Monoprotic Acid-Base Equilibria: Strong and Weak Acids/Bases, Equilibrium Calculations, and Buffers

Chapter 9: Monoprotic Acid-Base Equilibria

Overview

This chapter covers the fundamental principles of acid-base equilibria for monoprotic systems, focusing on strong and weak acids and bases, their equilibria, and the concept and calculation of buffer solutions.

• Strong Acids and Bases

• Weak Acids and Bases

• Weak-Acid Equilibria

• Weak-Base Equilibria

• Buffers

Strong Acids and Bases

Dissociation and pH Calculation

Strong acids and strong bases dissociate completely in aqueous solution. This means their concentration directly determines the concentration of hydronium ions (H+) or hydroxide ions (OH-).

• Example Reaction: HBr + H2O → H3O+ + Br-

• pH Calculation: For a strong acid, $\text{pH} = -\log[\text{H}^+]$

• Example: 0.10 M HBr: $\text{pH} = -\log(0.10) = 1.00$

pH and pOH Relationship

• $\text{pH} + \text{pOH} = 14.00$ (at 25°C)

• For a strong base: $\text{pOH} = -\log[\text{OH}^-]$

• To find pH from [OH-]: $\text{pH} = 14.00 - \text{pOH}$

Activity Coefficient in Strong Acid Calculations

At higher ionic strengths, the activity coefficient (γ) must be considered:

• $\text{pH} = -\log([\text{H}^+]\gamma_{+})$

• Example: For 0.10 M HBr, if $\gamma_{+} = 0.83$, $\text{pH} = -\log(0.10 \times 0.83) = 1.08$

Systematic Treatment of Equilibrium for Dilute Solutions

When acid or base concentrations approach $10^{-7}$ M, the contribution of water's autoionization becomes significant. In such cases, a systematic approach is required:

1. Write all relevant equilibrium reactions.

2. Predict the dominant species.

3. Set up mass and charge balance equations.

4. Simplify using reasonable assumptions.

5. Solve the resulting equations.

Example: For 1.0 × 10-8 M KOH, both the added base and water's autoionization contribute to [OH-].

Key Equations:

• Water dissociation: $\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$

• $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ at 25°C

Weak Acids and Bases

Weak Acid Dissociation and pH Calculation

Weak acids only partially dissociate in water. The equilibrium is described by the acid dissociation constant ($K_a$):

• General reaction: $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$

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

• If $[\text{H}^+] = [\text{A}^-] = x$ and $[\text{HA}] = F - x$ (where F is the formal concentration): $K_a = \frac{x^2}{F - x}$

Fraction of Dissociation (α)

The fraction of dissociation (α) quantifies the proportion of acid molecules that dissociate:

• $\alpha = \frac{[\text{A}^-]}{F} = \frac{x}{F}$

• For weak acids, α is typically much less than 1.

Weak Base Equilibria

Weak bases also partially react with water. The base hydrolysis constant ($K_b$) is used:

• General reaction: $\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-$

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

• If $[\text{OH}^-] = [\text{BH}^+] = x$ and $[\text{B}] = F - x$: $K_b = \frac{x^2}{F - x}$

The relationship between $K_a$ and $K_b$ for a conjugate acid-base pair is:

• $K_a \times K_b = K_w$

pK Values

• $pK = -\log K$

• As $K$ increases, $pK$ decreases (stronger acid or base).

Weak-Acid and Weak-Base Equilibria: Systematic Approach

Setting Up Equilibrium Problems

For weak acids and bases, equilibrium calculations often require:

• Writing all relevant equilibrium expressions

• Applying mass balance and charge balance equations

• Making justified approximations to simplify calculations

Example for Weak Acid:

• Mass balance: $F = [\text{A}^-] + [\text{HA}]$

• Charge balance: $[\text{H}^+] = [\text{A}^-] + [\text{OH}^-]$

• Equilibrium: $K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$

• Water: $K_w = [\text{H}^+][\text{OH}^-]$

Buffers

Definition and Function

A buffer is a solution containing a weak acid and its conjugate base (or a weak base and its conjugate acid). Buffers resist changes in pH upon addition of small amounts of strong acid or base.

• Buffer action is based on the equilibrium between the weak acid and its conjugate base.

• Buffer capacity is greatest when $[\text{acid}] = [\text{base}]$.

The Henderson-Hasselbalch Equation

This equation relates the pH of a buffer to the concentrations of acid and base:

• For a weak acid buffer: $\text{pH} = pK_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$

• For a weak base buffer: $\text{pH} = pK_a + \log\left(\frac{[\text{B}]}{[\text{BH}^+]}\right)$

Key Properties:

• When $[\text{A}^-] = [\text{HA}]$, $\text{pH} = pK_a$

• Changing the ratio $[\text{A}^-]/[\text{HA}]$ by a factor of 10 changes pH by 1 unit.

• Buffer capacity increases with total buffer concentration.

• The useful pH range for a buffer is $pK_a \pm 1$.

Buffer Preparation and Calculations

• To prepare a buffer, mix a weak acid and its conjugate base in desired proportions.

• Use the Henderson-Hasselbalch equation to calculate the required ratio for a target pH.

• Buffer pH is nearly independent of dilution, but buffer capacity depends on the absolute concentrations.

Example Buffer Calculation

• Mix 50 mL of 0.5 M sodium acetate with 25 mL of 0.1 M acetic acid. Total volume = 75 mL.

• Calculate new concentrations: $[\text{A}^-] = 0.33$ M, $[\text{HA}] = 0.033$ M.

• Given $pK_a = 4.757$, $\text{pH} = 4.757 + \log(0.33/0.033) = 5.757$

Buffer in Action

• Adding strong acid or base to a buffer results in only a small change in pH, as the buffer components react to neutralize the added species.

• Buffer effectiveness is lost if one component is depleted.

Choosing a Buffer System

• Select a weak acid/conjugate base pair with a $pK_a$ close to the desired pH.

• Example: For pH 10.0, use ammonia ($pK_a = 9.24$) as the buffer system.

Summary Table: Key Acid-Base Relationships

Concept	Equation/Relationship	Notes
Strong Acid/Base pH	$\text{pH} = -\log[\text{H}^+]$ $\text{pOH} = -\log[\text{OH}^-]$	Direct calculation from concentration
Weak Acid pH	$K_a = \frac{x^2}{F - x}$	Solve for x = [H+]
Weak Base pH	$K_b = \frac{x^2}{F - x}$	Solve for x = [OH-]
Conjugate Acid/Base	$K_a \times K_b = K_w$	For conjugate pairs
Henderson-Hasselbalch	$\text{pH} = pK_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$	Buffer calculations