Chapter 4 Type of Chemical Reactions and Solution Stoichiometric Water, Nature of aqueous solutions,...
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Transcript of Chapter 4 Type of Chemical Reactions and Solution Stoichiometric Water, Nature of aqueous solutions,...
Chapter 4Type of Chemical Reactions and
Solution Stoichiometric
Water, Nature of aqueous solutions, types of electrolytes, dilution.
Types of chemical reactions: precipitation, acid-base and oxidation reactions.
Stoichiometry of reactions and balancing the chemical equations.
The water molecule is polar.
A space-filling model of the water molecule.
Water is a dissolving media or solvent
Aqueous Solutions
Figure 4.2: Polar water molecules interact with the positive and negative ions of a salt assisting
in the dissolving process.
Cl-
Na+
Cl-
Na+
H2O
Some Properties of Water
Water is “bent” or V-shaped. The O-H bonds are covalent. Water is a polar molecule. Hydration occurs when salts dissolve in
water.
Figure 4.3: (a) The ethanol molecule contains a polar O—H bond similar to those in the water molecule. (b) The polar water molecule interacts strongly with the polar O—H bond in ethanol. This is a case of "like dissolving like."
Solvent
retains its phase (if different from the solute)
is present in greater amount (if the same phase as the solute)
Solute
dissolves in water (or other “solvent”)
changes phase (if different from the solvent)
is present in lesser amount (if the same phase as the solvent)
General Rule for dissolution
Like dissolve likewise
Polar dissolve polar (water dissolves in ethanol)
Non-polar dissolve nonpolar (Alkane H/C dissolves in fat)
Figure 4.5: When solid NaCl dissolves, the Na+ and Cl- ions are randomly dispersed in the water.
Example Example
Electrolytes
Strong - conduct current efficiently
NaCl, HNO3
Weak - conduct only a small current
vinegar, tap water
Non - no current flows
pure water (non-ionic or de-ionized), sugar solution
Figure 4.4: Electrical conductivity of aqueous solutions.
Acids
Strong acids - dissociate completely to produce H+ in solution
hydrochloric and sulfuric acid
HCl , H2SO4
Weak acids - dissociate to a slight extent to give H+ in solution
acetic and formic acid
CH3COOH, CH2O
Bases
Strong bases - react completely with water to give OH ions.
sodium hydroxide
Weak bases - react only slightly with water to give OH ions.
ammonia
Figure 4.6: HCl (aq) is completely ionized.
Figure 4.7: An aqueous solution of sodium hydroxide.
Figure 4.8: Acetic acid (HC2H3O2) exists in water mostly as undissociated molecules. Only a small percentage of the molecules are ionized such as 1 in 100 molecules.
Molarity
Molarity (M) = moles of solute per volume of solution in liters:
M
M
molaritymoles of soluteliters of solution
HClmoles of HCl
liters of solution3
62
Calculate the molarity (M) of the solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.
Find the concentration of each type of ion in 0.26 M Al(NO3)3 and 0.15 M CaCl2
Calculate the number of moles of Cl- ions in 1.75 L of 1.0 x 10-3 M ZnCl2.
Common Terms of Solution Concentration
Stock - routinely used solutions prepared in concentrated form.
Concentrated - relatively large ratio of solute to solvent. (5.0 M NaCl)
Dilute - relatively small ratio of solute to solvent. (0.01 M NaCl): (MV)initial=(MV)Final
Figure 4.10: Steps involved in the preparation of a standard aqueous solution.
Figure 4.12: Dilution Procedure (a) A measuring pipette is used to transfer 28.7mL of 17.4 M acetic acid solution to a volumetric flask. (b) Water is added to the flask to the calibration mark. (c) The resulting solution is 1.00 M acetic acid.
