Covalent (non metal- non metal)
    -electrons are shared between non-metals
3. metallic (metal)
    -holds pure metals together by electrostatic attraction

Electronegativity:
- Electronegativity is a measure of an atom's attraction for electrons in a bond
e.g. Fluorine=4.0, chlorine=3.0, cesium = 0.8
-atoms with greater Electronegativity attract electrons more
-polar covalent bonds form from an unequal sharing of electrons
-non-polar covalent bonds form from an equal sharing

Bonds:
-The type of bond formed can be predicted by looking at the difference in electronegativity of the elements
en>1.7 = ionic bond
en<1.7 = polar covalent bond
en=0 - non polar covalent bond
e.g.
Hydrogen - Oxygen
3.44-2.20= 1.24 (polar covalent bond)

Carbon - Hydrogen
2.55-2.20=0.35 (polar covalent bond)

Flourine - Potassium
3.98-0.82=3.16 (ionic bond)

Acid/Base Reactions!

Today in chemistry class we learned about pH and P0H, we also had visitors in class watching as we used our phones to text Mr. Doktor our answers for the in class problems.

Our 1st class problem was:


2000mL of 0.02 M NaOH represented our simulated lake and 150mL of 0.30 M HCl represented our simulated acid rain.

First we had to determine the pH of the acid rain and the pOH of the lake.

Determining the pH of the acid rain:

HCL -> H+ + Cl-

0.30 M x 1/1 = 0.30 M
-log(.30 = 0.52 = pH

Determining the pOH of the lake:

NaOH -> Na+ + OH-

0.02M x 1/1 = 0.02M
-log(0.02 = 1.7 = pOH

Determining the pH of the lake after acid rain has fallen on it:

HCl + NaOH -> NaCl + HOH

0.30mol/L x 0.150L = 0.045mol HCl

0.02mol/L x 2.00L = 0.04mol NaOH

0.045 - 0.040 = 0.005 mol

0.0050 x 1/2.15 = 0.002326M

-log(0.002326 = 2.6 = pH

Then we had to do another in class problem:

50.0mL of 0.150M H2SO4 reacts with 100mL of 0.100M Ba(OH)2



a) Determine the L.R.

H2SO4 -> 2H+ + SO4-2

0.150mol/L x 0.050L x 2/1 = 0.015mol of H, 0.015 x 1/2 = 0.0075mol of H2SO4

Ba(OH)2 -> Ba+ + 2OH-

0.100mol/L x 0.100L x 2/1 = 0.02mol of OH, 0.02 x 1/2 = 0.01mol of BaOH

H is L.R.


b) How much Ba(OH)2 is left after the reaction is completed

0.01 - 0.0075 = 0.0025 mol of BaOH is left

0.0025mol x 1/.150 = 0.017M


c) Determine the [Ba(OH)2] and [OH]

[Ba(OH)2] = 0.0025mol x 1/0.150 = 0.017M

[OH] = 0.017M x 2/1 = 0.034M


d) What is the pOH of the solution?

-log(0.034 = 1.48 = pOH


e) What mass of BaSO4 is produced?

0.0075mol x 1/1 x 233.4/1 = 17.5g of BaSO4

Ion Concentration

In this class we learned to determine the Ion concentration of compounds which is basically Dissociation.

-Ionic compounds are made up of 2 parts
-cation=positively charged particle
-anion=negatively charged particle
-When ionic compounds are dissolved in water the cation and anion separate from each other which is called dissociation
-When writing dissociation equations the atoms and charges must balance
-The dissociation of Sodium Chloride is:
NaCl (s) > Na+ (aq) + Cl- (aq)

-If the Volume does not change then the concentration of individual ions depends of the balanced coefficients in the dissociation equation.

Determine the ion concentration of a 0.56 M solution of Fe(OH)3
Fe(OH)3 > Fe3+ + 3OH-

0.56mol/L x 1/1= 0.56 M= FE3+
0.56mol/L x 3/1= 1.7M= OH-

Copper Chloride Lab

Today in class we had a lab where we had to create a solution of Copper Chloride with a concentration of O.1 M and then compare to samples Mr.Doktor set up. The solution that matched the solution you create (in terms of colour/shade) is the winner.

Materials:
Beaker
Graduated Cylinders
Stir Rod
Test Tube Stand
Water
Copper Chloride

Procedure:

Take about 5g of Copper Chloride and place it into a plastic dish
Take your plastic dish and pour the copper chloride onto a piece of wax paper
Take the wax paper and place it on your weighing scale
Weigh and record the total mass of the copper chloride
Put at least 25-50 mL of water in a graduated cylinder
Take an amount from the recorded amount of copper chloride with your scoopula
Put the same amount of copper chloride into the 25-50mL of water
Slowly stir the solution with a stir rod
Compare with Mr.Doktors solution and choose the closest shade of blue that matches yours
Repeat the procedure a few more times for consistent results

NECESSARY MATH:

0.050L/0.025L x (.1mol/L\) x (134.5g/mol) = ? g