CHEM 101 ppts Chapter 4 Part 2
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CHEM 101 ppts
Chapter 4Part 2
Acid Base Reactions
Neutralization Reaction
HCl (aq) + NaOH (aq) NaCl (aq) + H2O
H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O
H+ + OH- H2O
acid + base salt + water
When a strong acid reacts with a strong base, the net ionic equation is…
Neutralization Reaction Involving a Weak Electrolyte
weak acid + base salt + water
HCN (aq) + NaOH (aq) NaCN (aq) + H2O
HCN + Na+ + OH- Na+ + CN- + H2O
HCN + OH- CN- + H2O
Neutralization Reaction Producing a Gas
acid + base salt + water + CO2
2HCl (aq) + Na2CO3 (aq) 2NaCl (aq) + H2O +CO2
2H+ + 2Cl- + 2Na+ + CO32- 2Na+ + 2Cl- + H2O + CO2
2H+ + CO32- H2O + CO2
Metathesis (Exchange) Reactions• Metathesis comes from a Greek word that means
“to transpose”• It appears the ions in the reactant compounds
exchange, or transpose, ions
AgNO3 (aq) + KCl (aq) AgCl (s) + KNO3 (aq)
General reaction AX + BY AY + BX
Methathesis reactions are commonly represented with “molecular”, “ionic”(or “total ionic”), and “net ionic” equations. The above equation is anexample of a molecular equation
Metathesis Reactions• The reactions we’ve considered so far
– Formation of insoluble precipitate– Formation of a molecular compound (e.g. in an acid-
base neutralization)– Gas formation reactions
each drive metathesis reactions. The following combination of solutions would not result in a reaction:
NaC2H3O2(aq) + KNO3(aq) no rxn
Oxidation-Reduction Reactions(electron transfer reactions)
Oxidation Reduction (REDOX) Ledger
+4+3+2+10-1-2-3-4
Oxidation- a loss of electrons to go to a higher oxidation
state/number
Reduction- gain of
electrons to go to a lower
oxiation state/number Reducing Agent- aids in reduction by
giving up electrons(is oxidized)
Oxidizing Agent- aids in oxidation by gaining electrons (is reduced)
Oxidation Number(State)
The charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred.
1. Free elements (uncombined state) have an oxidation number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. In monatomic ions, the oxidation number is equal to the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. The oxidation number of oxygen is usually –2. In H2O2
and O22- it is –1.
4.4
4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.
6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.
5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.
HCO3-
O = –2 H = +1
3x(–2) + 1 + ? = –1
C = +4
What are the oxidation numbers of all the elements in HCO3
- ?
7. Oxidation numbers do not have to be integers. Oxidation number of oxygen in the superoxide ion, O2
-, is –½.
The Oxidation Numbers of Elements in their Compounds
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
What are the oxidation numbers of all the elements in each of these compounds? NaIO3 IF7 K2Cr2O7
Oxidation Numbers
• Elements in their elemental form have an oxidation number of 0.
• The oxidation number of a monatomic ion is the same as its charge.
“Elemental form” ofSodium: Na(s)
Iron: Fe(s)
Mercury: Hg(l)
Hydrogen: H2(g)
Oxygen: O2(g)
Chlorine: Cl2(g)
Iodine: I2(s)
Pretty muchany metal(except Hg) isa solid in itselemental form
i.e. as they would be found in nature
Oxidation Numbers• Elements in their elemental form have an
oxidation number of 0.• The oxidation number of a monatomic ion
is the same as its charge.
“Elemental form” ofSodium: Na(s)
Iron: Fe(s)
Mercury: Hg(l)
Hydrogen: H2(g)
Oxygen: O2(g)
Chlorine: Cl2(g)
Iodine: I2(s)
Several non-metal elements occuras diatomics (see chapter 2 notes)
Na(s) + Cl2(g) Na+(g) + Cl-(g)
oxidationnumber
0 0
Oxidation Numbers
• Elements in their natural, elemental form have an oxidation number of 0.
