Dilutions!

Dilutions are when two solutions are mixed making the conentration change. It is the process of decreasing the concentration by adding a solvent (usually water). Dilution can be expressed as the formula (n1 = n2) (C1V1 = C2V2.

To do a dilution problem, Mr. Doktor told us to always write down what we are given.

For Example:
"Determine the cocentration when 100mL of 0.10M Hcl is diluted to a final volume of 400mL"

Step 1: First write down all given data
Step 2: Manipulate the equation
Step 3: Plug in data to the variables
Step 4: Solve

V1 = 100mL V2=400mL C1=0.10M

(100)(0.1)/400 = 0.025M

Therefore the C2 is 0.025mol/L

<b> This video explains how to do indepth Dilution equation </b>

Titration of Vinegar Lab!

Today in Chemistry class, we did a lab to figure out the Molar Concentration of Acetic Acid in   household vinegar.

1. For the lab we set up a buret on a lab-stand with a clamp.

2. We filled the buret with 0.50M NaOH, and filled an Erlenmeyer flask with 10mL of vinegar and put two drops of Phenolphthalein Indicator in it. 

3. Then we recorded the initial volume of the NaOH in the buret, and then added the NaOH to the vinegar and swirled the flask until the solution turned light pink.

4. We then recorded the final volume to find the total amount of NaOH added in the trial.

   * We did this more than once.

Trial                                                1                           2

Initial Volume                              7.8                         24.2

 (mL)

Final Volume                               24.2                      40.8

 (mL)

NaOH Added                               17.0                    17.2

(mL)

 5. Then we got the average volume of NaOH added.

       17.0 + 17.2 / 2 = 17.1 mL

6. We finally determined the concentration of Acetic Acid using the balanced equation from the reaction of NaOH and CH3COOH.

NaOH + CH3COOH = H2O + NaCH3COOH

0.50mol/L x 17.1mL x 1/1 = 8.6mmol

8.3mmol x 1/10 mL = 0.86M

The accepted value for the concentration of household vinegar is 0.85M.

% error = 0.86 – 0.85 / 0.85 x 100 = 1% error.

For a better understanding of what we did today in class, here is a video demonstrating our lab.

Titration

In this class, We learned about titration which is an experimental technique used to determine the concentration of an unknwn solution.

In the Picture above:
-Buret= Contains the known solution to measure how much is added
-Stopcock= Valve used to control the flow of solution from the buret
-Pipet=used to accuratly measure the volume of unknown solution
-Erlenmeyer- container for unknown solution.
-Indicator- used to idetify the end point of the titration
-Stock Solution-known solution

e.g. Mike the mighty explorer completed a titration of 0.330M NaOH with 15.00 mL samgles of HI of unknown concetration. The data he gathered is below. Determine the concetration of HI.

0.330 mol/ L x 0.0112L x 1/1 x 1mol/ 0.015= 0.33M

Solution Stoichiometry

Solutions

Solutions are homogenous mixtures composed of a solute and a solvent
- Solute is the chemical present in lesser amount (whatever is dissolved)
- Solvent is the chemical present in greater amount (whatever does the dissolving)
Chemicals dissolved dissolved in water are aqueous
- NaCl(aq); H2SO4(aq)

Molarity

Concentration can be expressed in many different ways
- g/L, ml/L, % by volume, % by mass, mol/L
The most common (In Chemistry 11 & 12) is mol/L which is also called Molarity
- mol/L = M
- [HCl] = concentration of HCl

Molarity = Moles
Volume

Example:
- 100 mL of 0.250 M Iron (II) Chloride reacts with excess Copper. How many grams of Iron are produced?

FeCl2 + Cu ---> CuCl2 + Fe

0.250 mol/L x 0.100 L x 1/1 x 55.8g/mol = 1.40g

- How many moles of Copper Chloride are produced?

0.250mol/L x 0.100 L x 1/1 = 0.025 mol

- Determine [CuCl2]

0.0250 L x 1/0.100 L = 0.250 M

Limiting Reactants:

In today's class we learned how to determine the Limiting Reactant (L.R.)

- In chemical reactants, usually one chemical gets used up before the other
- the chemical used up first is called the limiting reactants
- one it is used up the reaction stops
- L.R. Determines the quantity of products formed
-To find the L.R. assume one reactant is used up. Determine how much of the reactant is required

e.g. 2.5 mol of CuSO4 reacts 2.5 mol of NaNO3. Determine the L.R. and calculate how many moles of NaSO4 are formed.

CuSO4+ 2NaNO3 > NaSO4 + Cu(NO3)2

2.5 mol x 2/1 = 5.0 mol NaNO3
soo the L.R> is NaNO3

2.5 mol x 1/2 = 1.25mol Na2SO4

Energy + Percent Yield

In today's class we learned how to calculate energy and also Percent Yield. Percent yield is basically the amount of products that should be formed in a reaction.

- Enthalpy is the energy stored in chemical bonds
- Symbol of Enthalpy is H
- units of Joules (J)
-Change in Enthalpy is ∆H
-In Exothermic reactions, enthalpy decreases
-In Endothermic reactions, enthalpy increases

Calorimetry:
-To experimentally determine the heat released we need to know 3 things:
1. Temperature change (∆T)
2. Mass (m)
3. Specific Heat Capacity (C)

They are all related by the equation:
∆H= mC∆T

e.g. Calculate the heat required to warm a cup of 700g of water (C= 4.18J/g°C) from 40.0° to 60.0°.
ΔH = mCΔT
= (700) (4.18) (20.0)
= 58 520 J / 1000
= 58.5 kJ


80.0 kJ of heat are added to a 700g glass of water initially at 40.0°C. Calculate the final temperature of the water (C= 4.18J/g°C).
ΔT = ΔH / Mc
= (80 000) / (700) (4.18)
= 27. 3°C
= 27. 3°C + 40.0°C
= 67.3 °C


Percent Yield
- The Theoretical yield of a reaction is the amount of products that should be formed
- Actual amount depends on the experiment
- The percent yield is like a measure of success
- How close is the actual amount to the predicted amount

Percent Yield = (Actual/ Theoretical) x100

e.g. Determine the percent yield for the reaction between 3.74 of Na and excess O² if 5.34 g of Na²O² is recovered

2Na + O2 > Na2O2

3.24 x 1mol/23.0g x 1/2 x 78.0/1mol = 6.34g

% yield = 5.34/6.34 = 84%