⚖️ Weak Acid Equilibrium

Part 1 of 7 — The $K_a$ Expression

Unlike strong acids that dissociate completely, weak acids only partially ionize in water. This partial ionization is an equilibrium process described by the acid dissociation constant, $K_a$.

Weak Acid Dissociation

A generic weak acid $HA$ in water:

$HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)$

The equilibrium expression is:

$K_a = \frac{[H^+][A^-]}{[HA]}$

Key Features

• $K_a$ is small (typically $10^{-2}$ to $10^{-12}$) because weak acids are mostly undissociated
• Larger $K_a$ = stronger weak acid (more dissociation)
• Smaller $K_a$ = weaker acid (less dissociation)
• Water is omitted from the expression (pure liquid)

Common Weak Acids and Their $K_a$ Values

Relative strength: $HF > CH_3COOH > H_2CO_3 > HCN$

The $pK_a$ Scale

Just as $pH = -\log[H^+]$, we define:

$pK_a = -\log K_a$

Interpreting $pK_a$

• Lower $pK_a$ → stronger acid (larger $K_a$)
• Higher $pK_a$ → weaker acid (smaller $K_a$)

This is the inverse relationship — don't mix it up!

Converting Between $K_a$ and $pK_a$

$K_a = 10^{-pK_a} \qquad pK_a = -\log K_a$

Weak Acid Concept Check 🎯

Strong vs. Weak Acids: Key Differences

Important

At the same concentration, a strong acid always has a lower pH (more acidic) than a weak acid because more $H^+$ is produced.

For example, 0.10 M $HCl$: $pH = 1.00$

But 0.10 M $CH_3COOH$: $pH = 2.87$ (we'll calculate this in Part 2)

Weak Acid Fundamentals 🔍

$K_a$ and $pK_a$ Conversions 🧮

1. Convert $K_a = 1.8 \times 10^{-5}$ to $pK_a$ (3 significant figures)

2. Convert $pK_a = 9.21$ to $K_a$ (Enter in scientific notation, e.g. 6.2e-10)

3. Rank by acid strength (enter strongest): Acid A ($pK_a = 2.1$), Acid B ($pK_a = 6.5$), Acid C ($pK_a = 4.3$). Enter A, B, or C.

Exit Quiz — Weak Acid Equilibrium ✅

🧊 ICE Tables for Weak Acids

Part 2 of 7 — Calculating pH of Weak Acid Solutions

The ICE table (Initial, Change, Equilibrium) is the essential tool for calculating the pH of weak acid solutions. Combined with the 5% approximation, it simplifies calculations dramatically.

Setting Up an ICE Table

For a weak acid $HA$ with initial concentration $C$ and acid dissociation constant $K_a$:

$HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)$

Substituting into $K_a$:

$K_a = \frac{x \cdot x}{C - x} = \frac{x^2}{C - x}$

This is a quadratic equation in $x$. But we can often avoid the quadratic formula!

The 5% Approximation

If $x \ll C$ (specifically, if $x < 5\%$ of $C$), we can approximate:

$C - x \approx C$

This simplifies the equation to:

$K_a \approx \frac{x^2}{C}$

$x = \sqrt{K_a \cdot C}$

$[H^+] = x = \sqrt{K_a \cdot C}$

$pH = -\log(\sqrt{K_a \cdot C})$

When Does the Approximation Work?

The approximation is valid when:

$\frac{C}{K_a} > 400 \quad \text{(conservative rule)}$

Or equivalently, when $\frac{x}{C} \times 100\% < 5\%$.

If the Approximation Fails

Use the quadratic formula:

$x^2 + K_a x - K_a C = 0$

$x = \frac{-K_a + \sqrt{K_a^2 + 4K_a C}}{2}$

(Take the positive root only — concentrations can't be negative!)

Worked Example

Find the pH of 0.10 M acetic acid ($CH_3COOH$, $K_a = 1.8 \times 10^{-5}$).

Step 1: Check if approximation works

$\frac{C}{K_a} = \frac{0.10}{1.8 \times 10^{-5}} = 5556 > 400 \quad \checkmark$

Step 2: Use the simplified equation

$x = \sqrt{K_a \cdot C} = \sqrt{(1.8 \times 10^{-5})(0.10)}$

$x = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \text{ M}$

Step 3: Verify the 5% check

$\frac{x}{C} \times 100\% = \frac{1.34 \times 10^{-3}}{0.10} \times 100\% = 1.3\% < 5\% \quad \checkmark$

Step 4: Calculate pH

$pH = -\log(1.34 \times 10^{-3}) = 2.87$

Compare: 0.10 M $HCl$ has $pH = 1.00$. Same concentration, but the weak acid has a much higher pH!