Diluted from 17.4 M to 1.00 M
Moles of solute after dilution = moles of solute before dilution
Prepare 2.00 L of each of the following solution:
a) 0.250 M NaOH from solid NaOH
b) 0.250 M NaOH from 1.00 M NaOH stock solution
M1V1 = M2V2
Before dilution after dilution
2.00 L × NaOHmol
NaOHg00.40
L
NaOHmol250.0 = 20.0 g NaOH
Types of Solution Reactions
Precipitation reactionsAgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
Acid-base reactionsNaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)
Oxidation-reduction (redox) reactionsFe2O3(s) + Al(s) Fe(l) + Al2O3(s)
Simple Rules for the Solubility of Salts in Water
1. Most nitrate (NO3) salts are soluble.
2. Most alkali (group 1A) salts and NH4+ are soluble.
3. Most Cl, Br, and I salts are soluble (NOT Ag+, Pb2+, Hg2
2+)
4. Most sulfate salts are soluble (NOT BaSO4, PbSO4, HgSO4, CaSO4)
5. Most OH salts are only slightly soluble (NaOH, KOH are soluble, and Ba(OH)2, Ca(OH)2 are marginally soluble)
6. Most S2, CO32, CrO4
2, PO43 salts are only slightly soluble.
Figure 4.13: When yellow aqueous potassium chromate is added to a colorless barium nitrate solution, yellow barium chromate precipitates.
Describing Reactions in SolutionPrecipitation
1. Molecular equation (reactants and products as compounds)
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
2. Complete ionic equation (all strong electrolytes shown as ions)
Ag+(aq) + NO3 (aq) + Na+ (aq) + Cl(aq)
AgCl(s) + Na+ (aq) + NO3 (aq)
Describing Reactions in Solution (continued)
3. Net ionic equation (show only components that actually reacts)
Ag+(aq) + Cl(aq) AgCl(s)
Na+ and NO3 are spectator ions.
Stoichiometry of Precipitation Reactions
How to calculate quantities of reactants and products involved in a chemical reaction?
1. Convert all quantities to moles.
2. Use the balancing coefficients to relate the moles of reactants &products in a balanced equation.
3. Determine the solid product (precipitate) and spectator ions from the rules of solubility.
4. Calculate the limiting reactant and theoretical yield.
5. Convert moles to mass if required.
Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl.
Hint: Na+ and NO3- are spectator ions, AgCl is the only product.
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.
Hint: Na+ and NO3- are spectator ions, PbSO4 is the only product.
Acid-Base Reactions
An acid produces H+ ions in water
A base produces OH- ions in water
An acid is a proton donor
A base is a proton acceptor
Arrhenius’s concept
Bronsted & Lowrys’ concept
Performing Calculations for Acid-Base Reactions
1. List initial species and predict reaction.
2. Write balanced net ionic reaction.
3. Calculate moles of reactants.
4. Determine limiting reactant.
5. Calculate moles of required reactant/product.
6. Convert to grams or volume, as required.Remember: n H+ = n OH- (MV) H+ = (MV) OH-
Neutralization Reaction
acid + base salt + water
HCl (aq) + NaOH (aq) NaCl (aq) + H2O
H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O
H+ + OH- H2O
4.3
What volume of a 0.100 M HCl solution is needed to neutralize 25.0 ml of 0.350 M NaOH?
In certain experiment, 28.0 ml of 0.0250 M HNO3 and 53.0 ml of 0.320 M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H+ or OH-
ions in excess after reaction goes to completion?
Neutralization Reaction of Strong Acid and Strong Base
M1V1f1 = M2V2f2
Where: M & V are molarities & volumes of respective acid & base
And f = number of H+ or OH- per formula unit of acid & base
For example in acids: in HCl, f = 1 while in H2SO4, f = 2 and in H3PO4, f = 3
Similarly in bases:In NaOH, f = 1, Ca(OH)2, f = 2 and so on
Key Titration Terms
Titrant - solution of known concentration used in titration
Analyte - substance being analyzed
Equivalence point - enough titrant added to react exactly with the analyte
Endpoint - the indicator changes color so you can tell the equivalence point has been reached.
movie
In one experiment, 1.3009 g sample of potassium hydrogen phthalate (KHC8H4O4 often abbreviated KHP) was titrated against NaOH solution. Exactly 41.20 mL of NaOH solution was required for complete titration of 1.3009 g of KHP. Calculate the concentration of the sodium hydroxide solution.
