Acids and Bases: Comprehensive Notes
Ch. 17 Acids & Bases Continued
The pH Scale
- pH is defined as the negative logarithm of the hydrogen ion concentration:
pH = -log[H^+] - It can also be expressed in terms of hydronium ion concentration:
pH = -log[H_3O^+] - The general formula for pX is:
pX = -log X - pOH is defined as the negative logarithm of the hydroxide ion concentration:
pOH = -log[OH^-]
- The pX formula can be applied to Kw, Ka, and Kb:
pX = -log X
pKw = -log(Kw)
pKa = -log(Ka)
pKb = -log(Kb)
Conjugate Acid-Base Pairs
- Example with formic acid (HCHO2) and water:
- HCHO2 + H2O ⇌ H3O+ + CHO2–
- CHO2– + H2O ⇌ HCHO2 + OH–
Ka vs. Kb
- As Ka increases, Kb decreases.
- The stronger the acid, the weaker its conjugate base.
- A "stronger weak acid" is still a weak acid.
- Problem: Determine the Ka and pKa of formic acid, given that a 0.100 M formic acid (HCHO2) solution has a pH of 2.38 at 25°C.
- Step 1: Use the definition of Ka and create an ICE (Initial, Change, Equilibrium) table.
- Ka expression includes [H+] in the products.
- Since we know pH, we can solve for [H+].
[H^+] = 10^{-pH} - Use \'x\' and the ICE table to solve for all concentrations needed for the Ka expression.
Example Problem: Methylamine Kb
- Problem: Methylamine (CH3NH2) is a weak base. In a 0.100 M solution, 6.30% of the base is ionized. What is its Kb?
- Use an ICE table to find the change in concentration.
- Percent ionization = (ionized concentration / initial concentration) * 100
- Calculate Kb using the equilibrium concentrations from the ICE table.
Key Reminder
- Ka will have [H+].
- Kb will have [OH-].
Exam Time Saver
- If the question specifies a base, use the general form: B + H2O ⇌ BH+ + OH-
- Practice figuring out the BH+ form, but on the exam, only determine it if needed to save time.
Determining Kb from Ka
- Problem: Determine Kb of the benzoate ion (C6H5COO-) given that the Ka of benzoic acid is 6.5 x 10^{-5}.
Hydrazine Example
- Problem: Hydrazine (N2H4) has a Kb = 1.7 x 10^{-6} at 25°C. What is the pH and % ionization of a 0.25 M solution?
- Approximation validity check:
- x < (0.05) [HA]initial
- [HA]initial ≥ (400) Ka
Approximation Validation
- How to check if the approximation is valid:
- x < (0.05) [HA]_{initial}
- [HA]_{initial} ≥ (400) Ka
Example Problem: Propionic Acid pH Calculation
- Problem: Calculate the pH at 25°C of a 0.10 M solution of propionic acid (HC3H5O2; Ka = 1.34 x 10^{-5}).
- pH = 2.94
When Simplifications Fail
- If the approximation is not valid, use:
- Quadratic equation
- Successive approximations
pH Calculation with Inadequate Simplification
- Problem: What is the pH of a 0.010 M solution of HC2H2O2Cl (Ka = 1.4 x 10^{-3})?
Diprotic and Polyprotic Acids
- Diprotic and polyprotic acids undergo a series of ionizations (loss of protons).
- Example: Carbonic acid (H2CO3)
- H2CO3 ⇌ H+ + HCO3– Ka1 = 4.2 x 10^{-7}
- HCO3– ⇌ H+ + CO32– Ka2 = 4.8 x 10^{-11}
- Example: Sulfuric acid (H2SO4)
- H2SO4 → H+ + HSO4– Ka1 = Large
- HSO4– ⇌ H+ + SO42– Ka2 = 1.2 x 10^{-2}
Simplifications for Polyprotic Acids
- Typically there is a large difference between Ka1 and Ka2 (~1000x).
- The first H+ is easier to lose from a neutral species, while the second H+ must be lost from an anionic species.
- Since Ka1 >> Ka2:
- [H+]total = [H+]1st + [H+]2nd
- [H+]1st >> [H+]2nd
- [H+]total ≈ [H+]1st
Concentration Approximations for Polyprotic Acids
- For H2CO3:
- H2CO3 ⇌ H+ + HCO3– Ka1 = 4.2 x 10^{-7}
- HCO3– ⇌ H+ + CO32– Ka2 = 4.8 x 10^{-11}
- [H+]total ≈ [H+]1st
- [HCO3–]total ≈ [HCO3–]1st
- [CO32–]: Ka2 = [H+][CO32–] / [HCO3–]
- [H+] = [HCO3–] (from 1st ionization), so Ka2 = [CO32–]
Example Problem: Polyprotic Acid
- Problem: You have a 0.050 M H2CO3 solution. What is the concentration of all species?
- Species: H2CO3, HCO3–, CO32–, H+, OH–
Acid-Base Properties of Salts
- When salts dissociate in solution, the resulting anions and cations can act like acids or bases.