ICE Table Concept Check 🎯

Weak Acid pH Calculations 🧮

1. Find the pH of 0.25 M $HF$ ($K_a = 6.8 \times 10^{-4}$). (2 decimal places)

2. Find $[H^+]$ for 0.050 M $HCN$ ($K_a = 6.2 \times 10^{-10}$). (Enter in scientific notation, e.g. 5.6e-6)

3. What is the percent ionization of 0.10 M acetic acid ($[H^+] = 1.34 \times 10^{-3}$ M)? (1 decimal place, enter number only)

ICE Table Reasoning 🔍

Exit Quiz — ICE Tables for Weak Acids ✅

🧴 Weak Bases and $K_b$

Part 3 of 7 — The Base Dissociation Constant

Weak bases partially react with water to produce $OH^-$ ions. The extent of this reaction is measured by the base dissociation constant, $K_b$.

Weak Base Equilibrium

A generic weak base $B$ in water:

$B(aq) + H_2O(l) \rightleftharpoons BH^+(aq) + OH^-(aq)$

The equilibrium expression is:

$K_b = \frac{[BH^+][OH^-]}{[B]}$

Key Points

• Water is omitted (pure liquid)
• $K_b$ is small → partial reaction only
• Larger $K_b$ = stronger weak base
• The base accepts a proton from water (Brønsted-Lowry)

Common Weak Bases

Relative strength: $CH_3NH_2 > NH_3 > C_5H_5N > C_6H_5NH_2$

ICE Table for Weak Bases

For $NH_3$ at concentration $C$:

$NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$

$K_b = \frac{x^2}{C - x} \approx \frac{x^2}{C}$

$x = [OH^-] = \sqrt{K_b \cdot C}$

Then: $pOH = -\log[OH^-]$ and $pH = 14 - pOH$

Worked Example

Find the pH of 0.15 M $NH_3$ ($K_b = 1.8 \times 10^{-5}$).

$[OH^-] = \sqrt{(1.8 \times 10^{-5})(0.15)} = \sqrt{2.7 \times 10^{-6}} = 1.64 \times 10^{-3} \text{ M}$

$pOH = -\log(1.64 \times 10^{-3}) = 2.79$

$pH = 14 - 2.79 = 11.21$

5% check: $1.64 \times 10^{-3}/0.15 = 1.1\% < 5\%$ ✓

Weak Base Concept Check 🎯

Conjugate Bases of Weak Acids

The conjugate base of a weak acid also acts as a weak base in water:

$A^-(aq) + H_2O(l) \rightleftharpoons HA(aq) + OH^-(aq)$

$K_b = \frac{[HA][OH^-]}{[A^-]}$

Example: Acetate Ion

$CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)$

This is why solutions of sodium acetate ($NaCH_3COO$) are basic — the acetate ion is a weak base!

Salts and pH

Weak Base Calculations 🧮

1. Find the pH of 0.20 M methylamine ($CH_3NH_2$, $K_b = 4.4 \times 10^{-4}$). (2 decimal places)

2. Find $[OH^-]$ for 0.10 M pyridine ($C_5H_5N$, $K_b = 1.7 \times 10^{-9}$). (Enter in scientific notation, e.g. 1.3e-5)

3. A solution of 0.25 M $NaCH_3COO$ is basic. If $K_b$ for $CH_3COO^-$ is $5.6 \times 10^{-10}$, find the pH. (2 decimal places)

Weak Base Reasoning 🔍

Exit Quiz — Weak Bases ✅

🔗 The $K_a \times K_b = K_w$ Relationship

Part 4 of 7 — Connecting Conjugate Pairs

One of the most powerful relationships in acid-base chemistry connects the strength of a conjugate acid-base pair through the ion product of water.

Deriving the Relationship

Consider acetic acid and its conjugate base, acetate:

Acid dissociation:

$CH_3COOH \rightleftharpoons H^+ + CH_3COO^- \qquad K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$

Base hydrolysis (conjugate base):

$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^- \qquad K_b = \frac{[CH_3COOH][OH^-]}{[CH_3COO^-]}$

Multiply $K_a \times K_b$:

$K_a \times K_b = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} \times \frac{[CH_3COOH][OH^-]}{[CH_3COO^-]}$

$K_a \times K_b = [H^+][OH^-] = K_w$

$\boxed{K_a \times K_b = K_w = 1.0 \times 10^{-14} \text{ at 25°C}}$

This is true for any conjugate acid-base pair!

