Bonding

Bonding and Electronegativity

There are three main types of bonds , Ionic, Covalent, and Mettalic bonds. Ionic bonds are bonds between metals and non-metals. Covalent bonds are bonds between non-metals. Metallic bonds are bonds between pure metals.

Electronegativity

Electronegativity (en) is a measure of an atom's attraction for electrons in a bond. Atoms with greater en attract more e⁻. Polar covalent bonds form from an unequal sharing of e⁻. Non-polar covalent bonds form from an equal sharing or e⁻.

Bonds

The type of bond formed can be predicted by looking at the difference in electronegativity (en) of the elements.

Predict the Type of Bond Formed

Identify the positiv and negative side of the polar bonds:

Myth Buster Break

As it was the last class before Spring Break, we watched Myth Busters: GRENADES AND GUTS. We also watched the James Bond episode. That was fun!

TEST: SOLUTION STOICHIOMETRY

Today we had our test on Solution Stoichiometry. All went well.

We look forward to starting our next unit: Bonding!

Review Class

Yup. Review Class. You know what that means; we have a test next class on solution stoichiometry. Good luck to all!

Make sure that you study hard and review the following (refer to the links on the right):

1. Molarity
2. Molarity & Stoichiometry
3. Titrations
4. Dilutions
5. Ion Concentration
6. Mixing Acids & Bases

Good luck to all!

Acid/Base Reactions!

This class what the pH of our simulated lake would be after simulated acid rain was added.

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) hiow 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

That concluded the lab!

Ion Concentration!

Dissociation!

Ionic compounds consist of two parts:

-Cation: a positively charged particle

-Anion: a negatively charged particle

When dissolved in water, ionic compounds separate into the Cation and Anion.

-This process is known as "Dissociation."

When writing dissociation equations the atoms and the charges must balance.

Example

Na3PO4(s)--> Fe 2+(aq) + Cl-(aq)

In some cases the volume doesn't change, in this case the concentration of the individual ions then depend on the balanced coefficients in the dissociation equation.

If [Cl-] = 0.365M in a solution of AgCl what is the solution's concentration?

AgCl--> Ag+ + Cl-

0.365 x 1/1= 0.365M

A 250ml solution of 0.760M Pbcl2 is added to 340ml solution of 0.178M NaOH. Determine the final [Cl-].

Pbcl2 + NaOh--> Pb2+ + 2Cl- + Na+ + OH-

0.760 x 0.250 ml= 0.19 mol of PbCl2

0.19 mol x 2/1= 0.38

0.38/0.590= 0.64M

Solutions and Creating Dilutions!

In today's class we did a lab. In this lab we had to find out which of the 5 test tubes with Copper (II) Chloride in it had a concentration of 2.0M.

The procedure we used was to first find out the molar mass of Copper (II) Chloride, then using the mass we found, we divide it by the concentration given, to find the amount of water needed to dilute the Copper (II) Chloride into.

Using these calculations, we diluted 3.0g of Copper (II) Chloride into 111.5 mL of water, giving us a concentration of 0.2M.