Dilutions!

On this day, we learned all about dilutions. Mr. Doktor took a "six month old pop bottle" and poured 50mL of Coke into a beaker. He then added 100mL of water to the beaker, and asked the class, "How much pop is there in the beaker now?" Some of us were quite surprised to learn that there was actually still 50mL of Coca-Cola in the beaker.

Dilution is the action of making a liquid more dilute; when we add a solvent (usually water), the concentration decreases. In fact, the amount of solute does not change, as Mr. Doktor demonstrated with the pop. So, we can conclude that when two solutions are mixed, the concentration changes.

Also, recall that concentration (M) is actually mol/L. So, we can write C = n/V and n = CV. Therefore, C₁V₁ = C₂V₂. Let's say that we wanted to cut a concentration of a solution in half (eg. 1.00 M glucose solution). If we mix 1.0 L of this solution with 1.0 L of distilled water, the volume is doubled to 2.0 L. The concentration is now cut in half to 0.50 M.

 

In the above diagram, the solutions become less and less concentrated (moving from the left to the right). Next class, we will need to determine which one of these has a specific concentration, without touching any of the test tubes.

EXAMPLES:

1. What is the final concentration if 0.045L of a 6.00M HCL solution are diluted to a final volume of 0.250L?

(C1)(V1) = (C2)(V2)

(6.00)(0.045) = (C2)(0.250)

C2 = [(6.00)(0.045)] / (0.250)

C2 = 1.08M

2. What volume of water should be added to 400mL of a 0.1M NaOH solution in order to change the concentration to 0.075M NaOH?

(C1)(V1) = (C2)(V2)

(0.1)(0.4) = (0.075)(V2)

V2 = [(0.1)(0.4)] / (0.075)

V2 = 0.5333 or 0.533L or 533mL

Vchange = Vf - Vi

Vchange = 533 - 400

Vchange = 133mL

3. How much water must be evaporated from 3.25L of 0.254M KCl solution for the final concentration to be 2.82M?

(C1)(V1) = (C2)(V2)

(0.254)(3.25) = (2.82)(V2)

V2 = [(0.254)(3.25)] / 2.82

V2 = 0.2927 or 0.293L

Vchange = Vf - Vi

Vchange = (0.293) - (3.25)

Vchange = -2.957 or -2.96L

4. If 200mL of 0.230M HCl is mixed with 300mL of 0.250M HCl, what concentration does this give?

0.230mol/L x 0.200L = 0.046mol

0.250mol/L x 0.300L = 0.075mol

0.046 + 0.075 = 0.121mol

0.200 + 0.300 = 0.500L

0.121mol / 0.500L = 0.242mol/L

5. What mass of HCl is needed to make 84.0mL of a 0.143M solution?

0.143mol/L x 0.084L x 36.5g/mol = 0.4384 or 0.44g

As usual, here's a video for more information on today's lesson:

 
Next time: Diluting & Creating Solutions!

Titration of Vinegar Lab!

In this lab we were trying to figure out the Molar Concentration of Acetic Acid in household vinegar.

For the lab we set up a buret on a lab-stand with a clamp. Then we filled the buret with 0.50M NaOH, and filled an Erlenmeyer flask with 10mL of vinegar and added two drops of Phenolphthalein Indicator to it. After we recorded the initial volume of the NaOH in the buret, we added NaOH to the vinegar and swirled the flask until the solution turned light pink. We then recorded the final volume to find the total amount of NaOH added in the trial. We repeated this step at least once more.





Next we had to determine the average volume of NaOH added.

16.9 + 17.1 = 34.0

34.0/2 = 17.0

Average volume = 17.0mL

Then we had to determine the concentration of Acetic Acid using the balanced eqaution for the reaction of Sodium Hydroxide with Acetic Acid.

CH3COOH + NaOH -> HOH + NaCH3COO

0.50mol/L x 17.0mL x 1/1 = 8.5mmol

8.5mmol x 1/10mL = 0.85M

The accepted vaule for the concentration of household vinegar is 0.85M

Calculating our percent error we get:

0.85 - 0.85 / 0.85 x 100 = 0% error

Here is a video to show you visually what we did for this lab!

Titration!

In today's class we were entertained not only by Mr.Holowka with the projector, but we also got to see an example of Titration done by Mr.Doktor!!