• The oxidation number of a monatomic ion is the same as its charge.
e.g. Fe3+ +3Cu+ +1S2- -2
monatomicion
oxidationnumber
Zn2+, H+ are examples of monatomic ions. NH4+ is an example of a polyatomic ion
Oxidation Numbers• The halogens (F, Cl, Br, I):
Fluorine always has an oxidation number of −1.The other halogens (Cl, Br, I) often have an
oxidation number of −1; they can have a positive oxidation number, however, most notably in oxyanions (e.g. ClO4
- Cl oxid. # = +7)
Determining Oxidation Numbers
• The sum of the oxidation numbers in a neutral compound is 0.
H2O
• The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.
+1 -2
2(+1) + -2 = 02H + O
ClO4-
+7 -2 to find oxid. # for Cl:x + 4(-2) = -1
oxid. # for Cl oxid. # for O
Oxidation-Reduction Reactions
One cannot occur without the other.
The electrons that come from theoxidized species are used to
reduce the other species
thisoxidizes
this
The species that gets reduced is called the oxidizing agentThe species that gets oxidized is called the reducing agent
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
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
Types of Oxidation-Reduction Reactions
Combination ReactionA + B C
2Al + 3Br2 2AlBr3
Decomposition Reaction
2KClO3 2KCl + 3O2
C A + B
0 0 +3 -1
+1 +5 -2 +1 -1 0
Types of Oxidation-Reduction Reactions
Combustion Reaction
A + O2 B
S + O2 SO2
0 0 +4 -2
2Mg + O2 2MgO0 0 +2 -2
Displacement Reaction
A + BC AC + B
Sr + 2H2O Sr(OH)2 + H2
TiCl4 + 2Mg Ti + 2MgCl2
Cl2 + 2KBr 2KCl + Br2
Hydrogen Displacement
Metal Displacement
Halogen Displacement
Types of Oxidation-Reduction Reactions
0 +1 +2 0
0+4 0 +2
0 -1 -1 0
Displacement Reactions
• In displacement reactions, ions oxidize an element (the ions, then, are reduced).
• The reaction
Fe(s) + Cu2+(aq) Fe2+
(aq) + Cu(s)
is thus a displacement reaction
Individually, the oxidation and reductionreactions can be written as “half-reactions”:
Oxidation: Fe(s) Fe2+(aq) + 2e-
Reduction: Cu2+(aq) + 2e- Cu(s)
notice: total charge on right = total charge on left
Displacement Reactions
In this reaction, silver ions oxidize copper metal.
Cu (s) + 2 Ag+ (aq) Cu2+ (aq) + 2 Ag (s)
2 here because each silver ion only gains one electron
Oxid: Cu(s) Cu2+(aq) + 2e-
Red: Ag+(aq) + e- Ag(s)
Displacement Reactions
Interestingly: The reverse reaction,
however, does not occur.
Cu2+ (aq) + 2 Ag (s) Cu (s) + 2 Ag+ (aq) x
Why?
So, apparently, Ag+(aq) is a better oxidizing agent than Cu2+
(aq)
Activity Series
An ion that lies beneath a metal in this table will be able to oxidize the metal.Example: Cu2+ can oxidize Fe
•Some metals are more
reactive than others
(they tend to lose
electrons more readily).
The activity series ranks
elements in order of
their ability to lose electrons
most reactive
least reactive
2Ag+(aq) + Cu(s) Cu2+
(aq) + 2Ag(s)
Cu2+(aq) + Fe(s) Fe2+
(aq) + Cu(s)
2Ag(s) + Cu2+(aq) no rxn
Oxidation-Reduction Reactions(electron transfer reactions)
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
•Oxidation and Reduction reactions occur together. •One reagent gives up electrons to be oxidized and the other gains electrons to become reduced.
The Activity Series for Metals
M + BC MC + B
Hydrogen Displacement Reaction
M is metalBC is acid or H2O
B is H2
Ca + 2H2O Ca(OH)2 + H2
Pb + 2H2O Pb(OH)2 + H2
The Activity Series for Halogens
Halogen Displacement Reaction
Cl2 + 2KBr 2KCl + Br2
0 -1 -1 0
F2 > Cl2 > Br2 > I2
I2 + 2KBr 2KI + Br2
The same element is simultaneously oxidized and reduced.
