Titration

A titration is an experimental technique used to determine the concentration of an unknown solution.

Example 1.

Mike the Mighty Chemist wants to determine the concentration of a Sodium Hydroxide sample so he does a titration with HCl. He gathers the following data. Use this information to determine [NaOH]

NaOH Samples = 10.00mL     [HCl] = 0.75M

Trial                                 1        2           3         4
Final Reading (mL)       13.3       26.0      38.8    13.4
Initiating Reading (mL)  0.2         13.3      26.0    0.60
Volume Used (mL)       13.1       12.7     12.8      12.8

So you're first step would to make a balanced equation. (Of course)

Our balanced equation would be...

NaOH + HCl  -->   HOH   +  NaCl

Since the formula is already balanced, would wouldn't have to put any numbers in front of the compounds.

We would have to gather the average between all of the Volume used, but you have to use the ones that are near each other. (12.7, 12.8 and 12.8) because 13.1 is a bit far off from those numbers. Since you know the average of the Volume which is, 12.85mL, you would want to convert that into L because you can't do anything when your volume is in mL while your equation is in L.

0.750mol/L x 0.01285L = 0.00964mol of HCl

Now you would want to use the ratio of WHAT YOU NEED OVER WHAT YOU HAVE. 


0.00964mol of HCl x 1 NaOH/ 1HCl = 0.00964mol of NaOH

Now, do you remember the NaOH samples that's in mL? Well, now you would want to convert that into L too so that you can finish your equation up.

So since it's 10.00 mL it would become 0.0100L. Now you can finish your equation!

0.00964mol x 1 / 0.0100L = 0.96mol/L

-Meldrick Mendoza

LIMITING REACTANTS

-In chemical reactions,usually one chemical gets used up before the other.
     -The chemical used up first is called the limiting reactant
     -Once 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

EXAMPLE
Determine the L.R when 1.4mol of hydrogen reacts with 0.89 mol of oxygen
2H2 + O2 -> 2H2O

1.4mol x 1 = 0.70mol of oxygen
              2
There is 0.70mol of oxygen used up out of the 1.4mol which makes hydrogen the L.R

OR

0.89mol x 2 = 1.78mol of H2
                1

1.78mol of hyrdrogen are needed when there is only 1.4mol, which makes hydrogen the L.R

HELPFUL LINKS
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm http://www.mikeblaber.org/oldwine/chm1045/notes/Stoich/Limiting/Stoich07.htm
http://www.youtube.com/watch?v=rESzyhPOJ7I

K.P

Mole to Mass and Mass to Moles

- Some questions will give you an amount of moles and ask you to determine the mass
- Converting moles to mass only requires one extra step

Example
How many grams of Bauxite (Al2O3) are required to produce 3.5 mol of pure Aluminum?

2Al2O3------> 4Al + 3O2


3.5 mol x 2/4mol =1.75 mol x 102g/1mol
                           =178.5g----> 1.8 x 10^2 g

How many grams of silver nitrate are needed to produce 1.02 mol of silver chloride according to the following unbalanced equation: 2AgNO3 + BaCl-----> Ba(NO3)2 + 2AgCl

1.02 mol x 2/2 =1.02g x 169.9g/1mol = 173.3g
                                                          = 173g

MASS TO MASS CONVERSIONS

-Mass to Mass problems involve one additional conversion
Grams of A ->Moles of A-->Chemical equation/multiply by the stoichiometric ratio-->Moles of B-->grams of B
Ex. Lead (IV) nitrate reacts with 5.0g of potassium iodide. How many grams of lead (IV) nitrate are require for a complete reaction?

Pb(NO3)4+4KI -> 4KNO3+ PbI4
5.0g x 1mol x 1 x 455.2 =3.4g
           166g   4    1mol

More Examples
http://www.800mainstreet.com/6/0006-004-Mole-Mass-Con.htm
http://www.youtube.com/watch?v=iF3C5bJ2UPM

K.P

Energy and Percent Yield

-Enthalpy is the energy stored in chemical bonds.
- Symbol of Enthalpy is H.
- Units of Joule(s).

-Change in Enthalpy is /\ H.
- In exothermic rxns enthalpy decreases.
- In endothermic rxns enthalpy increases.

EXOTHERMIC REACTION


Potential energy (KJ)

|               Reactants
|.............................
|             |                \
|             |                  \
|             |  /\H             \                      Products
|            \/                     \..............................
|
|_________________________________



ENDOTHERMIC REACTION

Potential Energy (KJ)

|                                          Products
| Heat is positive  ..............................                                      
|  /\ H is positive /                    
|                       /
|  Reactants    /             
|.................../
|_______________________________


Calorimetry

-To expirimentally determine the heat released we need to know three things

1. Temperature Change (/\T) Celcius
2. Mass (m) g
3. Specific Heat Capacity (C) J/ g Celcius

These are ralted by the equation: /\ H = mc /\ T
                                                               or
    
                                                /\ H = m C ( Tf - Ti)


Example:

Calculate the heat required to warm up a cup of 400g of water (C = 418 J/ g Celcius)
from 20.0 Celcius to 50.0 Celcius



/\H = mc/\T

= 400g x 30.0 Celcius x 418 J/ g Celcius

/\H = 50160

(sig figs) /\H = 5.02 x 10^4 J

Jomar Delos Santos