AP Chemistry - Chapter 4: Net Ionic Equations

AP Chemistry

Chapter 4: Aqueous Solutions

Chapter 4 Aqueous Solutions

Objectives:

Students will be able to predict the behavior of compounds placed in water using the solubility rules.
Students will be able to identify molecules as strong, weak, or non-electrolytes.
Students will be able to write and balance net ionic equations.
Students will be able to create solutions using of exact concentration.
Students will be able to change the concentration of a solution.
Students will be able to make stoichiometric calculations with Molarity.

Vocabulary:

Solubility

Solute

Solvent

Solution

Electrolytes

Strong, Weak, and Non-electrolyte

Precipitation Reactions

Neutralization Reactions

Ionic and Covalent compounds are different. Remember Ionic compounds contain a metal and a non-metal. While covalent compounds contain two non-metals.

Electrolytes: Strong, Weak, and Non

Strong, Weak, Non Electrolytes

Non-Electrolytes are solutes that do not conduct electricity. Theses are covalent compounds!

Strong Electrolytes are all ionics that completely break apart into individual ions.

Weak Electrolytes only partially break apart.

There is a list of ionics that complete break apart its known as the solubility rules.

In general we write Solid (s), Liquid (l) or Gas(g) if something does not break apart.

If it does break apart we call that substance aqueous.

Intro to Electrolytes.mp4

Net Ionic Equations.mp4

Net Ionic Equations

Check out this video on the basics of net ionic equations.

Then complete the worksheet on google classroom.

Intro to Conc.mp4

Introduction to Molarity

This video describes what concentration is, how to calculate it and how to make a solution.

Check Your Understanding:

On the three questions below, think about the question, then click on the down arrow when you have your answer to check to see if you are correct.

If an object begins at rest and experiences an acceleration of 2.5 m/s² at the end of the first second, it will be traveling at 2.5 m/s.

What concentration would a solution with 2.3moles of NaCl be if it were added to 4.1L of solution?

Molarity= moles/ Lsolution 2.3/4.1L=0.56M

2. What concentration would a solution of 33grams of NaCl being put into 400mL of solution be?

Convert grams to moles! 33grams/58.5g/mol=.564moles. Then divide by Liters of solution...400mL would be 0.4L, so you get .564moles/.4L=1.41M solution.

3. If you had a 6.0M solution how much would you require to make a 45mL of 2.5M solution?

M1V1=M2V2. 6M (V1)= 2.5M (45mL). V1=18.75mL This means you would take 18.75 mL of the original 6M solution and then add 26.25mL of water to make 45mL

You have finished Lesson 1!

Be sure to head over to google classroom and fill out the exit pass.

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