Chapter 4 Chemical Reactions. HW #3 Due by Monday, 5/5/2014 HW #4 due by Friday, 5/9/2014 Quiz #2: Monday, 5/12/2014 Exam #2: Wednesday, 5/14/2014. End-of-Chapter Homework: pp 167 - 177. 5, 6, 9-11 14-18 19 (a,d,g,h) 28 29 31 32 333841 43
Chapter 4Chemical Reactions
HW #3 Due by Monday, 5/5/2014
HW #4 due by Friday, 5/9/2014
Quiz #2: Monday, 5/12/2014
Exam #2: Wednesday, 5/14/2014
5, 6, 9-11 14-18 19(a,d,g,h) 28
29 31 32 333841 43
47 51 5557 59 61 65b 67
69 71 737981 8389 100
- Many Ionic compounds dissolve in water yielding ions.
-These soluble ionic compounds conduct electricity due to the ions. Are called electrolytes.
- Driving force is water being attracted to the ions, forming weak bonds, & releasing energy.
1) Nonelectrolyte (does not conduct electricity in water):
a) Molecular & Soluble - when dissolves, yields no ions - CH3OH
b) Ionic & Insoluble like AgCl or PbI2
2) Strong Electrolyte (Ionic, Soluble, Conducts well in water)
NH4Cl(s) -----) NH4+(aq) + Cl-(aq)
3) Weak Electrolyte (few % ionization in water; Weak Conductor)
NH3 + H2O ------) NH4+ + OH- (~ 1% ionized)
HC2H3O2 --------) H+ + C2H3O2- (~ 1 % ionized)
NaI in H2O – Soluble & Ionic; Many Ions
H2O – Molecular & No Ions
Nonelectrolyte – No Current
Electrolyte – Much Current
#1 Grp 1, NH4+, C2H3O2- &NO3- compds are soluble (except H).
#2 Cl- Br- I-compds are solubleexcept : Ag+ Pb2+ Hg22+
#3 SO4-2solubleexcept : Ca2+ Sr2+ Ba2+ Ag+ Pb2+ Hg22+
#4 OH-compds are insolubleexcept#1 & Ca2+ Sr 2+ Ba 2+
#5 Most CO3-2 (carbonates), PO4-3 (phosphates), & S-2 (sulfides) are insoluble except for #1, (rule 1).
[Note: Ag ion is always Ag+1 does not need (I); Solvent is H2O]
silver nitrate + sodium chloride --------) silver chloride + sodium nitrate
1) Molecular Equation (ME):Write all as if not ionized
AgNO3(aq) + NaCl(aq) --------) AgCl(s) + NaNO3(aq)
2) Complete Ionic Equation (CIE):Write soluble,
ionic compounds as ions; should use subscripts like (aq)
Ag+ + NO3- + Na+ + Cl- ----) AgCl(s) + Na+ + NO3-
3) Net Ionic Equation (NIE):Eliminate Spectator Ions
Ag+(aq) + Cl-(aq) -----) AgCl(s)
- Formation of an insoluble compound (precipitate, ppt) is a driving force in a chemical reaction.
- Can predict a product from the solubility rules.
Method:Write soluble ionic reactants as ions. Check solubility rules & see if insoluble product forms from ions.
Example: Predict product when mix K2SO4(aq) & BaCl2(aq) Write CIE & NIE.
2K+ + SO42- + Ba2+ + 2Cl- ---) BaSO4(s) + 2K+ + 2Cl-(CIE)
Ba2+(aq) + SO42-(aq) -----) BaSO4(s)(NIE)
1. Definitions(Note: Proton = H+)
Arrhenius:Acid =substance which releases H+ in water.
Base =substance which releases OH- in water.
Examples: HCl(aq) -----) H+(aq) + Cl-(aq)
NaOH(aq) -----) Na+(aq) + OH-(aq)
Bronsted-Lowry:Acid = Proton Donor
Base = Proton Acceptor
Example:(air is the solvent)
NH3(g) + HCl(g) -----) NH4Cl(s)
2. PropertiesNote: salt = any ionic compound
Acids: - Sour taste
- Turn Indicator dyes a specific color:
Phenolphthalein colorless; Litmus red
- React with bases to yield salt + H2O
Bases:- Bitter Taste, Feel Slippery
- Turn Indicator dyes a specific color:
Phenolphthalein red; Litmus blue
- React with acids to yield salt + H2O
- Strong Acids/Bases ionize 100% in water
- Weak Acids/Bases ionize < < 100% in water
3. Examples (Know these strong & weak acids & bases)
Strong Acids:HCl HBr HI HNO3 H2SO4 HClO4
Strong Bases: NaOH KOH Ba(OH)2 (but not too sol.)
