CHEM122 Acid & Bases

Acidity and Basicity Concepts

Definitions and Concepts

Acidity and basicity, pH, pKa, pKb, and thermodynamics
The nature of acids and bases
Acids produce hydrogen ions in aqueous solutions and bases produce hydroxide ion (restrictive)
Brønsted-Lowry concept: an acid is a proton (H+) donor, and a base is proton acceptor
Acids in aqueous solutions: water acts as a base to form a hydronium ion
An equilibrium - competition for the proton

Acid and Base Equilibria

Define an equilibrium for the general reaction in which: a proton is removed from HA to give the conjugate base A–
𝐾௔ᇱ= HଷOାAି / HA
Therefore, we can write 𝐾௔= HଷOାAି / HA

Acid Strength

Strong acid: the equilibrium lies far to the right (Ka is large)
Almost all the original HA is dissociated into ions at equilibrium
Strong acids yield weak conjugate bases (that is, they are weaker bases than water)
Weak acid: equilibrium lies far to the left (Ka is small)
Most of the original HA remains present as HA at equilibrium
Weak acids yield strong conjugate bases (a stronger base than water)

Strength of Bases

Strong base: Na+(aq) + OH−(aq) → NaOH(s)
General definition: B(aq) + H2O(l) → HO−(aq) + BH+(aq)
Ammonia and amines are bases: NH3(aq) + H2O(l) → HO−(aq) + NH4+(aq)

Water and pH Calculations

Water as Amphoteric

Water is amphoteric: it is both an acid and a base
Autoionization of water: transfer of a proton from one water molecule to another to produce a hydroxide anion and a hydronium cation
H2O(l) + H2O(l) → HO−(aq) + H3O+(aq)
KW: dissociation constant of water at 25°C, 𝐾ௐ= HଷOā × HO― = 10ିଵସ

pH Calculations for Strong Acids

Calculating the pH value of an aqueous solution
Strong acids (essentially completely ionized):
For example, what is the pH of a 0.01 M HCl solution?
Identify the major species in solution: solution contains H3O+ and Cl– ions and H2O
Which species can give H3O+ ions? H2O can also provide protons by autoionization
Thus, [H3O+] = 0.01 M and pH = –log [H3O+] = –log(0.01) = 2.00

pH Calculations for Weak Acids

Calculating the pH value of an aqueous solution
Weak acids (very little ionization):
What is the pH of a 1.00 M HF solution (Ka = 7.2×10–4)?
Identify the major species in solution: HF, H2O, H3O+ + F–
𝐾௔= HଷOāFି / HF
Sources of H3O+?

Acid-Base Equilibria

Weak Acids and Their Ionization

Although a weak acid, HF is the dominant source of protons (water is an even weaker acid).
Write the equilibrium expression:
Solve for F-, H3O+, HF:
Initial concentrations: 0, 10‒7 ≈ 0, 1.000
Change in concentrations: x, x, -x
Equilibrium concentrations: x, x, 1.000 - x

Calculating pH of Weak Acids

Calculating the pH value of an aqueous solution:
Weak acids (very little ionization):
Solve for K_a = 7.2 × 10–4 = H2O + F-
HF = x⋅x / (1.00 - x)
Use the approximation 1.00 - x ≈ 1.00
Then x² = 7.2 × 10–4 and x = 2.7 × 10–2
Is the approximation valid? Compare x with [HF]₀ (use the 5% rule).
x = 2.7 × 10–2 M, which is < 5%, so approximation is fine.
pH = –log(2.7 × 10–2) = 1.57

Breakdown of Approximations

When does this approximation breakdown, error is greater than 5%?
Weak acids (very little ionization):
Solve for K_a = 7.2 × 10–4 = H2O + F-
HF = x⋅x / (HF - x)
Use the approximation [HF] - x ≈ [HF]
Then x = sqrt(7.2 × 10–4 × [HF])
When does this approximation breakdown?
% error calculations for various concentrations.

