ION Concentration

Ionic compounds are made up of two parts
-Cation: positively charged particle
-Anion: negatively charged particle

when ionic compounds are dissolved in water the cation and anion separate from each other, this is known as dissociation.
when writing dissociation equations the charges and atoms must balance.
 Example: The dissociation of potassium sulfide is:  K2S —> K+ + S2-
 K2S —> 2K+ + S2-

If two solutions are added together, then the volume changes. The ions are diluted.

The equation is C1V1 = C2V2

Acid Base Reactions

 We learned about acid base reactions with strong acids and strong bases.

-Strong acids (SA) dissociate to produce H+ ions 

-Strong bases (SB) dissociate to produce OH- ions

-When a SA and SB are reacted with each other they form water and an ionic salt

-acid-base reactions always produce salt and water

NaOH(aq) + HCl(aq) —> NaCl(aq) + H₂O 

PH and pOH
-PH is a measure of the hydrogen ions present in a solution.
PH= -log[H+]
-pOH is a measure of the hydroxide ions present in a solution.
pOH= -log[H-]

Example:
150mL of .300M HCl is added to 250mL of .200M LiOH
Balanced equation: HCl + LIOH -> HOH + LiCl
                                                                                            
-Determine the L.R
.300mol x .150L = .0450 mol given
     L
.200mol x .250L x 1 = .0450 needed
    L                       1
L.R is HCl
                                                                                           
How much LiOH is left over after the reaction is complete.
.250L x .200mol = .0500 mol LiOH started with
                  L
.0450mol x 1 = .0450mol used
  (^of HCl given)1
(.0500) - (.0450) = .0050 mol of LiOH left over
                                                                                          
-Determine [LiOH] after the reaction is complete.
.0050mol x   1   = .0125M
                 .400L
[LiOH]= .0125M
                                                                                         

-Find the pOH of the final solution.
     LiOH -> Li+ + Li-
remember pOH = -log[OH-]
    so         pOH= -log(.0125) 
                                                                                           

Solution Stoichiometry

Solutions are homogeneous mixtures composed of a solute and a solvent. A solute is the chemical present in lesser amount. A solvent is the chemical present in greater amount. Chemicals that are dissolved in water are aqueous. 

Concentration can be expressed in different ways, such as g/L; mL/L; % by volume, % by mass; mol/L. The most common is mol/L which is called molarity. Another way of saying mol/L is by replacing with a capital 'M'. Another thing to note is when square brackets are involved [HCl], this is another symbol for concentration.  

Example:

• What is the concentration of sodium chloride made from 0.75mol of NaCl dissolved in 250mL of H2O?

0.75mol x     1   = 3.0mol/L = 3.0M

                 0.250L

• 25.0g of NaOH is added to 350mL of H2O. Determine [NaOH]

25.0g x 1 mol = 0.625mol = 1.79mol/L

             40.0g     0.350L

Stoichiometry - Limiting Reactants

In chemical reactions, usually one chemical is used up before the other. The chemical is known as the limiting reactant (L.R.). Once the limiting reactant is used up the reaction stops. The Limiting Reactant determines the quantity of products form+ed. To find the L.R. assume one reactant is used up. Then determine how much of this reactant is required.

Example:

1. Determine the L.R. when 1.4 of H2 reacts with 0.89mol of O2

• 2H2 + O2 -> 2H2O

1.4mol x 1 = 0.70mol of O2 needed
              1

• H2 is L.R.

                    

Stoichiometry - Percent Yield

Theoretical yield of a reaction is the amount of products that should be formed. Percent yield is like a measure of success. The equation for percent yield is:

% yield = experimental x 100
                  theoretical
Example:

• The production of Urea CO(HN2)2 is given by the equation 
• 2 NH3 + CO2 -> CO(NH2)2 + H2O

1. If 47.7g of Urea are produced, determine the theoretical (predicted) yield of CO2
2. What is the % yield of CO2 if 12.0g is produced?

1) 47.7g x 1 mol x 1 x 44,0g = 35g
                  60g     1    1 mol

2) 12g x 100 = 34%
    35g

Stoichiometry - Energy and Percent Yield

Enthalpy is the energy stored in chemical bonds. The Symbol of Enthalpy is H and the units is in Joules (J). In exothermic reactions, enthalpy decreases. In endothermic, the enthalpy increases.