- Cations:
- Conjugate acid of weak base → weak acid (e.g., ammonium ion NH4+)
- Hydrated metals can act like weak acids.
Key Question
- MUST ASK: What do I have in solution?
Charge Density
- The higher the charge density, the stronger the acid.
- Charge density = charge / unit volume
- Group IA metals: non-acids
- Group IIA metals: charge density too low (except Be2+)
- All other metals: weak acids
- Acidity increases as metal size decreases and charge increases.
Charge Density Explained
- Charge of ion / Size of ion
- Small charge, large size = low charge density
- High charge, small size = high charge density → Better at acting like an acid in water
Anions
- Conjugate bases of weak acids → weak bases
- Example: Sodium hypochlorite (NaOCl) is found in bleaches and disinfectants. Is it acidic, basic, or neutral?
NaOCl Example
- Problem: What is the pH of a 0.10 M NaOCl solution (Ka of HOCl = 3.0 x 10^{-8})?
The Role of Na+
- Na+ is neutral and will not affect pH.
Why Na+ is Neutral: Option 1
- Sodium has a low charge density as an alkali metal and is a neutral metal ion.
- Acidic properties of metal ions like Al3+ were previously discussed
Why Na+ is Neutral: Option 2
- Sodium is the counter ion of a strong base: NaOH
- The only way for sodium to act acidic (by protons) is for: Na+ + H2O → NaOH + H+
- This never happens because NaOH instantly dissociates and the reaction runs backwards.
- So there is no way for Na+ to produce protons.
Anion and Cation are Acid/Bases
- pH depends on the relative strengths of the conjugate acid and base.
- When Kb > Ka, the solution is basic.
- When Kb < Ka, the solution is acidic.
- When Kb = Ka, the solution is neutral.
Acidic, Basic, or Neutral Examples
- NH4C2H3O2: Kb (C2H3O2–) = 5.6 x 10^{-10}, Ka (NH4+) = 5.6 x 10^{-10}
- NH4CHO2: Kb (CHO2–) = 5.6 x 10^{-11}
Solutions for Previous Example
- NH4C2H3O2: neutral, since Ka = Kb
- NH4CHO2: acidic, since Ka > Kb
Polyprotic Acid Example with Na2CO3
- What is the pH of a 0.15 M Na2CO3 solution?
- Handle like a polyprotic acid.
- Only worry about the first ionization.
- Na2CO3 → 2 Na+ + CO32–
- CO32– + H2O ⇌ HCO3– + OH–
- HCO3– + H2O ⇌ H2CO3 + OH–
- Kb1 >> Kb2
- Get [CO32–] and [HCO3–] from the first equilibrium.
- [H2CO3] = Kb2
Lewis Acids and Bases
- Lewis base = substance that can donate a pair of electrons → must contain 1 or more lone pairs of e–
- Lewis acid = substance that can accept a pair of electrons → contains incomplete valence shell or multiple bonds that can shift to make room for lone pair
Lewis vs. Bronsted
- Lewis is a broader definition.
- If it was an acid by Bronsted, it is still an acid with Lewis.
- If it was a base by Bronsted, it is still a base with Lewis.
- Bronsted deals with protons, while Lewis deals with electrons.
Lewis Acid-Base Reaction
- A Lewis acid-base reaction involves the donation of a pair of electrons from one species to another (formation of a coordinate covalent bond).
- The Lewis concept includes all Arrhenius and Bronsted acids/bases.
- Example: H+ + OH–
Lewis Acid/Base Examples
- Be able to identify Lewis acid and base:
- Na+ + 6 H2O ⇌ Na(H2O)6 +
- Ag+(aq) + 2 NH3(aq) ⇌ Ag(NH3)2 +(aq)
- Cd2+(aq) + 4 I–(aq) ⇌ CdI4 2–(aq)
- Ni(s) + 4 CO(g) ⇌ Ni(CO)4(g)
Recognizing Lewis Acids
- Acids from previous definitions (strong and weak like HCl and HC2H3O2)
- Metals with a positive charge
- Usually neutral metals
- Compounds with a positive charge
- Draw the structure
More on Lewis Acids
- Lewis acid = substance that can accept a pair of electrons → contains incomplete valence shell or multiple bonds that can shift to make room for lone pair
Recognizing Lewis Bases
- Bases from previous definitions (strong and weak like NaOH and NH3)
- Compounds with a negative charge
- Draw the structure
More on Lewis Bases
- Lewis base = substance that can donate a pair of electrons → must contain 1 or more lone pairs of e–
Identifying Active Atoms
- Identify the most likely ATOM in a compound that can accept or donate electrons.
- Analyze the structure itself.
Active Atoms in Hydroxide and Carbon Dioxide
- In hydroxide, the oxygen carries the electrons and acts as a Lewis base.
- In carbon dioxide, the carbon accepts the electrons.
- The carbon is the atom that is DOING something.