Example in water: HCl(aq) -----) H+ + Cl-100% to the right
Weak Acids: HF HC2H3O2 HCN H2CO3
Weak Bases:NH3[= NH4OH (in water)]NaHCO3
Examples in water: NH4OH NH4+ + OH-~1% to the right
HC2H3O2H+ + C2H3O2-~1% to the right
4. Acid-Base or Neutralization Reactions
- Acids react with bases to give: salt(ionic compd) + H2O
- Driving force is to produce stable H2O& sometimes a gas.
- Examples:NaOH + HF -----) NaF + H2O
2KOH + H2SO4 ---) K2SO4 + 2H2O
H+ + HCO3- -----) CO2(g) + H2O
2H+ + CO32------)CO2(g)+ H2O
2H+ + SO32- -----) SO2(g) + H2O
2H+ + S2- -----) H2S(g)
H+ + CN- -----) HCN(g)
NH4+ + OH------) NH3(g)+ H2O
5. Write three rxn formats (ME, CIE & NIE) for reaction of potassium hydroxide with hydrochloric acid:
ME:KOH + HCl -----) KCl + H2O
CIE:K+ + OH- + H+ + Cl- -----) K+ + Cl- + H2O
NIE:H+(aq) + OH-(aq) -----) H2O(l)
Note: 1) The above net ionic equation is common for many acid-base reactions. The production of stable H2O is a driving force for the reaction.
2) H+ in water is better represented by H3O+
3) H3O+is a polyatomic ion = hydronium ion
5. ME, CEI and NIE
- Write the ME, CIE & NIE for: HCl + NaHCO3(in water)
ME: HCl + NaHCO3 -----) NaCl + H2O + CO2
CIE: H+ + Cl- + Na+ + HCO3- ---) Na+ + Cl- + H2O + CO2
NIE: H+(aq) + HCO3-(aq) -----) H2O(l) + CO2(g)
6. Titration - Process of adding known (amount & concentration) of Base (or Acid) in a burette to an unknown concentration of Acid (or Base). End point is determined with an indicator or a pH meter. Can now determine the amount of the unknown (quantitative analysis).
Known M & Volume
After titn can calculate moles & M.
(1) aprecipitate(solubility rules)
(2) agas(six gas equations)
(3) anonionized molecule like H2O or HC2H3O2
1) Ba+2 + I- + Ag+ + NO3- -----) AgI(s) + Ba+2 + NO3-
Ag+(aq)+ I-(aq) -----) AgI(s)(Balanced NIE)
2) H+ + Cl- + Na+ + CN- -----) HCN(g) + Na+ + Cl-
H+(aq) + CN-(aq) -----) HCN(g)(Balanced NIE)
1. Definitions & Introduction:
- Oxidation Reduction Reaction (Redox Rxn) is one in which electrons are transferred.
- Oxidation = Loss of Electrons
- Reduction = Gain of Electrons
2. Examples: (Note: electrons lost & gained must be same)
2Mg + O2 -----) 2MgO
[ 2Mgo ---) 2Mg2+ + 4e- = Oxidation ] Half Rxn
[ O2o + 4e- -----) 2O2- = Reduction ] Half Rxn
3Cu+2 + 2Al -----) 3Cu + 2Al+3
[ 2Alo -----) 2Al+3 + 6e- = Oxidation] Half Rxn
[ 3Cu+2 + 6e- -----) 3Cuo = Reduction] Half Rxn
Note: Oxidizing agent causes oxidation
Reducing agent causes reduction
3. Uses of RedoxRxns:a) Create new chemicals
b) Theoretical Importance (chem 123)
c) Battery – allow the e’stransferred to perform work
4. Oxidation Numbers (ON):
Definition:ON = the charge that an atom or group of atoms would have if they were ionic.
Uses:1)Redox Rxn(ID & what’s been oxid/red & #e-)
2) NomenclatureFeF2 = Iron(II)Fluoride; (II) = ON
3) Used in balancing redox rxns(chem 123)
Rules: Know the following rules [Order gives priority]
5. Rules for Determining ON (Know)
Note: When is a conflict, 1st rule controls result.
a) The sum of ON’s add up to give the charge
b) The ON of a neutral element by itself is 0
c) Group 1, 2, 13 ions are +1, +2 & +3 respectively;
(exception: H is -1 when combined with metal: KH , NaH)
d) F is -1
e) O is -2 (exception: O is -1 when found as a peroxide, O22-)
f) Cl, Br, I are -1 [when together like BrCl, most electronegative = -1]
Examples:Determine the ON of single N in each:
N2 N3-1 Na3N N2O NO NF3 NO2 NO3-1
0-1/3 -3 +1 +2 +3 +4 +5
- Which of the above forms of N can explosively react with organics? Why?