Example:
Disproportionation Reaction
Cl2 + 2OH- ClO- + Cl- + H2O
Types of Oxidation-Reduction Reactions
0 +1 -1
oxidized
reduced
Ca2+ + CO32- CaCO3
NH3 + H+ NH4+
Zn + 2HCl ZnCl2 + H2
Ca + F2 CaF2
Precipitation
Acid-Base
Redox (H2 Displacement)
Redox (Combination)
Classify each of the following reactions.
Chemistry in Action: Breath Analyzer
3CH3COOH + 2Cr2(SO4)3 + 2K2SO4 + 11H2O
3CH3CH2OH + 2K2Cr2O7 + 8H2SO4 +6
+3
What is the concentration of Na2SO4 in a solution prepared by diluting 0.010 mol Na2SO4 to 1.00 L?
• The answer is:… zero …
• WHY??• And … how do we describe the concentration of
this solution?
Ion Concentrations in Solution
• In 0.010 M Na2SO4:– two moles of Na+ ions are formed for each
mole of Na2SO4 in solution, so [Na+] = 0.020M.
– one mole of SO42– ion is formed for each mole
of Na2SO4 in solution, so [SO42–] = 0.010 M.
• An ion can have only one concentration in a solution, even if the ion has two or more sources.
Calculating Ion Concentrations in Solution
Quantitative Analysis
• Analytical chemistry deals with the determination of composition of materials – that is, the analysis of materials
■Quantitative analysis involves the determination of the amount of a substance or species present in a material
04/12/23 35
Quantitative Analysis
• Gravimetric Analysis– Gravimetric analysis is a type of quantitative
analysis in which the amount of a species in a material is determined by converting the species into a product that can be isolated and weighed
– Precipitation reactions are often used in gravimetric analysis
– The precipitate from these reactions is then filtered, dried, and weighed.
04/12/23 36
Gravimetric Analysis1. Dissolve unknown substance in water
2. React unknown with known substance to form a precipitate
3. Filter and dry precipitate
4. Weigh precipitate
5. Use chemical formula and mass of precipitate to determine amount of unknown ion
Quantitative Analysis
• Gravimetric Analysis– Consider the problem of determining the amount
of lead in a sample of drinking water• Adding sodium sulfate (Na2SO4) to the sample will
precipitate lead(II) sulfate
• The PbSO4 can then be filtered, dried, and weighed
04/12/23 38
2+ +2 4 4Na SO (aq) + Pb (aq) 2Na (aq) + PbSO (s)
Practice ProblemSuppose a 1.00 L sample of polluted water was analyzed for lead(II) ion, Pb2+, by adding an excess of sodium sulfateThe mass of lead(II) sulfate that precipitated was 229.8 mg.What is the mass of lead in a liter of the water?Express the answer as mg of lead per liter of solutionAns:
Thus, [PbThus, [Pb2+2+] = 157.0 mg/1.00 L = 157 mg/L] = 157.0 mg/1.00 L = 157 mg/L
04/12/23 39
2+ +2 4 4Na SO (aq) + Pb (aq) 2Na (aq) + PbSO (s)
2+ 2+4
4 2+4 4
1 mmol PbSO 1 mmol Pb 207.2 mg Pb229.8 mg PbSO × × ×
303.3 mg PbSO 1 mm PbSO 1 mmol Pb2+= 157.0 mg Pb (Mass of Lead in the 1.0 L sample)
Practice ProblemA soluble silver compound was analyzed for the percentage of silver by adding sodium chloride solution to precipitate the silver ion as silver chlorideIf 1.583 g of silver compound gave 1.788 g of silver chloride (FW = 143.321), what is the mass percentage of silver (AW = 107.868 g/mol) in the compound?Ans:
04/12/23 40
107.868 g / mol% Silver = x 100 = 85.07 % Ag
126.8 g / mol
+ - + -(aq) (aq) (aq) (s) (aq)AgX + Na + Cl = AgCl + Na + X ( )aq
1 mol Ag Cl1.788 g AgCl = 0.01248 mol AgCL = 0.01248 mol AgX
143.321 g AgCl
11.583 g AgX = 126.8 g AgX / 1 mol AgX = Mol Wgt AgX
0.01248 mol AgX
M ol W gt Ag = 107.868 g / mol M ol W gt AgX = 126.8 g / mol
Quantitative Analysis
• Volumetric Analysis– An important method for determining the amount
of a particular substance is based on measuring the volume of the reactant solution• Titration is a procedure for determining the amount of
substance A by adding a carefully measured volume of a solution with known concentration of B until the reaction of A and B is just complete
• Volumetric analysis is a method of analysis based on titration
04/12/23 41
• In a titration, two reactants in solution are combined carefully until they are in stoichiometric proportion.
