Acid/Base Theories

Acid/Base TheoriesAcids:Bases:Arrhenius’ TheoryAcids – Ionization – any process by which a neutral atom or molecule is converted into an ionBases –Dissociation – the separation of ions that occurs when an ionic compound dissolves in H2O.• this accounts for the fact that both acids and bases are ______• cannot account for acidic solutions that are not in water • does not explain acids that don’t have ___ or bases that don’t contain ______. The Brønsted-Lowry TheoryAcid –Base – • Acids and Bases have to react ______• there has to be an ____ present to donate to a _____ that accepts it.• now H2O is a ______not just the solvent eg. + • H3O is responsible for the acidic properties of H2O eg. − • OH is responsible for the basic properties of H2OWater is amphoteric, it can act as an acid or a base. − − − • some others include HCO3 , HSO4 , HS , NH3 eg. eg. Strong Acids/Bases• strong or weak refers to the electrolyte’s conductivity.• the stronger the acid/base, the weaker the conjugate• Percent ionization gives a measure of the acid/base strengthStrong:Acids:Bases:Weak :Acids: Bases:Relative Acid/Base Strength:1) Acids vs Bases • both NaOH and HClO have an ______• why is one a ______and the other an ______?• for NaOH, the ionic bond creates a stronger ______force than the polar covalent bond does with H2O• for HClO, the very polar HO bond has stronger ______interactions with H2O than the dipole – dipole forces of the O-Cl bond does.2) Binary Acids – HCl vs HI• even though HCl is ______polar than HI, I− ion is larger than Cl− ion which results in a ______ionic bond, so HI takes _____ energy to ionize.• even though ______energy is required to separate − more H2O molecules from each other, the larger I ion is surrounded by ______H2O molecules where each ______interaction releases energy (solvation energy)• the ______net energy release occurs with the ionization of HI and thus its acid strength is ______.3) Polyprotic Acids• have more than 1 ______H’s, eg. H2SO4.• the first ionization step is ______more acidic.− • once the the ______anion is formed (HSO4 ), the other O ─ H bond is not as ______due to the charge on the ion• as a result, the second or third H is harder to ______4) OxyAcids • as the # of O’s bonded to the central atom ______, the degree of ______(polarization) away from the O ─ H bond increases. • this makes the O ─ H bond ______polar and allows for a greater ______of the H.• HClO4 is the ______strong oxyacid5) Competition for H• Weak Acids and Bases are ______systems• the amount of ionization depends on which ______for the H is stronger, the original ______or H2OWater, pH and Strong Acids/BasesSelf – Ionization of Water1) System as H2O is amphoteric, (acts as acid or base), it does so even with itself, due to random collisions:+ − H2O (l) + H2O(l) ↔ H3O (aq) + OH (aq) acid1 base1 acid2 base2  neutral H2O conducts electricity,barely! measured using a pH meter the equilibrium exists in all aqueous sol’ns2) Addition of a Strong Acid/Base 2 systems operating+ − 1. H2O (l) + H2O(l) ↔ H3O (aq) + OH (aq)+ − 2. HCl(aq) + H2O(l)  H3O (aq) + Cl (aq)  HCl is a ______acid it produces much more ______than H2O the self ionization of water is ______ the major species is ______as the sol’n will be ______ the self ionization of water’s is important whenever + -7 a system’s [H3O ] approaches the Kw of 1.0 x10 Similar argument for Strong Bases an the OH− pH, pOH and pKw1) Definition for significant figures:# of decimal places in pH = # of s.f. in [ ]2) Measuring pH pH Meters are probes that measure the electrical conductivity + of [H3O ]Acid – Base Indicators Are weak acid/base equilibrium systems where the acid conjugate bases are different colors+ − HIn(aq) + H2O(l) ↔ H3O (aq) + In (aq) Colour1 Colour 2 these colour changes occur at the pH determined by their equilibrium constant, Kin eg. Determine the pH where the colour changes if the KIn = 2.00 x 10-10 .Weak Acids/Bases 1) General FormAcids:Bases:2) Percentage Ionization, pDefinition: eg.Calculate the percentage ionization of a 0.10 M solution of hydrofluoric acid whose pH = 3.96. + − HF + H2O ↔ H3O + F3) Effect of DilutionFor, + − HA + H2O ↔ H3O + A an  [H2O] • the acid strength does increase• but this doesn’t change the acidity greatly as the volume of solution has increased.4) Ka, Kb and Acid/Base StengthWeak Acid+ − HA + H2O ↔ H3O + A as  Ka, p, acid strengthWeak Base + − B: + H2O ↔ HB + OH as  Kb, p, base strength 5) Acid/Conjugate Base Strength eg. + − HClO4 + H2O  H3O + ClO4 ; and− − ClO4 + H2O ↔ HClO4 + OH ; ______acids have ______conjugate bases ______bases have ______conjugate acidsWeak ______have weaker ______Weak ______have weaker ______6) Ka and Kb for Conjugate Acid/Base PairsDerivation + − NH3 + H2O ↔ NH4 + OH ; + + NH4 + H2O ↔ NH3 + H3O ; • NH3 is classified as a base as Kb > Ka To find Ka from Kb (or the other way around): as from above: • if one value is known, then the other can be calculated. 