AP Chemistry 8.4 Acid-Base Reactions and Buffers Study Notes - New Syllabus Effective fall 2024

AP Chemistry 8.4 Acid-Base Reactions and Buffers Study Notes- New syllabus

AP Chemistry 8.4 Acid-Base Reactions and Buffers Study Notes – AP Chemistry – per latest AP Chemistry Syllabus.

LEARNING OBJECTIVE

Explain the relationship among the concentrations of major species in a mixture of weak and strong acids and bases.

Key Concepts:

Ionization of Weak Acids & Weak Bases

AP Chemistry-Concise Summary Notes- All Topics

8.4.A.1 pH Determination in Strong Acid–Strong Base Reactions:

1. Strong Acids and Bases:

i. Strong acids ionize completely in water.
Examples: HCl → H⁺ + Cl⁻
HNO₃ → H⁺ + NO₃⁻
H₂SO₄ → 2H⁺ + SO₄²⁻

ii. Strong bases ionize completely in water as well.
Examples: NaOH → Na⁺ + OH⁻
KOH → K⁺ + OH⁻
Ba(OH)₂ → Ba²⁺ + 2OH⁻

iii. Neutralization Reaction

When a base and an acid react, they form water and a salt:

$H^{+} + {OH}^{-} \to H_{2} O$

This is a neutralization reaction — the acid’s hydrogen ion reacts with the base’s hydroxide ion to produce water.

2. Mole Calculations and Limiting Reactant:

i. Mole Calculations:

Use the formula: $Moles (mol) = Concentration (mol/L) \times Volume (L)$

Note: If the volume is given in mL, convert it to L by dividing by 1000.

ii. Steps for Acid-Base Calculations:

a. Determine moles of acid and base using: $mol = c \times V$

b. Compare the mole ratio of the balanced chemical equation.

c. Find the limiting reactant (the one that gets consumed first).

d. Find the excess reactant (the one left over after the reaction).

3. pH and pOH Calculations:

i. pH and pOH (After Neutralization):

a. When acid is in excess: $\text{pH} = -\log[\text{H}^+]$ b. When base is in excess:

$pOH = - \log [{OH}^{-}], pH = 14 - pOH$

8.4.A.2 Weak Acid + Strong Base: pH Determination Summary:

1. Weak Acid + Strong Base Reaction:

i. Reaction: $HA (aq) + {OH}^{-} \to A^{-} + H_{2} O$

a. HA = weak acid (e.g., acetic acid)
b. OH⁻ = strong base (e.g., NaOH)
c. A⁻ = conjugate base of the weak acid

Key Point: While HA is a weak acid and does not fully dissociate on its own, it fully reacts with OH⁻ in this neutralization.

ii. When working problems with this reaction:

1. Calculate moles of HA and OH⁻ added.

Use: mol=Molarity×Volume (in L)

2. Identify the limiting reagent:

This tells you if OH⁻ fully neutralizes HA or if there is some leftover.

3. Use stoichiometry (1:1 mole ratio) to calculate:

Moles of HA reacted
Moles of A⁻ formed
Any remaining HA or OH⁻

4. Calculate final concentrations:

Based on total volume after mixing, calculate the concentration of all species in solution.

solution. $[species] = \frac{mol}{total volume after mixing}$

5. Calculate the pH:

i. If only A⁻ remains: Use Kb of A⁻ (derive from Ka of HA) to calculate pH (basic solution).
ii. If a buffer remains (both HA and A⁻): Use Henderson-Hasselbalch equation: $pH = pKa + \log (\frac{[A^{-}]}{[HA]})$ iii. If excess OH⁻ remains: Find pOH from residual OH⁻, then: $pH = 14 - pOH$

2. Case 1: Weak Acid in Excess:

i. Setup:

a. You are mixing a weak acid (HA) and a strong base (OH⁻)
b. Not all of the weak acid reacts—there is some excess
c. The strong base is the limiting reagent
d. The reaction produces A⁻, the conjugate base of the acid

ii. Result:

A buffer solution is created — both HA and A⁻ are present.

iii. Use the Henderson–Hasselbalch Equation: $pH = p K_{a} + \log (\frac{[A^{-}]}{[HA]})$ Where:

* [A-] : concentration of conjugate base produced (from OH⁻ reacting)
* [HA] : concentration of remaining unreacted weak acid
* Ka : acid dissociation constant of HA

iv. Steps to Solve:

1. Calculate moles of HA and OH⁻
2. Calculate limiting reactor→ OH⁻ is used up
3. Calculate moles of A⁻ formed = moles of OH⁻ added
4. Calculate moles of HA remaining= initial HA − OH⁻ added
5. Divide both by total volume to get concentrations
6. Substitute into Henderson–Hasselbalch equation

3. Case 2: Strong Base in Excess:

i. Situation:

* A strong base (OH⁻) is added in greater moles than the weak acid (HA)
* The whole weak acid is neutralized
* Excess OH⁻ remains in solution

ii. Result:

