Acid/Base Theories
Acids: ______
______
Bases: ______
______
Arrhenius’ Theory
Acids – ______
______
Ionization – any process by which a neutral ______or ______is converted into an ______
Bases – ______
______
Dissociation – the ______of ions that occurs when an ______compound dissolves in H2O.
• this accounts for the fact that both acids and bases are ______
• can’t account for acidic solutions that are ______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 ______properties of H2O
eg.
• OH− is responsible for the ______properties of H2O
Water is amphoteric, it can act as an ______or a ______.
• some others include HCO3−, HSO4−, HS−, NH3
eg.
eg.
Strong Acids/Bases
• strong or weak refers to the electrolyte’s ______
• the ______the acid/base, the ______the conjugate
• Percent ionization gives a ______of the acid/base ______
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 H-O bond has stronger ______interactions with H2O than the ______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 or hydration energy)
• the ______net energy release occurs with the ionization of HI and so 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
• so, 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 ______with ______, due to random collisions:
H2O (l) + H2O(l) ↔ H3O+(aq) + OH−(aq)
acid1 base2 acid2 base1
· neutral H2O conducts ______,barely!
· measured using a ______
· the ______exists in all ______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 major species is ______as the sol’n will be ______
· the self ionization of water is ______
· the self ionization of water’s is ______whenever a system’s [H3O+] ______the Kw of 1.0 x10-7
· Similar argument for ______Bases and the OH− ion
pH, pOH and pKw
1) Definition
· for significant figures:
# of decimal places in pH = # of s.f. in [ ]
and as
2) Measuring pH
pH Meters
· are probes that measure the ______of ______
Acid – Base Indicators
· Are weak ______equilibrium systems where the acid conjugate bases are ______colors
Phenolphthalein: C20H14O4
HIn(aq) + H2O(l) ↔ H3O+(aq) + In−(aq)
· these colour ______occur at the pH determined by their equilibrium ______, Kin
· when the colour changes, ______and
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 ______
• but this doesn’t change the ______greatly as the ______of solution has ______.
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− ;
ClO4− + H2O ↔ HClO4 + OH− ;
______acids have ______conjugate bases
______bases have ______conjugate acids
Weak ______have weaker conjugate ______
Weak ______have weaker conjugate ______
6) Ka and Kb for Conjugate Acid/Base Pairs
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.
eg. If for [Al(H2O)6]3+ the Ka is 1.4 x 10-5, calculate the Kb for [Al(H2O)5OH]2+.
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
Development of the “Shortcut”
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 [H2CO3]eq, [HCO3−]eq, [CO32−]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
NaHCO3 predicted pH? < 7 7 > 7
Al(NO3)3 predicted pH? < 7 7 > 7
NaC2H3O2 predicted pH? < 7 7 > 7
NH4C2H3O2 predicted pH? < 7 7 > 7
NaC2O4 predicted pH? < 7 7 > 7
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.
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 equivalence point (nA = nB) - Veq.pt.
d) After equivalence point - excess titrant, usually V1½ eq.pt.
e) Sketch the curve
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.
· 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.
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 equivalence point (nA = nB) - Veq.pt.
d) After equivalence point - excess titrant, usually V1½ eq.pt.
e) Sketch the curve
c) At Veq.pt. = ______
d) At V1½ eq.pt. = ______
e) Sketch the curve
STRONG ACID
a) Initial pH
b) At V½ eq.pt.
c) At Veq.pt.
d) At V1½ eq.pt.
Weak Acid
a) Initial pH
b) At V½ eq.pt.
c) At Veq.pt.
d) At V1½ eq.pt.
Buffers
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. Ka = 7.9 x 10-5
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)
c) What is the pH when 2.0 mL of 1.5 M HCl 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. Ka = 7.9 x 10-5
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)
c) What is the pH when 2.0 mL of 1.5 M HCl sol’n are added? (Vtot is the same, calculate na and nb)