KHC8H4O4 K+ (aq) + HC8H4O4- (aq)
NaOH Na+ (aq) + OH- (aq)
OH- (aq) + HC8H4O4- (aq) H2O (l) + HC8H4O4
2- (aq)
Practice Example
How many moles are in 18.2 g of CO2?
Practice Example
Consider the reaction
N2 + 3H2 2NH3
How many moles of H2 are needed to completely react 56 g of N2?
Practice Example
How many grams are in 0.0150 mole of caffeine C8H10N4O2
Practice Example
A solution containing Ni2+ is prepared by dissolving 1.485 g of pure nickel in nitric acid and diluting to 1.00 L. A 10.00 mL aliquot is then diluted to 500.0 mL. What is the molarity of the final solution?(Atomic weight: Ni = 58.70).
Practice Example
Calculate the number of molecules of vitamin A, C20H30O in 1.5 mg of this compound.
Practice Example
What is the mass percent of hydrogen in acetic acid HC2H3O2
Oxidation-Reduction Reactions(electron transfer reactions)
2Mg (s) + O2 (g) 2MgO (s)
2Mg 2Mg2+ + 4e-
O2 + 4e- 2O2-
Oxidation half-reaction (lose e-)
Reduction half-reaction (gain e-)
2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e-
2Mg + O2 2MgO
Reducing agent: Oxidizes it self
Oxidizing agent: Reduces it self
Redox Reactions• Many practical or everyday examples of
redox reactions:– Corrosion of iron (rust formation)– Forest fire– Charcoal grill– Natural gas burning– Batteries
– Production of Al metal from Al2O3 (alumina)
– Metabolic processes
combustion
Rules for Assigning Oxidation States
1. Oxidation state of an atom in an element = 0
2. Oxidation state of monatomic element = charge
3. Oxygen = 2 in covalent compounds (except in peroxides where it = 1)
4. H = +1 in covalent compounds
5. Fluorine = 1 in compounds
6. Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion
Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)
Zn is oxidizedZn Zn2+ + 2e-
Cu2+ is reducedCu2+ + 2e- Cu
Zn is the reducing agent
Cu2+ is the oxidizing agent
4.4
Copper wire reacts with silver nitrate to form silver metal.What is the oxidizing agent in the reaction?
Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s)
Cu Cu2+ + 2e-
Ag+ + 1e- Ag Ag+ is reduced Ag+ is the oxidizing agent
NaIO3
Na = +1 O = -2
3x(-2) + 1 + ? = 0
I = +5
IF7
F = -1
7x(-1) + ? = 0
I = +7
K2Cr2O7
O = -2 K = +1
7x(-2) + 2x(+1) + 2x(?) = 0
Cr = +6
Oxidation numbers of all the elements in the following ?
4.4
Balancing by Half-Reaction Method
1. Write separate reduction and oxidation half reactions.
2. For each half-reaction:
Balance other elements first (except H, O)
Balance O using H2O
Balance H using H+
Balance charges using electrons
Balancing by Half-Reaction Method (continued)