To see Images of the Titration example Mr.Doktor did in class today go to the links below.

http://imgur.com/GeomZ

http://imgur.com/IzM70

Here are some of the tools that Mr.Doktor used today in class to do the titration process:

These are Burets.

This is an Elenmeyer Flask.

Make sure you know the definition of a Meniscus and make sure you use it during a titration experiment.

Meniscus- the curved upper surface of a liquid in a tube/container.

Example:

John completed a titration of 0.471 M NaOH with 18ml samples of HI of unkown concentration. Here is the data he gathered. Determine the concentration of HI.

Trial 1 2 3 4

Final(ml) 11.9 22.8 34.2 43.1

Initial(ml) 0.7 11.9 22.8 34.2

Volume 11.2 10.9 11.4 8.9(do not include this one)

reading

Find the average of the Volume readings which in this case is 11.6ml

Write the balanced equation. NaOH + HI -----> NaI + HOH

Then Calculate.

0.471 Mol/L x 11.6ml x 1/1= 5.4636mmol

5.4636mmol/18ml =0.3035M=0.304M

Here is a diagram of a Titration.

Here is a very detailed video of the experiment and calculations:

SS: More Examples!

Determine the number of moles of H₂O produced when 0.250 L of 0.100 M NaOH is neutralized by H₂SO₄.

Determine the mass of Fe(OH)₂ if 50 mL of 0.250 M NaOH reacts with Fe.

75.0 mL of 0.250 M CaCO₃ is completely decomposed. Determine the volume of O₂ @STP.

2.5 g of Sn is completely reacted with 100 mL of HCL. Determine [SnCl₄].

Solution Stoichiometry!

On this day, we started our fifth unit: SOLUTION STOICHIOMETRY. What is solution stoichiometry you ask? This topic deals with quantities in chemical reactions taking place in solutions.

Solutions are homogeneous mixtures composed of a solute and a solvent. Remember that a solute is the chemical present in the lesser amount; it is whatever is dissolved. The solvent is the chemical present in the greater amount; it is whatever DOES the dissolving. Also recall that chemicals dissolved in water are called aqueous (NaCl_(aq);H₂SO_4(aq)). Such reactions look like so: NaCl_(s)-->NaCl_(aq). Note that Sodium chloride does not actually REACT with water.

CONCENTRATION
How do you explain the concentration of a solution? Well, the concentration of a solution is the amount of solute in a given amount of solvent or solution. A concentrated solution contains a large amount of solute for the amount of solvent. This concentrated solution can be diluted by adding more solvent - solutions that have a relatively small amount of solute for the amount of solvent are called dilute.

Is there a limit to how concentrated a solution can be? The answer is yes, there is, because if you add too much sugar to water, some will sink to the bottom of the cup. A solution is called saturated when no more solute can be dissolved in it, that is, when it contains as much solute as can possible be dissolved under the existing conditions of temperature and pressure. An unsaturated solution is a solution that has less than the maximum amount of solute that can be dissolved. A supersaturated solution is a solution that contains a greater amount of solute than that needed to form a saturated solution; these are not stable solutions because, in some cases, some of the solute particles can escape from the solution and reform the pure solute.

Concentration can be expressed in many different ways: g/L, mL/L, % by volume, % by mass, and mol/L. However, the most common is mol/L, which is also called molarity (M). To determine molarity, divide the number of moles of solute by the number of liters of solution. MOLARITY = MOLES / VOLUME Molality is the number of moles of solute dissolved in each kilogram of solvent.Mole Fractions are the number of moles of one component divided by the total number of moles in the solution.

Molarity was the focus of today's lesson. Note that the capital -m, M, means mol/L. Also, square brackets, [], mean "concentration of." For example, [BaCl₂] = 3.2M, or the concentration of Barium chloride is 3.2mol/L.

To explain the effect of concentration, Mr. Doktor mixed two different solution of calcium carbonate (chalk) and hydrochloric acid. One reacted faster than the other, because of a higher concentration.