Water as an Amphoteric Substance

Properties of Water

Water is amphoteric: it is both an acid and a base.
H2O(l) + H2O(l) ⇌ HO−(aq) + H3O+(aq)
In acidic solutions, pH < 7.00, [H3O+] > [OH–]
In basic solutions, pH > 7.00, [H3O+] < [OH–]
Autoionization of water: transfer of a proton from one water molecule to another to produce a hydroxide anion and a hydronium cation.

Dissociation Constant of Water

K_w: dissociation constant of water
K_w = H2O + OH− = 10–14 at 25°C
K_w = H2O × OH− = 10–14

Calculating pH of Strong Acids

Example with Strong Acids

Calculating the pH value of an aqueous solution:
Strong acids (essentially completely ionized):
For example, what is the pH of a 0.01 M HCl solution?
Identify the major species in solution: H3O+ and Cl– ions and H2O.
What is the pH of a 1.00 M HF solution (K_a = 7.2 × 10–4)?
Identify the major species in solution: HF, H2O, H3O+, and F–.

Sources of H3O+

Sources of H3O+: although a weak acid, HF is the dominant source of protons (water is an even weaker acid).
Write the equilibrium expression:
Solve for F-, H3O+, HF:
Initial concentrations: 0, 10‒7 ≈ 0, 1.000
Change in concentrations: x, x, -x
Equilibrium concentrations: x, x, 1.000 - x.

Worked Problem Example

3-Chlorobutanoic Acid (CBAH)

Worked problem – sample:
3-Chlorobutanoic acid (CBAH) (C4H7ClO2) is a weak acid with K_a = 8.91 × 10–5 and it has a molar mass of 122.6 g•mol–1.
A solution of CBAH is made that contains 0.625 g in 100 mL of water at 25 °C.
Part (a): What is the concentration of CBAH in this solution?
Part (b): What is the pH of this solution of CBAH at 25 °C?
Hint: You can assume that less than 5% of the CBAH is ionized.

Approximations for Equilibrium Constants

Small Equilibrium Constant

Using an approximation for cases where the equilibrium constant is very small
now check your solution
[N2] = 0.0482 – 2x 0.0482 mol/L
[O2] = 0.0933 – x 0.0933 mol/L
[N2O] = 2x 6.6 10–9 mol/L
𝐾஼= NଶO ଶ / (0.0482 ଶ * 0.0933) = 2.0 × 10ିଵଷ

Large Equilibrium Constant

Using approximations for cases where the equilibrium constant is very large
now look at the reverse reaction:
𝐾஼= NଶଶOଶ / NଶO ଶ = 5.0 × 10ଵଶ
ICE TABLE
0 0 0.06 concinitial
2x x -2x concchange
0.06 0.03 y concequilibrium
0.06−2x
so we insert 0.06-2x = y into the equation
Note, x ~ 0.03
𝐾஼= NଶଶOଶ / (0.06 −2x) ଶ= 5.0 × 10ଵଶ
𝐾஼= NଶଶOଶ / NଶO ଶ = 2x ଶ× x / 𝑦ଶ = 5.0 × 10ଵଶ
NଶO ଶ = 0.06 ଶ× 0.03
𝑦ଶ= 0.06 ଶ× 0.03
5.0 × 10ଵଶ = 2.16 × 10ିଵ, 𝑦= 4.65 × 10ିଽ

Setting Up Equilibrium Calculations

Problem Statement

How to set up calculations to determine equilibrium composition
question:
31.2 g of PCl5 is placed in a 500 mL reaction vessel and allowed to reach equilibrium with its decomposition products phosphorus trichloride and chlorine at 250 C
given that Kp = 78.3
what are the equilibrium compositions in mol/L? what is the percentage decomposition of PCl5?
read the question carefully
you are asked for concentrations (mol/L), but are given the equilibrium constant for pressure (Kp)