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

1. Temperature Change (ΔH)
2. Mass (m)
3. Specific Heat capacity (c)

These 3 are related by the equation

• ΔH = mCΔT
• ΔH = mC(Tf-Ti)

Stoichiometry - Other Conversions (Volume and Heat)

Heat can be included as a separate term in chemical reactions (also known as Enthalpy). Reactions that release heat are called exothermic reactions and reactions that absorb heat are called endothermic reactions.
Examples:

1. Volume: In the formation of Copper (II) Oxide 3.5g of Copper reacts. How many liters of Oxygen (@STP) are needed?
2. Heat: Find the amount of heat released when 5.0mol of H2 are consumed according to the reaction:

Stoichiometry - Mass to Mass

This conversion is almost exactly the same as the other conversions, except that we add an additional step. We go from Mass -> Moles -> Mole Ratio -> Mass. REMEMBER 2 RULES: 1. Balanced equation and 2. What you need over what you have.

Examples:

1. How many grams of Oxygen are produced from the decomposition of 3.0g of Potassium chlorate?
2. Solid Zinc reacts with Hydrochloric Acid. What mass of Hydrogen is produced when 2.5g of Zinc react?

Stoichiometry - Mole to Mass (and vice versa)

Some questions in Stoichiometry will give you an amount of moles and will ask you to determine the mass. When converting moles to mass, 1 extra step is required, which is using the Molar Mass of that element. REMEMBER 2 RULES) 1. Balanced equation and 2. What you need over what you have.

Examples:

1. How many grams of Al2O3 are required to produce 3.5 mol of pure aluminum?
2. How many grams of water are produced if 0.84mol of Phosphoric Acid is neutralized by Barium Hydroxide?

* the numbers circled in red is the Molar Mass. signifying the extra step needed in order to complete the conversion

...Dilution

When two Solutions are mixed, the concentration changes.
To dilute is the process of decreasing the concentration by adding a solvent (usually water)
*The amount of solute does not change
Because concentration is mol/Lwe can write:
C= n/v or n=(c)(v) meaning (C)(V)=(C2)(V2)
C= Concentration (M)
V= Volume (L)

Example:
Determine the concentration when 200mL of 0.13M H Cl is diluted to a final volume of 300mL
CV=C2V2
(.13M)(.200L)=C2(.300L)
C2(concentration)=0.087M

Stoichiometry - Mole to Mole Conversion

Coefficients in balanced equations tells us the number of moles reacted or produced. They can also be used as conversion factors. With mole conversions, remember: WHAT YOU NEED OVER WHAT YOU HAVE.
Examples:

1. If 0.15mol of methane are consumed in a combustion reaction, how many moles of CO2 are produced? (refer to picture for answer)
2. How many moles of bauxite (Aluminum Oxide) are required to produce 1.8 moles of pure Aluminum?

Stoichiometry - Quantitative Chemistry

Stoichiometry is a branch of chemistry that deals with the quantitative analysis of chemical reactions. It is a generalization of mole conversions to chemical reactions. Understanding the 6 types of chemical reactions is the foundation of stoichiometry.
There are 6 types of reactions:

1. Synthesis (formation)
2. Decomposition
3. Single Replacement
4. Double Replacement
5. Neutralization
6. Combustion

1. Synthesis
Synthesis is the combination of 2 elements (a cation and an anion) together to make a single element
Examples:

1. 2H2 + O2 = 2H2O
2. 4K + O2 = 2K2O

2. Decomposition 
Decomposition is generally the opposite of Synthesis. Remember, always assume the compounds decompose into elements during decomposition
Examples:

1. 4H3PO4 -> 6H2 + P4 + 8O2
2. 2C12H22O11 -> 24C + 22H2 + 11O2

3. Single Replacement (A + BC -> B + AC)
Example:

1. Ca + 2KCl -> 2K + CaCl2   

4. Double Replacement (AB + CD -> AD + BC)
Examples:

1. MgCl2 + K2SO4 -> MgSO4 + 2KCl
2. AgNO3 + NaCl -> AgCl + NaNO3

5. Neutralization
Neutralization is a reaction between an acid and a base
Example: 

1. H2SO4 + 2KOH -> 2HOH + K2SO4

6. Combustion
Combustion is a reaction with something (usually hydrocarbon) with air. Hydrocarbon combustion always produces CO2 and H2O
Example:

1. CH4 + 2O2 -> CO2 + 2H2O 

 

...Titrations

Today we learned how to use the experimental technique Titration to determine the concentration of an unknown solution.

Equipment:
Burrette- contains the known solution. used to measure how much is added.
Stopcock- Valve used to control the flow of solution from the Burrette.
Pipet-used to accurately  measure the volume of the unknown solution.
Erlenmeyer Flask- Container for the unknown solution
Indicator- used to identify the end point of the titration
stock solution- known solution (what your given the concentration of)

Stopcock

Pipet

Erlenmeyer Flask

burrette