CH 15 Summary Equilibrium is a balance between products and reactants Use stoichiometry to determine reactant or product ratios, but NOT reactant to product ratios. Capital K is used to represent the equilibrium constant Products over reactants raised to their stoichiometric coefficients Calculated from balanced equation, subscript designates units used, K c , K p No units used in final written K Equlibrium Calculations and Reaction Quotients ICE tables used to manipulate initial and equilibrium concentrations. Factors influencing K Concentration of chemicals and temperature affect all equilibria T, P, V changes affect K p in gases Catalysis Effect of catalysts and inhibitors, reasons they are used. 1
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CH 15 SummaryEquilibrium is a balance between products and reactants
Use stoichiometry to determine reactant or product ratios, but NOT reactant to product ratios.
Capital K is used to represent the equilibrium constantProducts over reactants raised to their stoichiometric coefficientsCalculated from balanced equation, subscript designates units used, Kc, KpNo units used in final written K
Equlibrium Calculations and Reaction QuotientsICE tables used to manipulate initial and equilibrium concentrations.
Factors influencing K Concentration of chemicals and temperature affect all equilibriaT, P, V changes affect Kp in gases
CatalysisEffect of catalysts and inhibitors, reasons they are used.
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Polyatomic IonsMemorize These!
Ammonium NH4+ Nitrate NO3
-
Hydronium H3O+ Nitrite NO2-
Acetate CH3COO- Phosphate PO43-
Carbonate CO32- Cyanide CN-
Permanganate MnO4- PerchlorateClO4
-
Hydroxide OH- Sulfate SO42-
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Chapter 4: Solution Chemistry
MolarityConversions between moles and liters
DilutionsCalculate concentration of diluted solution from a stock solution M1V1=M2V2Calculate mass of solid needed to make a solution
Acid Base TitrationsIdentifying acids and basesKnow properties of acids and basesKnow list of specific acids both names and formulasDetermine concentration of unknown solutions using titration
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Common Acids to MemorizeStrongHydrochloric Acid: HClSulfuric Acid: H2SO4Nitric Acid: HNO3Perchloric Acid: HClO4Hydrobromic Acid: HBrHydroiodic Acid HI
Bronsted Acids and BasesAcid Compound that loses H+ to a base
Base Compound that gains H+ from an acid
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SA WB CA CB
WB WA CA CB
Conjugate Acid-Base PairAcid-Base pair exchanging the H+Conjugate acid: Product with the extra H+
Conjugate base: Product with 1 less H+ ion than reactant
Carboxylic Acids: -COOHWeak organic acids:
COOH group on molecule is acidicCreates resonance structureStabilizes anion
Never fully dissociate in waterWill always be an equilibrium reaction
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Reactivity of Weak Acid and Bases
Strong Acids and Bases: Full dissociation Conjugate bases and acids form spectator ionsCan use Chem 101 stoichiometry in calculationsNo original reactant or product left in solution
Weak Acids and Bases: Partial dissociationForms an equilibrium: Ka or Kb
Acid/Base strength in aqueous solutionsH3O+ is the strongest acidOH- the strongest baseWater acts as weak acid or base in the reaction
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Acid-Base Properties of Water: Autoionization
Water is slightly conductive due to the following reaction:
2H2O(l) H3O+(aq) + OH–(aq)Process is called Autoionization
Acid-Base Reaction between identical molecules1 molecule acts as an acid, the other a base
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Calculation of [H3O+] and [OH–] in WaterTreat as an equilibrium reaction at 25° C
H2O(l) + H2O(l) H3O+(aq) + OH–(aq) WB WA SA SB
Kw = [H3O+][OH–] = 1.0×10–14 Ion-Product constant Kw is very small: Favors weaker acid-base pair (H2O)
Make ICE tableProducts [H3O+] [OH–]Initial 0 0Change +x +xEquilibrium x x
Solve for [H3O+] and [OH-]1.0 x 10-14 = [H3O+][OH–] = [x][x] = x2
x = [H3O+] = [OH–] = 1.0×10–7 M pH = 7 of pure water
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pHMethod of measuring acidity (p)ower of the (H)ydrogen ion
CalculationspH = -log[H3O+] [H3O+] = 10(-pH)
pOH = -log[OH-] [OH-] = 10(-pOH)
Kw = [H3O+][OH-] = 1x10-14MpKw=pH + pOH = 14
Effects in 1M Strong Acid (pH = 0)[H3O+]= 1M then [OH-]= 1x10-14M
Effects in 1M Strong Base (pH= 14)[OH-]= 1M then [H3O+]= 1x10-14M
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pH and pOHCalculations
Strong Acids and Bases
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pH and pOH CalculationspOH: (p)ower of the (OH-) ion = 14.00-pH1. Find pH and pOH of an 0.0050M HBr at 25°C.