Calculating Kc from Kp

answer:
first calculate Kc from Kp, recall
𝐾஼= 𝐾௣(𝑅𝑇)ି୼௡
Kp = 78.3 and n = moles products – moles of reactants = 1
𝐾஼= 78.3 × (𝑅𝑇)ିଵ
R = 0.0821 L atm mol–1 K–1, T = 523 K (250 C)
therefore, RT = 42.94 L atm mol–1
𝐾஼= 78.3
𝐾஼= PCl²Cl² / PClହ
42.9 = 1.82

Initial Concentrations

Kc = 1.82
𝐾஼= PCl²Cl² / PClஹ
now calculate initial concentrations:
31.2 g of PCl5 is placed in a 500 mL reaction vessel
31.2 𝑔 / 208.22 𝑔/𝑚𝑜𝑙= 0.150 𝑚𝑜𝑙
thus, we have 0.150 mol PCl5 in 0.5 L
[PCl5]0 = 0.30 mol/L
at the start Q = 0 so reaction will shift to the right

Setting Up the ICE Table

now set up quadratic equation
Cl2 PCl3 PCl5
sometimes
0 0 0.300 concinitial
x x -x concchange
called an ICE table
x x 0.300 −x concequilibrium
= 𝑥ଶ
𝐾஼= PCl²Cl² / PClஹ
0.300 −𝑥= 1.82

Solving the Quadratic Equation

now solve quadratic equation
𝑥= −𝑏±√(𝑏ଶ−4𝑎𝑐) / 2𝑎
0 = 1.82 0.300 −𝑥−𝑥ଶ= −𝑥ଶ−1.82𝑥+ 0.546
x= 1.82 ± √(1.82 ଶ+ 4 × 1 × 0.546) / −2
x = 0.262 (or –2.08)

Final Concentrations and Percentage Decomposition

so
[PCl5] = 0.300 – x = 0.038 mol/L
[PCl3] = x = 0.262 mol/L
[Cl2] = x = 0.262 mol/L
now check your solution
[PCl5] = 0.300 – x = 0.038 mol/L
[PCl3] = x = 0.262 mol/L
[Cl2] = x = 0.262 mol/L
= 0.262ଶ
𝐾஼= PCl²Cl² / PClஹ
0.038 = 1.81
last, what is the percentage decomposition of PCl5?
number of moles PCl5 dissociated = (0.262 mol/L)(0.500 L) = 0.131 mol
percent dissociation is
0.131𝑚𝑜𝑙 / 0.150𝑚𝑜𝑙× 100% = 87.3%

Acid-Base Properties of Salts

Understanding Salt Solutions

• When a salt dissolves in water, it dissociates into its constituent ions, which can affect the pH of the solution.

• The resultant solution can be acidic, neutral, or basic depending on the nature of the cations and anions produced.

• Example: Sodium chloride (NaCl) produces Na+ (weak acid) and Cl- (weak base), resulting in a neutral solution.

• Conversely, ammonium chloride (NH4Cl) produces NH4+ (weak acid) and Cl- (weak base), leading to an acidic solution.

• The strength of the original acid or base from which the salt is derived plays a crucial role in determining the pH.

Factors Influencing pH of Salt Solutions

• The acidity or basicity of the solution is influenced by the conjugate acids and bases of the ions present.

• Strong acids yield weak conjugate bases, while weak acids yield stronger conjugate bases.

• Example: Cl- is a weak conjugate base from the strong acid HCl, while CN- is a stronger conjugate base from the weak acid HCN.

• The cation's nature also matters; for instance, Na+ from NaOH is a weak associated acid, while NH4+ from NH3 is a stronger conjugate acid.

• The interplay between the cation and anion determines the overall pH of the solution.

Polyprotic Acids

Characteristics of Polyprotic Acids

• Polyprotic acids can donate more than one proton (H+), leading to multiple dissociation steps.