EXAMPLES

1. A solution contains 0.45 moles of NaCl in 4.2L. What is the molarity of the solution?
   0.45mol/4.2L = 0.1071 or 0.11M
   
   
2. 5.6g of NaCl is dissolved in enough water to make 170mL of solution. What is the molarity of the solution?
   5.6g x 1mol/58.5g x 1/0.170L = 0.5631 or 0.56M
   
   
3. Determine [NaCl] if 6.62g of NaCl is dissolved in 100mL of water.
   6.62g x 1mol/58.5g x 1/0.100L = 1.1316 or 1.13mol/L
   
   
4. To produce 32.0g of NaCl, what volume of 14.2M NaCl would be needed?
   1L/14.2mol x 1mol/58.5g x 32.0g = 0.03852L or 0.0385L

Next time: Molarity & Stoichiometry!

TEST: STOICHIOMETRY

We had our unit four test today. All went well.

TOPICS COVERED:

-Six Types of Reactions

-Mole to Mole Problems

-Mole to Mass & Mass to Mole Problems

-Mass to Mass Problems

-Other Types of Problems

-Percent Yield & Energy in Stoichiometry

-Limiting & Excess Reactants

Review Class

Last class we went over a few examples for Limiting Reactants and reviewed the rest of the unit in preparation for the test on Friday! Good Luck to all!

Limiting Reactants!

In this class we learned about limiting reactants, which is defined as the chemical that determines how far the reaction will go before the substance that you are focusing on is used up.

Most commonly in chemical reactions, one chemical ends up getting completely used up before the other one. This chemical that gets used up is called the "Limiting Reactant." This Chemical determines the quantity of the product formed. To find the Limiting Reactant you, must assume that one of the reactants is used up, so that you can determine how much of the reactant is used. In a Chemical reaction , the reactants that are left over after the reaction has taken place are called the "Excess Reactants."

 

In class Mr.Doktor gave us the example of making sandwiches. Lets say that we are making 10 sandwiches containing one slice of meat, two buns, one piece of lettuce, and one slice of cheese to make one sandwich. So if you plan on making ten sandwiches, and you have 10 slices of meat, 11 pieces of cheese, 20 buns, and nine pieces of lettuce, you will run out of lettuce before you finish, so the lettuce will be the limiting reactant. And, because we have eleven pieces of cheese, the cheese will be an Excess Reactant, since we will only use 10 pieces of cheese.

Examples:

Solving for Limiting Reactants:If a reaction such as 2NaOH + CO2 -----> Na2CO3 + H20 were used in an experiment containing 1.70g of NaOH and 1.00g of CO2, then what would be the limiting reactant? The first step you would take would be to make sure that everything is in moles if not then convert into moles. This must be done because reactants react in a mole ratio, which is given in the chemical equation, you can't use mass to find the limiting reactant.NaOH's molar mass is [23.0g/mole Na + 16.0g/mole O +1.01g/mole H] = 40.0g/mole NaOH

CO2's molar mass is [12.0g/mole + (16.0g/mole x 2)] = 44.0g/mole of CO2

Here are two YouTube videos explaining what we learned in class today!

Energy and Percent Yield!

Enthalpy is the energy stored in chemical bonds. The symbol representing empalthy is H and change in empalthy is ΔH. Empalthy increases in endothermic reactions and decreases in exothermic reactions.
Exothermic Reaction



Endothermic Reaction



Calorimetry

To experimentally determine the heat released we need to know 3 things

1. Temperature change (ΔT)
2. Mass (m)
3. Specific Heat Capacity (C)

These are related by the question:
ΔH = mCΔT

Example:

Calculate the heat required to warm a cup of 800g of water (C= 8.36J/g°C) from 30.0° to 80.0°.

ΔH = mCΔT
= (800)(8.36)(50.0)
= 334 400J
= 3.34 x 10^5

Percent Yield

The theoretical yield of a reaction is the amount of products that should be formed. The actual amount depends on the experiment. So to get percent yield you have to put actual amount over the theoretical amount, or, put what you got over what you should have got.

Example:

1.98/2.00 = 99%

The percent yield is like a measure of success. It's how close the actual amount is to the predicted amount.

Example:

Find out the percent yield for the reaction between 4.78g of K and excess Se if 9.99g of K2Se is recovered

2K + Se -> K2Se

4.78g x 1mol/39.1 x 1/2 x 157.2/1mol = 9.61

9.99/9.61 = 104%

Lastly, here is a helpful video on determining percent yield!