Dissociation
-Ionic compounds are made up of 2 parts:
→Cation: positively charged particle
→Anion: negatively charged particle
-When ionic compounds are dissolved in water the cation & anion separate from eachother
-This process is called dissociation
-When writing dissociation equations, the atoms & charges must balance.
The dissociation of Sodium Chloride is:
NaCl(s) → Na+(aq) = Cl-(aq)
*have to write charges*
-Paulo Santillan
Wednesday, 14 March 2012
Dilutions
Diluting Solutions
-When 2 solutions are mixed the concentration changes
-Dilution is the process of the decreasing the concentration by adding a solvent (usually water)
-The amount of solute does not change
n1 = n2
C1V2 = C2V2
-Because concentration is mol/L we can write:
C = n/v
&
n = CV → mol/L x L = mol
so
C1V1 = C2V2
-EXAMPLE
-Determine the concentration when 100ml of 0.10M of HCL is diluted to a final volume of 400ml.
V2= 400mL
V1= 100mL
C1= 0.10M
C2= ?
C1V1 = C2V2
C1V1 / V2 = C2
(100)(0.1) / (400) = C2
C2 = 0.025M
-Paulo Santillan
-When 2 solutions are mixed the concentration changes
-Dilution is the process of the decreasing the concentration by adding a solvent (usually water)
-The amount of solute does not change
n1 = n2
C1V2 = C2V2
-Because concentration is mol/L we can write:
C = n/v
&
n = CV → mol/L x L = mol
so
C1V1 = C2V2
-EXAMPLE
-Determine the concentration when 100ml of 0.10M of HCL is diluted to a final volume of 400ml.
V2= 400mL
V1= 100mL
C1= 0.10M
C2= ?
C1V1 = C2V2
C1V1 / V2 = C2
(100)(0.1) / (400) = C2
C2 = 0.025M
-Paulo Santillan
Titrations
-A titration is an experimental technique used to determine the concentration of an unknown solution
Terms and Equipment
-Buret: contains the known solution. Used to measure how much is added
-Stopcock: Valve used to control the flow of the solution from the Buret
-Pipet: Used to accurately measure the volume of the unknown solution
-Erlenmeyer Flask: Container for unknown solution
-Indicator: Used to identify the end point of the titration
-Stock Solution: Known solution
-Paulo Santillan
Terms and Equipment
-Buret: contains the known solution. Used to measure how much is added
-Stopcock: Valve used to control the flow of the solution from the Buret
-Pipet: Used to accurately measure the volume of the unknown solution
-Erlenmeyer Flask: Container for unknown solution
-Indicator: Used to identify the end point of the titration
-Stock Solution: Known solution
-Paulo Santillan
Solution Stoichiometry
Solutions
- Solutions are homogenous mixtures composed of a solute and a solvent
- solute is the chemical present in the lesser amount (whatever is dissolved)
- solvent is the chemical present in the greater amount (whatever is dissolving)
- Chemicals dissolved in water are aqueous
NaCl(s) --H20--> NaCl(aq) : to show something dissolved by water
Molarity
- Concentration can be expressed in many different ways
- g/L, mL/L, % by volume, % by mass, mol/L
- The most common ( in Chem 11/12) is mol/L which is also called Molarity
- mol/L = M
- [HCl] = concentration of HCl
- JanCarlo Paysan
- Solutions are homogenous mixtures composed of a solute and a solvent
- solute is the chemical present in the lesser amount (whatever is dissolved)
- solvent is the chemical present in the greater amount (whatever is dissolving)
- Chemicals dissolved in water are aqueous
NaCl(s) --H20--> NaCl(aq) : to show something dissolved by water
Molarity
- Concentration can be expressed in many different ways
- g/L, mL/L, % by volume, % by mass, mol/L
- The most common ( in Chem 11/12) is mol/L which is also called Molarity
- mol/L = M
- [HCl] = concentration of HCl
- JanCarlo Paysan
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. Determine how much of this reactant is required
ex. 2.5 mol of CuSO4 reacts with 2.5 mol of NaNO3. Determine the L.R. and calculate how many moles of Na2SO4 are formed
CuSO4 + 2NaNO3 -> Na2SO4 + Cu(NO3)2
2.5 mol * 2/1 = 5 mol of NaNO3 needed
2.5 mol * 1/2 = 1.25 mol of CuSO4 needed
- NaNO3 is the L.R.
- 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. Determine how much of this reactant is required
ex. 2.5 mol of CuSO4 reacts with 2.5 mol of NaNO3. Determine the L.R. and calculate how many moles of Na2SO4 are formed
CuSO4 + 2NaNO3 -> Na2SO4 + Cu(NO3)2
2.5 mol * 2/1 = 5 mol of NaNO3 needed
2.5 mol * 1/2 = 1.25 mol of CuSO4 needed
- NaNO3 is the L.R.
- JanCarlo Paysan
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