• Example: Phosphoric acid (H3PO4) has three dissociable protons, resulting in three equilibria constants: Ka1, Ka2, and Ka3.

• The first dissociation is typically the strongest, with subsequent dissociations being weaker.

• The charge of the conjugate base increases with each proton lost, making it harder to remove additional protons.

• The equilibria constants decrease with each step, indicating weaker acid strength.

Calculating pH of Polyprotic Acids

• To calculate the pH of a polyprotic acid, consider the contributions from each dissociation step.

• For example, when 0.1 M H3PO4 is added to water, the primary contribution to [H3O+] comes from the first dissociation.

• The pH can be calculated using the formula: pH = -log[H3O+].

• Subsequent contributions from H2PO4- and HPO4^2- can be considered negligible for strong acids.

• Example calculation: For H3PO4, Ka1 = 7.5 × 10^-3, leading to a calculated pH of 1.62.

Weak Bases and Their pH Calculations

Understanding Weak Bases

• Weak bases partially accept protons from water, resulting in a limited increase in [OH-].

• Example: Methylamine (CH3NH2) has a Kb of 3.6 × 10^-4, indicating its weak basicity.

• The pH of a weak base can be calculated using the formula: Kb = [OH-]^2 / (C - [OH-]), where C is the initial concentration.

• The percentage protonation can also be determined from the equilibrium concentrations.

• Example calculation: For a 0.20 M solution of methylamine, the pH and percentage protonation can be derived from Kb.

Mathematical Relationships in Acid-Base Chemistry

pKa and pKb Relationships

• The relationship between pKa and pKb is given by the equation: pKa + pKb = 14.0 at 25 °C.

• This relationship holds true for any acid and its conjugate base or a base and its conjugate acid.

• The equilibrium constants for acids (Ka) and bases (Kb) are related by the ion product of water (Kw): Ka * Kb = Kw.

• Example: For a weak acid HA, the dissociation can be expressed as: HA ⇌ H+ + A-, leading to Ka = [H+][A-]/[HA].

• Similarly, for a weak base B, the expression is: B + H2O ⇌ BH+ + OH-, leading to Kb = [BH+][OH-]/[B].

Understanding Acid-Base Pairs and Salt Solutions

Conjugate Acid/Base Pairs

• Sodium acetate (NaCH3COO) is a salt formed from a weak acid (acetic acid) and a strong base (sodium hydroxide).

• Na+ is a weak acid compared to water (H2O) because its conjugate base (NaOH) is strong, while CH3CO2- (acetate ion) is a stronger base than water due to its weak conjugate acid (acetic acid).

• Conclusion: Sodium acetate is classified as a basic salt due to the nature of its components.

Calculating pH of Sodium Acetate Solutions

• To calculate the pH of a 0.1 M sodium acetate solution, we first need the Ka of acetic acid, which is 1.8 × 10–5.

• Using the relationship Ka × Kb = Kw, we find Kb for acetate: Kb = 10–14 / 1.8 × 10–5 = 5.56 × 10–10.

• Set up ICE (Initial, Change, Equilibrium) tables to analyze the dissociation of acetate in water.

ICE Table Setup for Sodium Acetate

• Initial concentrations: [CH3CO2H] = 0, [CH3CO2-] = 0.1, [H2O] = 10–7 (approx.)

• Change: Let x be the change in concentration due to dissociation: [CH3CO2H] = x, [CH3CO2-] = 0.1 - x, [H2O] = 10–7 + x.

• Equilibrium: [CH3CO2H] = x, [CH3CO2-] = 0.1 - x, [H2O] = 10–7 + x.

pH Calculations for Ammonium Chloride Solutions

Understanding Ammonium Chloride (NH4Cl)

• Ammonium chloride is a salt formed from a weak base (ammonia) and a strong acid (HCl).

• To find the pH of a 0.25 M NH4Cl solution, we first calculate Ka for NH4+ using Kb for NH3: Ka = 10–14 / 1.8 × 10–5 = 5.56 × 10–10.