• The objective of a titration is to determine the number of moles, or the number of grams, or the percentage, or the concentration, of the analyte (the sought-for substance in an analysis, the substance we are looking for).
Titrations
• In a titration, one reactant(the titrant) is placed in a buret. The other reactant is placed in a flask along with a few drops of an indicator.
• The titrant is slowly added to the contents of the flask until the indicator changes color (the endpoint or equivalence point).
• If the indicator has been chosen properly, the endpoint tells us when the reactants are present in stoichiometric proportion.
• A titration may be based on any of the previously discussed types of reactions …
Titrations (cont’d)
Acid Base Titrations
• A solution of accurately known concentration, called a standard solution, is added gradually to another solution of unknown concentration, until the chemical reaction between the two solutions is complete (the equivalence point).
• Indicators are substances that have distinctly different colors in a basic or acidic environment which are used to indicate the equivalence point.
04/12/23 44
Acid Base Titration Theory
• If we know the volumes to the standard and the unknown solutions….
• Along with the concentration of the standard solution
• We can calculate the concentration of the unknown.
04/12/23 47
• … are not new to us.• We simply apply the method of stoichiometry
calculations (that we have already done) to the titration.
• Titration calculations for acid–base, precipitation, redox, and other types of titrations are very similar.
• Recall that the objective of a titration is to determine the number of moles, or the number of grams, or the percentage, or the concentration, of the analyte.
Titration calculations …
An Acid–Base TitrationA measured portion of acid solution is placed
in the flask, and an indicator is added.
Base solution of known concentration is slowly added from the buret.
When the indicator changes color, we have added just enough base to
react completely with the acid.
A Precipitation Titration
An unknown concentration of chloride ion is being titrated …
… with silver nitrate solution.
The indicator is orange dichromate ion; white AgCl precipitates.
When the chloride has reacted completely …
… the next drop of Ag+ solution produces brick-red silver dichromate.
A Redox TitrationDeep-purple MnO4
– is the titrant …
… and Fe2+ is being titrated.
During titration, Mn2+ and Fe3+ (nearly colorless) are produced.
After the Fe2+ has been consumed, the next drop of MnO4
– imparts a pink color.
WRITE THE CHEMICAL EQUATION!
volume red moles red moles oxid M oxid
0.1327 mol KMnO4
1 Lx
5 mol Fe2+
1 mol KMnO4
x1
0.02500 L Fe2+x0.01642 L = 0.4358 M
M
red
rxn
coef.
V
oxid
5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2O
16.42 mL of 0.1327 M KMnO4 solution is needed to oxidize 25.00 mL of an acidic FeSO4 solution. What is the molarity of the iron solution?
16.42 mL = 0.01642 L 25.00 mL = 0.02500 L
What volume of a 1.420 M NaOH solution is required to titrate 25.00 mL of a 4.50 M H2SO4 solution?
WRITE THE CHEMICAL EQUATION!
volume acid moles red moles base volume base
H2SO4 + 2NaOH 2H2O + Na2SO4
4.50 mol H2SO4
1000 mL solnx
2 mol NaOH
1 mol H2SO4
x1000 ml soln
1.420 mol NaOHx25.00 mL = 158 mL
M
acid
rxn
coef.
M
base
End of Chapter 4
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