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

+ • H3O is responsible for the acidic properties of H2O

− • 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. 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

+ − 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.

 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.

 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)