3. If necessary, multiply by integer to equalize number of electrons on both sides.
4. Must cancel electrons on both sides
5. Add both half-reactions
6. Check that number of elements and charges are balanced on both sides.
7. Cancel same number of H+ from both sides
8. This will be your balanced redox reaction in acidic medium.
Half-Reaction Method - Balancing in Base
1. Start balancing as in acidic medium.
2. Then add OH ions equal to H+ ions present on both sides of the reaction.
3. Form water by combining H+ and OH.
4. Cancel same number of H2O from both sides
5. Check number of elements and charges are balanced on both sides of the reaction..
Balancing Redox Equations
•Example: Balance the following redox reaction:
•Cr2O72- + Fe2+ Cr3+ + Fe3+ (acidic soln)
1) Break into half reactions:
Cr2O72- Cr3+
Fe2+ Fe3+
Balancing Redox Equations
2) Balance each half reaction:
Cr2O72- Cr3+
Cr2O72- 2 Cr3+
Cr2O72- 2 Cr3+ + 7 H2O
Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O
6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O
Balancing Redox Equations
2) Balance each half reaction (cont)
Fe2+ Fe3+
Fe2+ Fe3+ + 1 e-
Balancing Redox Reactions
3) Multiply by integer so e- lost = e- gained
6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O
Fe2+ Fe3+ + 1 e- x 6
Balancing Redox Reactions
3) Multiply by integer so e- lost = e- gained
6 Fe2+ 6 Fe3+ + 6 e-
6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O
4) Add both half reactions
Cr2O72- + 6 Fe2+ + 14 H+ 2 Cr3+ + 6 Fe3+ + 7 H2O
Balancing Redox Reactions
Cr2O72- + 6 Fe2+ + 14 H+ 2 Cr3+ + 6 Fe3+ + 7 H2O
5) Check the equation
2 Cr 2 Cr
7 O 7 O
6 Fe 6 Fe
14 H 14 H
+24 + 24
Balancing Redox Reactions
• Procedure for Basic Solutions:– Divide the equation into 2 incomplete half
reactions• one for oxidation
• one for reduction
Balancing Redox Reactions– Balance each half-reaction:
• balance elements except H and O
• balance O atoms by adding H2O
• balance H atoms by adding H+
• add 1 OH- to both sides for every H+ added
• combine H+ and OH- on same side to make H2O
• cancel the same # of H2O from each side
• balance charge by adding e- to side with greater overall + charge
different
Balancing Redox Equations– Multiply each half reaction by an integer so that
• # e- lost = # e- gained
– Add the half reactions together.• Simply where possible by canceling species
appearing on both sides of equation
– Check the equation• # of atoms
• total charge on each side
Balancing Redox Reactions
Example: Balance the following redox reaction.
NH3 + ClO- Cl2 + N2H4 (basic soln)
NH3 N2H4
ClO- Cl2
1) Break into half reactions:
Balancing Redox Reactions
NH3 N2H4
2) Balance each half reaction:
2 NH3 N2H4
2 NH3 N2H4 + 2 H+
2 NH3 + 2 OH- N2H4 + 2 H2O
+ 2 OH- + 2 OH-
2 H2O
2 NH3 + 2 OH- N2H4 + 2 H2O + 2 e-
Balancing Redox Reactions
2 ClO- Cl2
2) Balance each half reaction:
2 ClO- Cl2 + 2 H2O2 ClO- + 4 H+ Cl2 + 2 H2O
+ 4 OH- + 4 OH-
2 ClO- + 4 H2O Cl2 + 2 H2O + 4 OH-
2 ClO- + 2 H2O Cl2 + 4 OH-
2 e- + 2 ClO- + 2 H2O Cl2 + 4 OH-
ClO- Cl2
Balancing Redox Reactions
3) Multiply by integer so # e- lost = # e- gained
2 NH3 + 2 OH- N2H4 + 2 H2O + 2 e-
2 e- + 2 ClO- + 2 H2O Cl2 + 4 OH-
4) Add both half reactions
2 NH3 + 2 OH- + 2ClO- + 2 H2O N2H4 + 2 H2O + Cl2 + 4 OH-
Balancing Redox Reactions5) Cancel out common species
2 NH3 + 2 OH- + 2 ClO- + 2 H2O N2H4 + 2 H2O + Cl2 + 4 OH-
2
2 NH3 + 2 ClO- N2H4 + Cl2 + 2 OH-
6) Check final equation:
2 N 2 N
6 H 6 H
2 Cl 2 Cl
2 O 2 O
-2 -2
Practice ExampleIn the following the oxidizing agent is:
5H2O2 + 2MnO4- + 6H+ 2Mn2+ + 8H2O + 5O2
a. MnO4-
b. H2O2
c. H+
d. Mn2+
e. O2
Practice ExampleDetermine the coefficient of Sn in acidic solution
Sn + HNO3 SnO2 + NO2 + H2O
1
Practice ExampleThe sum of the coefficients when they are
whole numbers in basic solution:
Bi(OH)3 + SnO22- Bi + SnO3
2-
13
http://www.chemistrycoach.com/balancing_redox_in_acid.htm#BalancingRedoxEquationsinAcidicorBasicMedium
http://www.chemistrycoach.com/tutorials-5.htm#Oxidation-Reduction