Acid/Base Theories
Acids:
Bases:
Arrhenius’ Theory
Acids –
Ionization – any process by which a neutral atom or molecule is converted into an ion
Bases –
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 Theory
Acid –
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 H2O
Water 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 strength
Strong:
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 oxyacid
5) Competition for H
• Weak Acids and Bases are ______systems
• the amount of ionization depends on which ______for the H is stronger, the original ______or H2O
Water, pH and Strong Acids/Bases
Self – Ionization of Water
1) 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’ns
2) 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 pKw
1) 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 Form
Acids:
Bases:
2) Percentage Ionization, p
Definition:
eg.Calculate the percentage ionization of a 0.10 M solution of hydrofluoric acid whose pH = 3.96. + − HF + H2O ↔ H3O + F
3) Effect of Dilution
For,
+ − 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 Stength
Weak Acid
+ − HA + H2O ↔ H3O + A
as Ka, p, acid strength
Weak 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 acids
Weak ______have weaker ______Weak ______have weaker ______
6) Ka and Kb for Conjugate Acid/Base Pairs
Derivation + − 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 pH
Solving Acid/Base Problems 1. List the major species in solution
2. 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 Kb
From 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 > 7
NaC2H3O2 predicted pH? < 7 7 > 7
NH4C2H3O2 predicted pH? < 7 7 > 7
NaC2O4 predicted pH? < 7 7 > 7 NaHCO3 predicted pH? < 7 7 > 7
Al(NO3)3 predicted pH? < 7 7 > 7
Lewis Acid/Base Theory
Lewis Acids:
Lewis Bases: Acid/Base Titrations
1) Terminology
Titration: 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 ______added
Equivalence Point: the titre at the ______addition point
End Point: the point during a titration at which a sharp visible characteristic ______changes
- the indicator colour change
- large change in the pH meter readings
Strong (weak) acid/ strong (weak) base: acid is the ______base is the titrant
Strong (weak) base/ strong (weak) acid: base is the sample, acid is the ______
2) Symbols
A, 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 Effect
Buffers:
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)