3+ -5 eg. If for [Al(H2O)6] the Ka is 1.4 x 10 , calculate 2+ the Kb for [Al(H2O)5OH] .Determination of Ka, Kb and pHSolving Acid/Base Problems 1. List the major species in solution2. Look for reactions that are not equilibrium systems.+ − − 3. For above, determine [H3O ] or [OH ] or [A ].4. Determine the equilibrium systems and pick the equilibrium that will control the pH. + − − 5. Calculate the [H3O ]i or [OH ]i or [A ]I from [HA]i or [B:]i and the values from #3 above.6. Write the equation for the rxn and the Ka or Kb.7. Create an I.C.E. table, calculate 100 Rule.8. Substitute [ ]eq into Ka or Kb.9. Solve for the pH or [ ]eq or Ka or Kb as required. 1) Determining Ka or KbFrom pH eg. Calculate the Ka for formic acid (HCO2H) if a 0.100 M solution has a pH of 2.38 From p + − • different determination of [H3O ]eq or [OH ]eq. eg. Calculate the Kb for NH3 if a 0.150 M solution has a p of 7.8 %.2) Determining pH from Ka or Kb eg. Calculate the pH of a 0.200 M acetic acid sol’n. 3) Polyprotic Acids• ionize only ______at a time.• each step has its own Ka, designated Ka1, Ka2, etc.• Ka1 is the ______and sets the pH of the sol’n. eg. Calculate the pH of 2.500 M H2CO3 and the − 2− [H2CO3]eq, [HCO3 ]eq, [CO3 ]eq. Hydrolysis of Salts• when salts ______in water, the resultant hydrated ions may ______further with Water• the pH of the sol’n may ______• this reaction with water is called ______For the following 0.100 M solutions:Na2CO3 predicted pH? < 7 7 > 7 NH4Cl predicted pH? < 7 7 > 7NaC2H3O2 predicted pH? < 7 7 > 7NH4C2H3O2 predicted pH? < 7 7 > 7NaC2O4 predicted pH? < 7 7 > 7 NaHCO3 predicted pH? < 7 7 > 7Al(NO3)3 predicted pH? < 7 7 > 7Lewis Acid/Base TheoryLewis Acids:Lewis Bases: Acid/Base Titrations1) TerminologyTitration: the precise addition of a solution in a ______into a measured volume of a sample sol’n Titrant: the solution in the ______Sample: the solution being ______Titre: the volume of ______addedEquivalence Point: the titre at the ______addition pointEnd Point: the point during a titration at which a sharp visible characteristic ______changes- the indicator colour change- large change in the pH meter readingsStrong (weak) acid/ strong (weak) base: acid is the ______base is the titrantStrong (weak) base/ strong (weak) acid: base is the sample, acid is the ______2) SymbolsA, B are subscripts used for ______acids, ______bases. a, b are subscripts used for _____ acids, ______bases 3) Strong /Strong Titrations 1st – the equivalence point is calculated using stoichiometry.For: then pH is determined at 4 stages of the titration: a) Initial pH – before addition of any titrant b) At volume of titrant added that is half way to the equivalence point - V½ eq.pt. c) At equivalence point (nA = nB) - Veq.pt. d) After equivalence point - excess titrant, usually V1½ eq.pt. e) Sketch the curve eg. Create a titration curve for the neutralization of 50.0 mL of a 0.100 M HCl sample with 0.250 M NaOH solution.Equivalence point: a) Initial pH b) At V½ eq.pt. = ______c) At Veq.pt. = ______ d) At V1½ eq.pt. = ______ e) Sketch the curve 4) Weak/Strong Titrations 1st – the equivalence point is calculated using stoichiometry.For: then pH is determined at 4 stages of the titration: a) Initial pH – before addition of any titrant b) At volume of titrant added that is half way to the equivalence point - V½ eq.pt. c) At equivalence point (nA = nB) - Veq.pt. d) After equivalence point - excess titrant, usually V1½ eq.pt. e) Sketch the curve eg. Create a titration curve for the neutralization of 50.0 mL of a 0.100 M HC2H3O2 sample with 0.250 M NaOH solution.Equivalence point: a) Initial pH b) At V½ eq.pt. = ______c) At Veq.pt. = ______ d) At V1½ eq.pt. = ______ e) Sketch the curve STRONG ACID WEAK ACID a) Initial pH b) At V½ eq.pt. c) At Veq.pt. d) At V1½ eq.pt. Buffers and the Common Ion EffectBuffers: contain ______amounts of HA and A− , (B: and HB+) as both components are ______, shifting of the equilibrium is ______. these soln’s act as ______(buffers) against + − large pH changes when H3O or OH are added. uses the Henderson – Hasselbalch eq’n which is a ______of the Ka eq’n (when the 100 Rule applies)Derivation: Now the Vtot is the same, so: Example of the effectiveness of a Buffer: 1) Normal solution, not a Buffer: a) Calculate the pH of 50.0 mL of a 0.200 M ascorbic acid, HC6H7O6 sol’n. b) What is the pH when 2.0 mL of 1.5M NaOH sol’n are added? (Vtot is the same, calculate na and nb) 2) Buffer sol’n – Comparison a) Calculate the pH of 50.0 mL of a 0.200 M ascorbic acid, HC6H7O6 sol’n and 0.180 M sodium ascorbate NaC6H7O6 sol’n. b) What is the pH when 2.0 mL of 1.5 M NaOH sol’n are added? (Vtot is the same, calculate na and nb)

Acid/Base Theories

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