chapter 4 types of chemical reactions and solution stoichiometry
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Chapter 4 Types of chemical reactions and Solution Stoichiometry. What are aqueous solutions? Substances dissolved in water Solvent Why is water considered the most common solvent? It can dissolve many different compounds because it is not a linear molecule bent at an angle of 105 °. - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 4Types of chemical reactions and
Solution Stoichiometry• What are aqueous solutions?• Substances dissolved in waterSolvent
• Why is water considered the most common solvent?
• It can dissolve many different compounds because it is not a linear molecule bent at an angle of 105°
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What type of bond occurs in water?
• Polar covalent• Water is a polar molecule has an unequal
charge distribution • What happens to a Salt when dissolved in water?• It breaks into cations which are attracted to the
negative ends of water and the anions which are attracted to the pos. ends of water
• This process is called Hydration
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The nature of aqueous solutions
• What is a solution?
• Homogeneous mixture
• Has 2 parts
• A. Solute- lesser amount of substance
• B. Solvent- greater amount of substance
How can we characterize a solution?
One way is by Electrical Conductivity
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What is electrical conductivity?
• Ability to conduct an electric current
• Electrolyte- substance that when dissolved in water can conduct electricity
• Can be classified on its ability to conduct electricity
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Electrolyte Conductivity Degree of
Dissociation
Examples
strong High total Strong acids and bases
weak Low to
moderate
partial Weak organic acid and bases
non none Close to
zero
Sugar, ethanol
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The Composition of Solutions
• The most commonly used expression of concentration is Molarity
• The moles of solute per volume of solution in liters
or
M= moles of solute/ liters of solution
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Examples
• Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.
• Calculate the mass of NaCl needed to prepare 175ml of a .500 M NaCl solution.
• How many ml of solution are necessary if we are to have a 2.48 M NaOH solution that contains 31.52 g of the dissolved solid?
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Molarity of Ions in Solution
Calculate the molarity of all ions in .25 M Ca(ClO)2.
H20(l)
Ca(ClO)2 (s) Ca2+ + 2ClO-
Molarity of Ca2+ = .25 M
Molarity of ClO- = 2 • .25 M = .50 M
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Another Example
Determine the molarity of Cl- ions in a solution prepared by dissolving 9.82 g of CuCl2 in enough water to make 600. ml.
1. Determine the molarity.
2. Write an equation to determine the mole ratio.
3. Set-up a mole problem.
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Dilutions
- adding a solvent ( usually water) to lower the concentration of a solute.
M1V1= M2V2
What volume of 16 M sulfuric acid must be used to prepare 1.5 L of .10 M H2SO4.
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Types of Chemical Reactions
1. Precipitation reactions( double displacement)
2. Acid- Base reactions
3. Oxidation-reduction reactions
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Precipitation reactions
- 2 solutions are mixed, sometimes a precipitate forms
What is a precipitate?
- an insoluble substance
Ba(NO3)2(aq) + K2CrO4(aq) ?
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Predict what will happen?
1. Na2SO4(aq) + Pb(NO3)2(aq) ?
2NaNO3(aq) + PbSO4(s)
2. 3KOH(aq) + Fe(NO3)3(aq) ?
3 KNO3(aq) + Fe(OH)3(s)
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Describing Reactions in Solutions
3 Types of Equations1. Molecular equation
K2CrO4(aq) + Ba(NO3)2(aq) BaCrO4(s) +
2KNO3(aq)2. Complete Ionic equation- represents the actual forms of
reactants and products
2K+(aq) + CrO4-(aq) + Ba+2(aq) + 2NO3
-(aq) BaCrO4(s) + 2K+
(aq) + 2NO3-(aq)
Spectator Ions- ions that do not participate directly in the reaction
K+ and NO3- are spectator ions
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Con.
3. Net Ionic Equation- includes only those components that undergo changes in the reaction
Ba2+(aq) + CrO42-(aq) BaCrO4(s)
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Example
Aqueous potassium chloride is added to aqueous silver nitrate to form a silver chloride precipitate plus aqueous potassium nitrate. Write the three types of equations.
Molecular equation
KCl(aq) + AgNO3(aq) AgCl(s) + KNO3(aq)
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Con.