* No buffer (HA is neutralized)
* The solution is basic, and the pH is regulated by the excess OH⁻

iii. Steps to Solve:

1. Determine initial moles of HA and OH⁻
2. Reaction:

All HA reacts:

$HA + {OH}^{-} \to A^{-} + H_{2} O$

3. Determine [OH⁻]

4. Calculate pOH
5. Calculate pH

4. Case 3: Equimolar Amounts:

i. Case 3: Equimolar Amounts of Weak Acid and Strong Base

Circumstance:

* You mix equivalent moles of a weak acid (HA) and a strong base (OH⁻)
*The reaction goes to completion:

$HA + {OH}^{-} \to A^{-} + H_{2} O$

* All HA is converted to A⁻ (its conjugate base)

ii. What’s Left in Solution?:

Only A⁻ remains.

Since A⁻ is the conjugate base of a weak acid, it is hydrolyzed with water:

$\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-$

This makes the solution slightly basic.

iii. Steps to Solve:

1. Find moles of HA and OH⁻ (same)

2. All HA is converted to A⁻

3. Use total volume to find [A⁻]

4. Use Kb to find [OH⁻]:

First determine Kb from Ka: $K_{b} = \frac{K_{w}}{K_{a}} where K_{w} = 1.0 \times 10^{- 14}$

Use ICE table for hydrolysis: $A^{-} + H_{2} O ⇌ H A + O H^{-}$

Solve for $[OH^-]$ using: $K_{b} = \frac{x^{2}}{[A^{-}]} \Rightarrow x = \sqrt{K_{b} \cdot [A^{-}]}$

5. Find pOH:

Convert to pH:
$pH = 14 - pOH$

8.4.A.3 pH Determination: Weak Base + Strong Acid Reaction:

1. Neutralization Reaction:

Weak base + strong acid forms a conjugate acid (HB⁺) and water:

$B (a q) + H_{3} O^{+} (a q) ⇌ H B^{+} (a q) + H_{2} O (l)$

$B(aq) + H_3O^+(aq) \rightleftharpoons HB^+(aq) + H_2O(l)$

2. Buffer Solution (Weak Base in Excess):

i. When weak base (B) is in excess, a buffer is created (B and HB⁺).
ii. Use the Henderson-Hasselbalch equation to determine pH: $\text{pH} = \text{p}K_b + \log\left(\frac{[\text{B}]}{[\text{HB}^+]}\right)$

3. Excess Strong Acid:

If strong acid is in excess, calculate [H₃O⁺] from excess acid and total volume, then find pH:

$pH = - \log [H₃O⁺]$

4. Equimolar Amounts (Weak Base + Strong Acid):

i. When equimolar amounts are mixed, HB⁺ is produced, and the solution is slightly acidic.
ii. Use the equilibrium: $\text{HB}^+ (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{B} (aq) + \text{H}_3\text{O}^+ (aq)$

Determine pH via the Ka of HB⁺ using an ICE table to solve.

5. Ka and Kb:

For a conjugate acid-base pair (e.g., HB⁺ and B): $K_{a} \cdot K_{b} = K_{w}$

* Ka = dissociation constant of acid HB⁺
* Kb = dissociation constant of base B
* Kw = water ionization constant = $1.0 \times 10^{-14}$ at 25°C

If you know Kb of the base, you can find Ka of its conjugate acid from: $K_{a} = \frac{K_{w}}{K_{b}}$

8.4.A.4 Weak Acid + Weak Base: Equilibrium Reaction:

1. Equilibrium Reaction Behavior: $\text{HA} + \text{B} \rightleftharpoons \text{A}^- + \text{HB}^+$

Partial reaction: This one doesn’t go to completion like a strong acid/base reaction.

Reaction sets up an equilibrium, so both reactants and products exist together in solution.

i. Key Points:

1. Reversible: Forward and reverse reactions take place at the same time.
2. No complete neutralization: Partial reaction of only a fraction of HA and B.
3. Equilibrium expression: $K = \frac{[\text{A}^-][\text{HB}^+]}{[\text{HA}][\text{B}]}$

4. Amount of reaction depends on the relative acidity of the acid and the basicity of the base (i.e., their Ka and Kb).
5. pH is a function of the position of equilibrium, not a quantity of reaction.

2. Ka, Kb, and pH Prediction:

* Ka: Measures acid strength (how well an acid donates H⁺)
* Kb: Measures base strength (how well a base accepts H⁺)

i. pH Prediction Rules:

* If Ka > Kb → solution is acidic
* If Kb > Ka → solution is basic
* If Ka = Kb → solution is neutral

Example:

For a salt like NH₄CH₃COO (ammonium acetate):

* NH₄⁺ (Ka = 5.6 × 10⁻¹⁰)
* CH₃COO⁻ (Kb = 5.6 × 10⁻¹⁰)

Since Ka = Kb, the solution is neutral.