• Set up ICE tables to analyze the dissociation of NH4+ in water.

ICE Table Setup for Ammonium Chloride

• Initial concentrations: [NH4+] = 0.25, [NH3] = 0, [H3O+] = 10–7 (approx.)

• Change: Let x be the change in concentration due to dissociation: [NH4+] = 0.25 - x, [NH3] = x, [H3O+] = x.

• Equilibrium: [NH4+] = 0.25 - x, [NH3] = x, [H3O+] = x.

Comparing Acids and Bases

Weak Acid and Weak Base Salt Analysis

• To determine if a salt is acidic or basic, compare the Ka of the cation with the Kb of the anion.

• Example: For NH4CN, compare Ka of NH4+ (5.56 × 10–10) with Kb of CN– (1.6 × 10–5).

• If Ka > Kb, the solution is acidic; if Ka < Kb, the solution is basic.

Molecular Structure and Acidity

• Acidity is influenced by electronegativity and bond dissociation energy (BDE).

• As electronegativity increases across a period, acidity generally increases due to stronger polar bonds.

• BDE also affects acidity; weaker bonds (lower BDE) lead to stronger acids.

Trends in Acidity and Basicity

Acid Strength Trends

• The strength of acids can be qualitatively analyzed based on molecular structure and electronegativity.

• Example: HF < HCl < HBr < HI in terms of acid strength due to bond strength and electronegativity differences.

• The presence of electronegative atoms attached to the central atom lowers the pKa value, indicating stronger acidity.

Comparing Acids with Different Acidities

• pKa values provide a logarithmic scale to compare acid strengths: pKa = -log10(Ka).

• A lower pKa indicates a stronger acid; for example, if HA1 has a pKa of 3.0 and HA2 has a pKa of 4.0, HA1 is ten times stronger than HA2.

Fundamental Concepts of Acids and Bases

Production of H3O+ Ions

• Acids are known to dominate the production of hydronium ions (H3O+) in aqueous solutions, significantly affecting pH levels.

• In dilute solutions of weak acids, the contribution of H3O+ ions may be minimal compared to that of water, which has an inherent H3O+ concentration of 1 × 10–7 M.

• When an acid is added to water, the solution cannot become basic; the pH must remain below or equal to 7.0, indicating an acidic environment.

• For example, a 1 × 10–8 M HCl solution does not yield a pH of 8, as the initial concentration of H3O+ from water must be considered. The actual pH is calculated as follows: pH = -log(1 × 10–8 + 1 × 10–7) = 7.0.

• The relationship between H3O+ concentration and pH is crucial for understanding acid-base equilibria, especially in weak acids where [H3O+] is less than 10–6 M.

• The equilibria of weak acids can be analyzed using the acid dissociation constant (Ka) to determine the extent of ionization.

Buffer Solutions and Their Importance

• Buffered solutions are designed to resist changes in pH upon the addition of acids or bases, maintaining homeostasis in biological systems, such as blood.

• Blood pH is typically maintained around 7.4, with venous blood slightly lower at 7.35 due to increased CO2 levels, which forms carbonic acid in solution.

• A common buffer system consists of a weak acid and its conjugate salt, such as acetic acid (CH3CO2H) and sodium acetate (CH3CO2Na).

• The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]).

• For example, a buffer solution with 0.10 M CH3CO2H and 0.10 M CH3CO2Na has a pH calculated as follows: pH = -log(1.8 × 10–5) = 4.74.

• The addition of a strong acid like HCl to a buffer solution results in a slight decrease in pH, demonstrating the buffer's capacity to resist drastic changes.

Calculating pH in Buffer Solutions

Weak Acid Buffer Calculations

• To calculate the pH of a buffer solution containing 0.10 M CH3CO2H and 0.10 M CH3CO2Na after adding 0.01 M HCl, we set up ICE tables to analyze the changes in concentration.