- Determine ON of each atom in FeSO3(Note: SO3-2 ON = -2)
[+2 +4 -2]
1) Mg + Cl2 -----) MgCl2
2) CH4 + 2 O2 -----) CO2 + 2 H2O
3) HCl + NaOH -----) H2O + NaCl
6. Redox Reaction Types
Combination: 2Al + 3F2 -----) 2AlF3
Decomposition: 2HgO -----) 2Hg + O2
Combustion: CH4 + 2O2 -----) CO2 + 2H2O
Single Replacement:2Na + 2HCl -----) 2NaCl + H2
[ Can predict reaction from table 4.6, Pg 153;
The more active metal will replace the cation ]
7. Balancing redox rxns: Can become difficult; so, rules were generated & will cover in Chemistry 123.
A. Molarity, M
- Many reactions take place in solution & we need a
way to measure concentration; Mis common way.
- M = moles of Solute M = m(use as eqn)
L Solution L
solvent = substance present in largest amount solute = substances dissolved in solventsolution = solute + solvent (everything present)
B. Molarity Calculations
1) Calculate the M if 4.5 moles of HCl are dissolved in a total volume of 500 mL (0.500 L)
M = m = 4.5 moles = 9.0 mHCl = 9.0 MHCl L 0.500 L L
2) Calculate M if 0.020 g of CaCO3 are dissolved in 40 L
a) convert g CaCO3to mols; b) plug into M eqn; M = m/L
0.020 g CaCO3 x 1 mole CaCO3/100. g CaCO3 = 2.0x10-4 mol CaCO3
M = moles = 2.0x10-4 mol CaCO3 = 5.0x10-6M CaCO3
L 40 L
3) How many moles of HCl are present in 2.0 L of 0.30 MHCl? M = moles/Lmoles = M x L
moles= 0.30 m HCl x 2.0 L = 0.60 moles HCl
4) How many L of 12 MHCl will provide 2.0 moles of HCl ?
M = m L = m = 2.0 mole HCl = 0.17 L HCl
LM 12 mole/L
- Frequently we are given a concentrated solution & we need to add solvent to get a more dilute solution. This is a dilution problem.
M1 x V1 = M2 x V2 Note: 1) Initial moles = Final moles
2) Works with any conc. or vol. units
3) Use eqn ONLY for dilution problem
Example: How many mL of 12 M HCl are needed to make 100. mL of 3.0 M? M1 x V1 = M2 x V2
V1 = M2 x V2= (3.0 M x 100. mL) = 25 mL
- Take 25 mL of 12 M HCl and add enough water to make 100 mL.
D. Gravimetric Analysis - A type of analysis where one converts the analyte to an insoluble compound which is then isolated, dried & weighed.
Example: A 1960 dime (2.50g) is dissolved in nitric acid. HCl is then added to form insoluble AgCl, which is collected, washed, dried & weighed. Calculate a) the g of Ag & b) w/w % Ag in the dime if 2.98 g of AgCl was obtained.Ag ---) Ag+1 ---) AgCl
a) g Ag in dime:
2.98 g AgCl 1 mol AgCl 1 mol Ag 108 g Ag = 2.25 g Ag
143 g AgCl1 mol AgCl1 mol Ag
b) % Ag in dime:
w/w % Ag = g Ag x 100%= 2.25 g Ag x 100% = 90.0%
g Dime 2.50 g
E. Volumetric Analysis
- An analysis technique where the M & volumeof a reagent in a burette is used to calculate the mols or M of an unknown.
- NOTE: This is a typical stoichiometric problem except that we calculate mols of reagent from M & Linstead of from grams.
Example: 25.0 mL of HCl is titrated with 45.5 mL of 0.433 MNaOH. Calculate the a) mols of HCl present & b)M of the original HCl.
1 HCl + 1 NaOH -----) 1 NaCl + 1 H2O
a) m 0.433 m NaOH 0.0455 L NaOH 1 m HCl = 0.0197 m HCl
L NaOH 1 m NaOH
b) M = moles HCl = 0.0197 m HCl = 0.788 M HCl
L 0.0250 L
1) Moles HA (12.1 mL NaOH = 0.0121 L)
0.500 m NaOH x 0.0121 L = 0.00605 m NaOH
0.00605m NaOH x 1m HA/1m NaOH = 0.00605 mHA
2) M HA(10.0 mL HA = 0.0100 L HA)
M= m = 0.00605 m HA = 0.605 m/L HA
L 0.0100 L HA
3) w/w % HA = (g HA / g Soln) x 100%
g HA = 0.00605m x 60.05 g/m HA = 0.3633 g HA
w/w% HA = (0.3633g HA /10.3 g soln) x100 = 3.53%
Pb(NO3)2 + HI ----)
Pb+2 + NO3- + H+ + I- ----) PbI2(s) + H+ + NO3-
Pb+2(aq) + 2I-(aq) ----) PbI2(s)
-? Mols PbI2 from 0.10 m HI. 0.10 m I-1 m PbI2 = 0.050 m PbI2
2 m I-
- ? Mols PbI2 from 2.0 L of 0.050 m/L HI?
0.050 m HI 2.0L 1 m PbI2 = 0.050 m PbI2
L 2 m HI