Acidity Calculations for Dilute SolutionsCalculate the H3O+(aq) + OH– (aq) of a 1.00×10–6M
solution of NaOH at 25°C?Write reactions:
NaOH(aq) Na+(aq) + OH–(aq) [OH–] = 1.00×10-6M 2H2O(l) H3O+(aq) + OH–(aq) [OH–] = 1.00×10-7M[OH–] < 1.00×10-7M because of Le Chatelier’s Principle
Total concentrations:[OH–] = 1.00×10–6 + 0.10×10-6M (10%)NOT less than 1.00×10–7 not negligibleMust use an ICE table if solution less than 10-6M
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Strength of Acids and Bases
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Strength of Acids and BasesStrong acids and bases are strong electrolytes
Completely ionized in water, no original compound leftGood conductors of electricity.Directional arrow (→) indicates dissociation is complete
Weak acids and bases are weak electrolytesPartial ionization in water, original compound remainingPoor conductors of electricityDouble arrow (↔) indicates dissociation is incompleteGoverned by an equilibrium constant, Ka or Kb
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Strong vs. Weak Acids
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Weak Acids and Bases
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Stronger acids will dominate over weaker acidsHNO2(aq)+ CN-(aq)HCN (aq)+ NO2
- (aq) K>1
If paired with group 1 cation, will be strong base
Weak Acids and Ka
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Weak Acids, Ka and pKaALWAYS write a reaction of weak acid HA in water:HA(aq) + H2O(l) H3O+(aq) + A–(aq)
The equilibrium constant for this reaction is:
Ka is the acid ionization constant.Quantitative measure of acid strengthLarge Ka :Stronger acid
pKa = -log Ka
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ac KHA
AOHK ==−+
][]][[ 3
Ka Values of Common Weak Acids
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Ka Calculations0.100 mole of HF is dissolved in 1.00 L of water at 25 °C. The pH at equilibrium was found to be 2.08. Calculate Ka.HF(aq)+ H2O(l) H3O+(aq) + F–(aq)Make table [HF] [H3O+] [F-]
More double bonded oxygens: Stronger acidH2SO4 H2SO3
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Two terminal oxo atomsMore resonance stabilizes ionMore e- withdrawal from O-H
H is more positive: more acidic
One terminal oxo atomFewer resonance structuresLess e- withdrawal from O-H H is less positive: less acidic
Effect of ElectronegativityCharge:
No differenceBoth neutral
Structure: No difference in terminal oxo atoms1 oxo atom each
Electronegativity: S more electronegative than P
H2SO3 more acidic than H3PO4
S more electronegative than SeH2SO4 more acidic than H2SeO4
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Acid Strength
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Bond lengthvs
Electronegativity
Charge Difference
Structure:Terminal O
Electronegativity
Acid-Base Properties of Salts
Hydrolysis
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Hydrolysis of Ionic Salts
Ionic salts dissolved in water affect pHNaNO3 (s) + H2O(aq) Na+(aq) + NO3
- (aq)
Undergo HydrolysisHydrolysis: Reaction of an ionic salt with waterMay change the pH of the solution
Both cations and anions may undergo hydrolysis Not all ions hydrolyzeExamine both ions to determine acid/base character
3 step process to predict acidity of a salt solution1. Write the reaction that dissociates salt into its ions2. Check the cation for acid hydrolysis: produce H+
3. Check the anion for base hydrolysis: produce OH-
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Is NaCl (aq) Acidic, Basic or Neutral
1. Dissociate the salt into ions:NaCl(aq) Na+(aq) + Cl–(aq)
2. Check the cation for hydrolysis:Na+(aq) + H2O(l) NaOH(aq) + H+(aq) No Hydrolysis: NaOH strong base, not acidic
3. Check the anion for hydrolysis:Cl–(aq) + H2O(l) HCl (aq) + OH-(aq)
No Hydrolysis: HCl is a strong acid, not basicIf neither acidic or basic, solution is neutral
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Is Na2CO3 Acidic, Basic or Neutral1. Dissociate the salt into ions:
Na2CO3 (aq) 2Na+ (aq) + CO32- (aq)
2. Check the cation for hydrolysis:Na+(aq) + H2O(l) NaOH(aq) + H+(aq) No Hydrolysis: NaOH would be a strong baseNo H+ generated: Solution is not acidic
3. Check the anion for hydrolysis:CO3
2-(aq) + H2O(l) HCO3- (aq) + OH- (aq)
Anion hydrolyzes: HCO3- (aq) weak acid
Some OH- is generated: Solution may be basicIf not acidic but possibly basic, solution is basic 63
Is Fe(NO3)2 Acidic, Basic or Neutral?1. Dissociate the salt into ions:
Fe(NO3)2(aq) Fe2+(aq) + 2NO3–(aq)
2. Check the cation for hydrolysis:Fe2+(aq) + H2O(l) FeOH+(aq) + H+(aq) Reaction occurs: FeOH+ is a weak baseH+ generated: Acidic solution
3. Check the anion for hydrolysis:NO3
–(aq) + H2O(l) HNO3 (aq) + OH-(aq)No OH- generated: Nitric is a strong acid, not
basicOverall: the solution is acidic
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Is ZnF2 Acidic, Basic or Neutral
1. Dissociate the salt into ions:ZnF2(aq) Zn2+(aq) + 2F–(aq)
2. Check the cation for hydrolysis:Zn2+(aq) + H2O(l) ZnOH+(aq) + H+(aq) Reaction Occurs: ZnOH+(aq) is a weak
baseH3O+generated: Acidic Solution
3. Check the anion for hydrolysis:F–(aq) + H2O(l) HF (aq) + OH-(aq)Reaction Occurs: HF is a weak acidOH- is generated: Basic Solution
Can’t tell acidity : Both cation and anion hydrolyze
Both H3O+(aq) and OH-(aq) possible65
Is ZnF2 Acidic, Basic or Neutral1. Dissociate the salt into ions:
ZnF2(aq) Zn2+(aq) + 2F–(aq)2. Cation Hydrolysis: Ka based