• Initial concentrations before adding HCl: [H3O+] = 0, [CH3CO2Na] = 0.10, [CH3CO2H] = 0.10. After adding HCl, the concentrations change accordingly.

• The equilibrium expression for the weak acid dissociation is: Ka = [H3O+][CH3CO2-]/[CH3CO2H].

• After calculations, the new pH is found to be approximately 4.66, indicating a slight decrease due to the addition of HCl.

• This example illustrates the buffer's ability to maintain pH despite the addition of a strong acid, showcasing its effectiveness in biological systems.

• The buffer capacity is determined by the concentrations of the weak acid and its conjugate base, which can be adjusted to achieve desired pH levels.

Weak Base Buffer Calculations

• Similar to weak acid buffers, weak base buffers can also stabilize pH. For instance, a solution containing 0.1 M CH3NH2 and 0.05 M CH3NH3Cl can be analyzed.

• The base dissociation constant (Kb) is used to determine the pH of the solution, with Kb = 4.38 × 10–4 and Ka = 2.28 × 10–11.

• Setting up the ICE table for the weak base, we can calculate the equilibrium concentrations and subsequently the pH.

• The calculated pH for this buffer solution is approximately 10.94, demonstrating the ability of weak bases to maintain a stable pH.

• The Henderson-Hasselbalch equation can also be applied here: pH = pKa + log([conjugate base]/[weak acid]).

• This highlights the versatility of buffer systems in both acidic and basic environments.

Solubility Equilibria and Ksp Calculations

Understanding Solubility Product (Ksp)

• Solubility is a critical concept in chemistry, particularly in understanding how salts dissolve in water and affect pH.

• The solubility product constant (Ksp) is defined for salts in equilibrium with their ions in solution, expressed as Ksp = [Ca2+][OH-]^2 for calcium hydroxide, for example.

• A saturated solution of calcium hydroxide can be prepared, and its Ksp can be calculated based on the concentrations of the ions present.

• For instance, if Ksp = 1.3 × 10–6, we can derive the concentration of hydroxide ions and subsequently calculate the pH of the solution.

• The pH of a saturated solution of calcium hydroxide at 25 °C is found to be 12.14, indicating a strongly basic solution.

• Understanding Ksp is essential for predicting the solubility of various salts and their impact on pH in different environments.

Typical Exam Questions on Ksp

• Students should be familiar with typical exam questions regarding Ksp calculations, such as determining the Ksp for Ag2CO3 given its solubility in water.

• The expression for Ksp can vary based on the dissociation of the salt, and students must be able to identify the correct expression from multiple choices.

• For example, if 0.032 g of Ag2CO3 dissolves in 1 L, students should calculate the concentration and then use it to find Ksp.

• The correct Ksp expression for Ag2CO3 is Ksp = 4x^3, where x is the concentration of Ag+ ions in solution.

• This reinforces the importance of stoichiometry and understanding the relationships between solubility and ion concentrations in solution.

• Practice with these types of questions can enhance problem-solving skills and prepare students for exams.

Lewis Acids and Bases

Fundamental Concepts of Lewis Acids and Bases

• The definitions of acids and bases extend beyond the traditional Brønsted-Lowry definitions, introducing the concepts of Lewis acids and bases.

• A Lewis acid is defined as an electron pair acceptor, often possessing an incomplete valence shell, such as BF3 and AlCl3.

• Conversely, a Lewis base is an electron pair donor, typically containing a lone pair of electrons, such as NH3 and H2S.

• The interaction between Lewis acids and bases involves the donation of an electron pair from the base to the acid, forming new bonds.

• An example of this interaction is the formation of calcium carbonate from carbon dioxide and calcium oxide, showcasing the practical applications of Lewis acid-base theory.

• Understanding these concepts is crucial for advanced studies in organic and inorganic chemistry, as they provide a broader framework for acid-base reactions