Chapter 4 Types of chemical reactions and Solution Stoichiometry
Solution Stoichiometry and Types of Reactions...4.4 Types of Chemical Reaactions 4.5 Precipitation...
Transcript of Solution Stoichiometry and Types of Reactions...4.4 Types of Chemical Reaactions 4.5 Precipitation...
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Solution Stoichiometry and Types of Reactions
Chapter 4
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C6H5Cl + C2HOCl3 → C14H9Cl5 + H2O
1) Balance the equation
2) How many grams of water will be produced from the reaction of 45.0 g of C6H5Cl and 40.0 g of
C2HOCl3?
2, 1, 1, 170.9g
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4.3 Stoichometric Analysis of Solutions
4.4 Types of Chemical Reaactions
4.5 Precipitation Reactions
4.6 Acid-Base Reactions
4.7 Oxidation-Reduction Reactions
4.8 Fresh Water-Issues of Quantitative Chemistry
4.2 The Concentration of Solutions
4.1 Water-A Most Versatile Solvent
Solution Stoichiometry and Types of Reactions
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4.1 Water – A Most Versatile Solvent
• A soluteconsists of atoms, molecules or ions that dissolve in
the solvent.• A solvent
is the substance that dissolves the solute.• A solution
is a homogenous mixture made by dissolving a solute in a solvent.
• If a solute is dissolved in water, it forms an aqueous solution
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Solution
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Hydration Process
• Water molecules surround the ions.
• Note the positions of the positive and negative charges.
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Electrolytes
• Strong Electrolytes– Good conductors of electricity in solution.– Strong acids, strong bases, soluble ionic compounds.– Completely ionized in solution.
• Weak Electrolytes– Poor conductors of electricity in solution.– Partially ionized in solution. CH3COOH, NH3
• Non-electrolytes– Do not conduct electricity in solution.– Do not form ions in solution.
: C6H12O6,
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Electrolytes
• A strong electrolyte is completely dissociated in water.KCl(s) KCl(aq) K+(aq) + Cl−(aq)
• A weak electrolyte is partiallydissociated in water.NH3(aq) + H2O(l) NH4
+(aq) + OH−(aq)
• Non-electrolytes do not form ions.C12H22O11(s) C12H22O11(aq)
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Electrolytes
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Electrolytes
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Sample Problem
Classify each of the following compounds as strong electrolyte, weak electrolyte or non-electrolyte.
• CaCl2• C6H12O6 (glucose)
• HF
• HNO3
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4.2. The Concentration of Solutions
• The concentration of a solution is the amount of solute per volume of solution.
moles of soluteMolarity ( ) = liter of solution
M
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Sample Problem
What is the molarity of 100 mL solution containing 15.6g of NaOH?
( ) 1 mol NaOH15.6g NaOH 0.390 mol NaOH40.0g NaOH
0.390 mol NaOH 3.90 NaOH1 L100 mL
1000 mL
=
=
M = M
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Sample Problem
How many moles of HCl are contained in 250 mL of a 0.600 molar solution?
( ) =
1 L 0.600 mol HCl250. mL 0.150 mol HCl1000 mL 1 L
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Other Concentration Units
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Other Concentration Units
ppm ppb ppt
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Sample Problem
What is the molarity of a solution that is 4.0 ppb Cr6+?
6+ 6+8 6+
6 6+
4.0 g4.0 ppbL soln.
4.0 g 1 g Cr 1mol Cr 7.7 10 CrL soln. 10 g 52.0 gCr
−
µ=
µ = × µ M
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Diluting Solutions
moles in the concentrated solution = moles in the dilute solution
Cinitial × Vintial = Cfinal × Vfinal
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Sample Problem
How many mL of 15.0 M nitric acid solution should be dilute to make 250. mL of a 0.50 M solution?
×=
×=
=
final finalinitial
initial
C VVC
0.50 250. mL15.0
8.3 mL
MM
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4.3. Stoichiometric Analysis of Solutions
TitrationControlled addition of a solution of known
concentration to react with a solution of unknown concentration.
Equivalence pointThe point when all reactants are completely
consumed.End point
Close to the equivalence point marked by an indicator.
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Titration
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4.4. Types of Chemical Reactions
• Molecular equation (components written as compounds):
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
• Complete Ionic Equation (electrolytes written as ions):Ag+(aq) + NO3
−(aq) + Na+(aq) + Cl−(aq) →AgCl(s) + Na+(aq) + NO3
−(aq)
• Net ionic equation (ions participating in the reactionAg+(aq) + Cl−(aq) → AgCl(s)
(Na+ and NO3− are spectator ions.)
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4.5. Precipitation Reactions
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Precipitation Reactions
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Solubility Rules
• How can a precipitation reaction be predicted ?
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Sample Problem
How much AgCl will be formed from mixing 1.50 L of 0.500 M AgNO3 with 1.75 L of 0.300 M NaCl?
( )
( )
+
-
0.500 mol Ag1.50 L 0.750 mol Ag1L
0.300 mol Cl1.75 L 0.525 mol Cl
+
−
=
=
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Sample Problem (cont)
The net ionic equation is:
Ag+(aq) + Cl−(aq) → AgCl(s)
Thus Cl- is the limiting reactant and 0.525 mol AgCl will be formed. In grams,
( ) 143.4g0.525 mol AgCl 75.3g AgCl1mol
=
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4.6. Acid-Base Reactions
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Acid-Base Reactions
Drain decloggers –contain s.base
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Acid-Base Reactions
• Neutralization reaction for strong acids and strong bases:
H+(aq) + OH-(aq) → H2O(l)
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Sample Problem
What volume of 0.200 M HCl solution is need to neutralize 25.0 mL of a 0.450 MKOH solution?
The balanced molecular equation is:
HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)
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Sample Problem (cont)
( )
( )
-
-
- +
++
0.450 mol KOH 1mol OH 1 L25.0 mL1 L 1 mol KOH 1000 mL
0.01125 mol OHat neutralization, mol OH mol H
1mol HCl 1 L 1000 mL0.01125 mol H1 mol H 0.200 mol HCL 1 L
56.3 mL HCl
=
=
=
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4.7. Oxidation-Reduction Reactions
Zn(s) + Cu2+(aq)→ Zn2+(aq) + Cu(s)
oxidation reaction (loss of electrons)
Zn → Zn2+ (aq) + 2e-
reduction reaction (gain of electrons)
Cu2+ (aq) + 2e- → Cu
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Sample Problem
Assign oxidation numbers to all atoms in the following:
CO2, K2Cr2O7, PCl5
CO2 – oxygen is -2×2 = -4, making C +4.
K2Cr2O7 – oxygen -2×7 = -14.
K (group 1A) is +1×2 = +2
2Cr must equal +12 = -14+2
each Cr = +6
PCl5 – Cl (group VIIA) = -1×5 = -5
P = +5
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Zinc and Iodine
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Dry Ice and Magnesium
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Reaction of Phosphorous