Complete ionic
K+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq)
AgCl(s) + K+(aq) + NO3-(aq)
Net ionic
Cl-(aq) + Ag+(aq) AgCl(s)
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Selective Precipitation
- separating cations by precipitating them out one at a time
- Use a Solubility chart to make a Flow Chart
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Solving Stoichiometry Problems for reactions in solutions
1. Write the balanced equation.2. Set-up 2 mole problems to determine limiting and
excess.3. Change to grams if needed.
Example: When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed.
Na2SO4(aq) + Pb(NO3)2(aq) PbSO4(s) + 2NaNO3(aq)
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Acid-Base Reactions
• Bronsted-Lowry Concept of Acids and Bases
• - an acid is a proton donor
• - a base is a proton acceptor
• Also called a neutralization reaction
• Produces a salt and water
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Example 4.12p. 159
• What volume of a 0.100 M HCl solution is needed to neutralize 25.0 ml of .350M NaOH?
HCl(aq) + NaOH (aq) NaCl(s) + H20(l)
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Acid-Base Titrations
• Volumetric analysis- a technique for determining the amount of a certain substance by doing a titration
• Titration- the use of a buret to deliver the measured volume of a known concentration(titrant) into the substance being analyzed(analyte)
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Con.
• Equivalence point or Stoichiometric point- the point where enough titrant has been added to react with a substance
• Often marked by an indicator• Indicator- substance added at the beginning of a
titration that changes color at or very near the equivalence point
• Endpoint- point where the color change occurs• Ex: phenophthalein(colorless in acid, pink in
base)
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Example 4.14p. 162
• A student carries out an experiment to standardize(determine the exact concentration of) a sodium hydroxide solution. To do this, the student weighs out a 1.3009 g of KHP(KHC8H4O4). Its molar mass is 204.22g/mol. The student dissolves the KHP in distilled water, adds phenolphthalein as an indicator, and titrates the resulting solution with the sodium hydroxide solution to the endpoint. The difference between the final and initial buret readings is 41.20 ml. Calculate the concentration of the NaOH solution.
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Oxidation-Reduction Reactions
• Oxidation States
-imaginary charges the atoms would have if shared electrons were divided equally
Oxidation- increase in the oxidation state: a loss of electrons
Reduction- decrease in oxidation state: gain of electrons
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A helpful Mnemonic
OIL RIGOxidation ReductionInvolves InvolvesLoss GainExample:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) -4 +1(+4) 0 +4 –2(-4) +1(+4) -2Carbon -4 to +4 oxidizedOxygen 0 to –2 reduced
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What else is involved in Redox?
1. Oxidizing Agent- electron acceptor
2. Reducing Agent- electron donor
In the previous example what is the oxidizing and reducing agent?
O2 is oxidizing agent
CH4 is reducing agent
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Examples
2Al(s) + 3I2(s) 2AlI3(s)
2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g)
What atoms are oxidized/reduced and which compounds are the oxidizing and reducing agents?
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Balancing Redox equations in Acid Solutions
1. Assign oxidation states2. Write ½ cells( half-reactions)
3. Balance all elements except H and O2
4. Balance O by adding H2O5. Balance H by adding H+6. Balance Charges by adding e-7. If necessary, multiply all coefficients in 1 or both ½
cells by an integer to equalize the # of e-8. Add ½ cells and cancel identical species, check that
charges and atoms are balanced
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Examplep. 173
MnO4-(aq) + Fe2+(aq) Fe3+(aq) + Mn2+(aq)
acid
Example 4.19 p. 175
H+(aq)+ Cr2O72-(aq)+C2H5OH(l)Cr3+(aq) +
CO2(g) + H2O(l)
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Balancing Redox equations in Basic Solution
1. Assign oxidation states2. Write ½ cells(half-reactions)
3. Balance all elements except H and O2
4. Balance O by adding H2O5. Balance H by adding H+
6. Balance charges by adding e-
7. If necessary multiply all coefficients in 1 or both ½ cells by an integer to equalize # of e-
8. Add ½ cells and cancel identical species. Check that charges and atoms are balanced
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Con.
9. Count # of H+ and add same # of OH- to both sides of the equation
10. If H+ and OH- are on the same side, form H20.
11. Cancel out H20 molecules on both sides
12. Check that charges and atoms are balanced
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Balancing Redox in basic solution example
I-(aq) + OCl-(aq) I2(s) + Cl-(aq) +H20(l)