The $pK_a + pK_b = 14$ Relationship

Taking $-\log$ of both sides of $K_a \times K_b = K_w$:

$-\log K_a + (-\log K_b) = -\log K_w$

$\boxed{pK_a + pK_b = pK_w = 14 \text{ at 25°C}}$

Applications

If you know $K_a$ for an acid, you can find $K_b$ for its conjugate base:

$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{K_a}$

Example

$CH_3COOH$ has $K_a = 1.8 \times 10^{-5}$. What is $K_b$ for $CH_3COO^-$?

$K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$

$pK_b = 14 - 4.74 = 9.26$

Key Insight

• Strong acid ($K_a$ very large) → very weak conjugate base ($K_b$ very small)
• Weak acid ($K_a$ small) → relatively stronger conjugate base ($K_b$ less small)

$K_a \times K_b$ Concept Check 🎯

$K_a$/$K_b$ Conversion Drill 🧮

1. $HF$ has $K_a = 6.8 \times 10^{-4}$. Find $K_b$ for $F^-$. (Enter in scientific notation, e.g. 1.5e-11)

2. $NH_3$ has $K_b = 1.8 \times 10^{-5}$. Find $K_a$ for $NH_4^+$. (Enter in scientific notation, e.g. 5.6e-10)

3. A weak acid has $pK_a = 3.75$. Find $pK_b$ for its conjugate base. (3 significant figures)

Using $K_a/K_b$ to Predict Salt Solutions

For salts of weak acid + strong base (e.g., $NaCH_3COO$):

1. Identify the ion that reacts with water ($CH_3COO^-$)
2. Find its $K_b$ using $K_b = K_w/K_a$
3. Use ICE table with $K_b$ to find $[OH^-]$
4. Convert to pH

Example: pH of 0.20 M NaCN

$K_a(HCN) = 6.2 \times 10^{-10}$ → $K_b(CN^-) = 1.6 \times 10^{-5}$

$[OH^-] = \sqrt{(1.6 \times 10^{-5})(0.20)} = \sqrt{3.2 \times 10^{-6}} = 1.8 \times 10^{-3} \text{ M}$

$pOH = 2.74 \qquad pH = 11.26$

Conjugate Pair Strength 🔍

Exit Quiz — $K_a \times K_b = K_w$ ✅

📈 Percent Ionization and Polyprotic Acids

Part 5 of 7 — Advanced Weak Acid Concepts

This part covers two important topics: how concentration affects the degree of dissociation, and how acids with multiple ionizable protons behave.

Percent Ionization

$\text{Percent ionization} = \frac{[H^+]_{eq}}{[HA]_0} \times 100\%$

Key Trend

For a given weak acid, diluting the solution increases percent ionization.

Why? Le Chatelier's principle: dilution shifts the equilibrium $HA \rightleftharpoons H^+ + A^-$ to the right (toward more ions, since there are more moles of product than reactant).

Mathematical Proof

$[H^+] = \sqrt{K_a \cdot C}$

$\text{\% ionization} = \frac{\sqrt{K_a \cdot C}}{C} \times 100 = \frac{\sqrt{K_a}}{\sqrt{C}} \times 100$

As $C$ decreases, $\frac{1}{\sqrt{C}}$ increases, so percent ionization increases!

Example: 0.10 M vs 0.010 M Acetic Acid

Polyprotic Acids

Polyprotic acids can donate more than one proton. Each dissociation has its own $K_a$.

Diprotic Acid Example: $H_2SO_3$

$H_2SO_3 \rightleftharpoons H^+ + HSO_3^- \qquad K_{a1} = 1.5 \times 10^{-2}$

$HSO_3^- \rightleftharpoons H^+ + SO_3^{2-} \qquad K_{a2} = 6.3 \times 10^{-8}$

Triprotic Acid Example: $H_3PO_4$

$H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^- \qquad K_{a1} = 7.5 \times 10^{-3}$

$H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-} \qquad K_{a2} = 6.2 \times 10^{-8}$

$HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-} \qquad K_{a3} = 4.8 \times 10^{-13}$

Critical Rule

$K_{a1} \gg K_{a2} \gg K_{a3}$

Each successive dissociation is much weaker because it's harder to remove $H^+$ from an increasingly negative ion.

Practical Consequence

For pH calculations, only the first dissociation matters (in most cases). The second and third contribute negligible additional $[H^+]$.

