Chemistry Study Notes
Topic: Ionic Equilibrium & Acid–Base Concepts
Program: INSPIRE Scholarship
Audience: Junior‑level chemistry students
1. Simple Explanation
-
Ionic Equilibrium
- An equilibrium that involves ions in solution.
- Written as a reversible reaction:
[ \text{A}^+ + \text{B}^- \rightleftharpoons \text{AB}(s) \quad \text{or} \quad \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- ] - The equilibrium constant (K) quantifies how far the reaction proceeds in one direction.
- For ionic equilibria involving solutes, the ionic strength and common‑ion effect influence the activity coefficients of ions, but for most teaching purposes we treat activities ≈ concentrations.
-
Acid–Base Equilibria
- Brønsted–Lowry definition: an acid donates a proton (H⁺); a base accepts a proton.
- Equilibrium in water:
[ \text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{A}^- + \text{H}_3\text{O}^+ ] - The acid dissociation constant (Ka) or base dissociation constant (Kb) describes the position of this equilibrium.
- pH is the negative logarithm of the hydronium ion concentration:
[ \text{pH} = -\log[H_3O^+] ] - pKa = –log Ka. Small pKa → strong acid.
-
Le Chatelier’s Principle in Ionic Equilibria
- Adding or removing ions shifts equilibrium to counter the change.
- Common‑ion effect: adding a species that is a product pushes the equilibrium to the left (less dissociation).
- Dilution shifts equilibria towards dissociation.
-
Solubility Product (Ksp)
- For sparingly soluble salts:
[ \text{MX}\text{s} \rightleftharpoons \text{M}^{n+} + \text{X}^{m-} \quad K{sp} = [\text{M}^{n+}]^m [\text{X}^{m-}]^n ] - Determines how much solid can dissolve in a given solution.
- For sparingly soluble salts:
-
Buffer Systems
- Mixture of weak acid and its conjugate base (or weak base and its conjugate acid).
- Maintains pH upon addition of small amounts of acid or base.
- Henderson–Hasselbalch equation gives the buffer pH: [ \text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]} ]
2. Key Points
| Concept | Symbol | Typical Expression |
|---|---|---|
| Acid dissociation constant | (K_a) | (\displaystyle K_a = \frac{[H^+][A^-]}{[HA]}) |
| Base dissociation constant | (K_b) | (\displaystyle K_b = \frac{[OH^-][BH^+]}{[B]}) |
| Solubility product | (K_{sp}) | (\displaystyle K_{sp} = [M^{n+}]^m [X^{m-}]^n) |
| pH | – | (\displaystyle \text{pH} = -\log[H^+]) |
| pKa | – | (\displaystyle \text{p}K_a = -\log K_a) |
| pKb | – | (\displaystyle \text{p}K_b = -\log K_b) |
| Henderson–Hasselbalch | – | (\displaystyle \text{pH} = \text{p}K_a + \log \frac{[\text{A}^-]}{[\text{HA}]}) |
Important rules
- Strong acids/bases: essentially fully dissociated → Ka or Kb large → pKa or pKb small (negative).
- Weak acids/bases: Ka or Kb small → pKa or pKb large.
- Common‑ion effect: Adding a salt that contains one of the ions of an equilibrium reduces the ion concentration and shifts the equilibrium toward the undissociated form.
- Dilution: Increases ( [\text{HA}] ) relative to ( [\text{A}^-] ) → shifts equilibrium rightward for acids (more dissociation).
- Solubility: If (Q_{sp} > K_{sp}), precipitation occurs; if (Q_{sp} < K_{sp}), dissolution occurs.
3. Worked Example
Problem
A 0.100 M solution of acetic acid (CH₃COOH) is buffered with 0.050 M sodium acetate (CH₃COONa).
- Calculate the pH of this buffer solution.
- Predict the effect on pH if 0.010 M HCl is added.
Step 1 – Identify constants
- Ka (CH₃COOH) = 1.8 × 10⁻⁵ → pKa = 4.74
- CH₃COONa dissociates completely → provides CH₃COO⁻ (conjugate base).
Step 2 – Use Henderson–Hasselbalch
Given:
([HA] = 0.100 , \text{M})
([A^-] = 0.050 , \text{M})
[ \text{pH} = 4.74 +