Aqueous Ionic Equilibrium: Conjugate Acids/Bases, Buffers, and Solubility

Chapter 18: Aqueous Ionic Equilibrium

Overview

This chapter covers the principles of aqueous ionic equilibrium, focusing on conjugate acid-base pairs, the common-ion effect, buffer solutions, acid-base reactions, and solubility product equilibrium. These concepts are essential for understanding chemical equilibria in aqueous solutions, particularly in General Chemistry II.

Conjugate Acids and Bases

Bronsted-Lowry Definition

The Bronsted-Lowry definition classifies acids as proton donors and bases as proton acceptors. A conjugate acid-base pair consists of two species that differ by a single proton (H+).

• Conjugate acid: The species formed when a base gains a proton.

• Conjugate base: The species formed when an acid loses a proton.

Example:

• $\mathrm{CH_3COOH} + \mathrm{H_2O} \rightleftharpoons \mathrm{CH_3COO^-} + \mathrm{H_3O^+}$

• Here, CH3COOH (acetic acid) and CH3COO- (acetate) are a conjugate acid-base pair.

Identification of Conjugate Acid-Base Pairs

To identify conjugate pairs in reactions, look for species that differ by one proton:

• $\mathrm{HCl} + \mathrm{H_2O} \rightarrow \mathrm{H_3O^+} + \mathrm{Cl^-}$

• $\mathrm{NH_3} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_4^+} + \mathrm{OH^-}$

Common-Ion Effect

Definition and Application

The common-ion effect occurs when the addition of an ion common to two solutes reduces the ionization of a weak electrolyte. This effect is used to control the concentrations of ions in solution and is important in buffer preparation.

• Adding sodium acetate to acetic acid solution suppresses the ionization of acetic acid.

• Example: $\mathrm{CH_3COOH} + \mathrm{H_2O} \rightleftharpoons \mathrm{CH_3COO^-} + \mathrm{H_3O^+}$; addition of $\mathrm{NaCH_3COO}$ increases $\mathrm{CH_3COO^-}$, shifting equilibrium left.

Buffer Solutions

Properties and Function

Buffer solutions resist changes in pH when small amounts of acid or base are added. They contain significant quantities of both a weak acid and its conjugate base, or a weak base and its conjugate acid.

• Weak acid neutralizes added base.

• Conjugate base neutralizes added acid.

• Buffers can also be made from a weak base and its conjugate acid.

Example:

• $\mathrm{CH_3COOH} + \mathrm{H_2O} \rightleftharpoons \mathrm{CH_3COO^-} + \mathrm{H_3O^+}$

• Addition of acid: $\mathrm{CH_3COO^-} + \mathrm{H^+} \rightarrow \mathrm{CH_3COOH}$

• Addition of base: $\mathrm{CH_3COOH} + \mathrm{OH^-} \rightarrow \mathrm{CH_3COO^-} + \mathrm{H_2O}$

Characteristics of Buffer Solutions

• Acidic buffers have a pH higher than the original acid alone.

• Basic buffers have a pH lower than the original base alone.

• Buffer capacity depends on the concentrations of acid and conjugate base.

Henderson-Hasselbalch Equation

Calculating pH of Buffer Solutions

The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentrations of the acid and its conjugate base:

• For acidic buffers: $\mathrm{pH} = \mathrm{p}K_a + \log \left( \frac{[\text{conjugate base}]}{[\text{acid}]} \right)$

• For basic buffers: $\mathrm{pOH} = \mathrm{p}K_b + \log \left( \frac{[\text{conjugate acid}]}{[\text{base}]} \right)$

Example Calculation:

• Given 0.20 M $\mathrm{NH_3}$ and 0.10 M $\mathrm{NH_4Cl}$, $K_b = 1.8 \times 10^{-5}$

• Calculate $[OH^-]$ and then pOH and pH:

• $\mathrm{pOH} = -\log[OH^-]$

• $\mathrm{pH} = 14.00 - \mathrm{pOH}$

Preparation of Buffer Solutions

Steps for Buffer Preparation

• Mix a weak acid (or base) with its conjugate base (or acid).

• Calculate the concentrations after dilution.

• Use the Henderson-Hasselbalch equation to determine the pH.

• Adjust the ratio of acid/base to achieve the desired pH.

Example: Preparing a buffer with acetic acid and sodium acetate, calculate the final concentrations and use the equation above to find pH.

Acid-Base Reactions

Mixing Strong Acids and Bases

When strong acids and bases are mixed, the resulting pH depends on the stoichiometry and concentrations of the reactants.

• Write the balanced chemical equation.

• Calculate the moles of acid and base.

• Determine the limiting reactant and the concentration of excess species.

• Calculate the final pH using $\mathrm{pH} = -\log[H^+]$.

Example: Mixing 25.00 mL of 0.1500 M $\mathrm{HNO_3}$ with 30.00 mL of 0.1000 M $\mathrm{NaOH}$.

Solubility Product Equilibrium (Ksp)

Solubility Rules

Solubility rules help predict whether an ionic compound is soluble in water. Compounds containing alkali metals, ammonium, and nitrates are generally soluble.

Solubility Product Constant (Ksp)

The solubility product constant ($K_{sp}$) is the product of the concentrations of the constituent ions, each raised to the power of its coefficient in the balanced equation, for a saturated solution.

• Example: $\mathrm{BaSO_4 (s) \rightleftharpoons Ba^{2+} (aq) + SO_4^{2-} (aq)}$

• $K_{sp} = [\mathrm{Ba^{2+}}][\mathrm{SO_4^{2-}}]$

Determining Ksp Expressions

Compound	Dissociation Reaction	Ksp Expression
AgCl	$\mathrm{AgCl (s) \rightleftharpoons Ag^+ (aq) + Cl^- (aq)}$	$K_{sp} = [\mathrm{Ag^+}][\mathrm{Cl^-}]$
Zn3(PO4)2	$\mathrm{Zn_3(PO_4)_2 (s) \rightleftharpoons 3Zn^{2+} (aq) + 2PO_4^{3-} (aq)}$	$K_{sp} = [\mathrm{Zn^{2+}}]^3[\mathrm{PO_4^{3-}}]^2$
Ca(OH)2	$\mathrm{Ca(OH)_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2OH^- (aq)}$	$K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{OH^-}]^2$

Molar Solubility

Molar solubility is the maximum amount of a compound (in mol/L) that can dissolve in water at a specific temperature. It can be calculated from $K_{sp}$ or by dimensional analysis.

• Example: For $\mathrm{PbCl_2}$, $K_{sp} = 1.17 \times 10^{-5}$

• Dissociation: $\mathrm{PbCl_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2Cl^- (aq)}$

• Let $x$ be the molar solubility: $K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{Cl^-}]^2 = x \cdot (2x)^2 = 4x^3$

• Solve for $x$ to find molar solubility.

Calculating Molar Solubility and Ksp

Given the mass of a compound dissolved in a known volume, calculate molar solubility and use it to determine $K_{sp}$.

• Example: Dissolving 0.25 g of $\mathrm{BaSO_4}$ in 1.00 L water.

• Calculate moles of $\mathrm{BaSO_4}$, then use $K_{sp} = [\mathrm{Ba^{2+}}][\mathrm{SO_4^{2-}}]$.

Additional info: Some calculations and tables were inferred and expanded for clarity and completeness.