Percent Ionization & Polyprotic Acids 🎯

Percent Ionization & Polyprotic Calculations 🧮

1. What is the percent ionization of 0.050 M $HF$ ($K_a = 6.8 \times 10^{-4}$)? (1 decimal place)

2. Find the pH of 0.10 M $H_3PO_4$ ($K_{a1} = 7.5 \times 10^{-3}$). Use only the first dissociation. (2 decimal places)

3. For $H_2CO_3$ ($K_{a1} = 4.3 \times 10^{-7}$, $K_{a2} = 4.7 \times 10^{-11}$), what is $[CO_3^{2-}]$ in a 0.10 M solution? (Enter in scientific notation, e.g. 4.7e-11)

Advanced Concepts 🔍

Exit Quiz — Percent Ionization & Polyprotic Acids ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Weak Acids, Bases, and $K_a$/$K_b$

This workshop brings together ICE tables, $K_a$/$K_b$ relationships, percent ionization, and polyprotic acid concepts in multi-step AP-style problems.

Problem 1: Complete Weak Acid Analysis

A 0.20 M solution of benzoic acid ($C_6H_5COOH$, $K_a = 6.3 \times 10^{-5}$) is prepared.

Step 1: Check the approximation

$C/K_a = 0.20/(6.3 \times 10^{-5}) = 3175 > 400$ ✓

Step 2: Calculate $[H^+]$

$[H^+] = \sqrt{K_a \cdot C} = \sqrt{(6.3 \times 10^{-5})(0.20)} = \sqrt{1.26 \times 10^{-5}} = 3.55 \times 10^{-3} \text{ M}$

Step 3: pH

$pH = -\log(3.55 \times 10^{-3}) = 2.45$

Step 4: Percent ionization

$\% = (3.55 \times 10^{-3}/0.20) \times 100 = 1.8\%$ ✓ (under 5%)

Step 5: $K_b$ of conjugate base

$K_b(C_6H_5COO^-) = K_w/K_a = 1.0 \times 10^{-14}/(6.3 \times 10^{-5}) = 1.6 \times 10^{-10}$

Your Turn: Complete Analysis 🧮

Perform the same analysis for 0.15 M $HNO_2$ ($K_a = 4.5 \times 10^{-4}$):

1. What is $[H^+]$? (Enter in scientific notation, e.g. 8.2e-3)

2. What is the pH? (2 decimal places)

3. What is the percent ionization? (1 decimal place, enter number only)

Problem 2: Salt Solution pH

What is the pH of 0.30 M sodium fluoride ($NaF$)?

Analysis

$NaF$ dissociates completely: $Na^+$ (spectator) + $F^-$ (conjugate base of $HF$)

$F^-$ is a weak base: $F^- + H_2O \rightleftharpoons HF + OH^-$

Find $K_b$

$K_a(HF) = 6.8 \times 10^{-4}$

$K_b(F^-) = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} = 1.47 \times 10^{-11}$

ICE Table

$[OH^-] = \sqrt{K_b \cdot C} = \sqrt{(1.47 \times 10^{-11})(0.30)} = 2.10 \times 10^{-6} \text{ M}$

$pOH = 5.68 \qquad pH = 14 - 5.68 = 8.32$

Salt Solution Practice 🎯

Problem 3: Determining $K_a$ from pH 🧮

A 0.25 M solution of an unknown weak acid has a pH of 2.72.

1. What is $[H^+]$? (Enter in scientific notation, e.g. 1.9e-3)

2. What is the $K_a$ of the acid? (Enter in scientific notation, e.g. 1.5e-5)

3. What is the $pK_a$? (2 decimal places)

Workshop Synthesis 🔍

Exit Quiz — Problem-Solving Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Weak Acids, Bases, and $K_a$/$K_b$

This final part provides comprehensive AP-style review covering all weak acid/base concepts: $K_a$/$K_b$ calculations, ICE tables, the 5% approximation, percent ionization, polyprotic acids, and salt solutions.

Complete Summary

Key Equations

Decision Flowchart

1. Strong acid/base? → Use concentration directly
2. Weak acid? → ICE table with $K_a$
3. Weak base? → ICE table with $K_b$, find $[OH^-]$ first
4. Salt? → Identify hydrolyzable ion, use $K_b = K_w/K_a$ or $K_a = K_w/K_b$
5. Always check the 5% approximation!

AP-Style Questions — Set 1 🎯

AP Calculation Practice 🧮

1. Calculate the pH of 0.35 M $NH_3$ ($K_b = 1.8 \times 10^{-5}$). (2 decimal places)

2. What is $K_a$ for $NH_4^+$? (Enter in scientific notation, e.g. 5.6e-10)

3. A solution of 0.10 M $NH_4Cl$ has what pH? (2 decimal places)

AP-Style Questions — Set 2 🎯

Comprehensive Review 🔍

Final Exit Quiz — Weak Acids & Bases ✅