Chapter 17: Additional Aspects of Aqueous Equilibrium

Additional Aspects of Aqueous Equilibrium

Common-Ion Effect

The common-ion effect refers to the shift in ionic equilibrium caused by the addition of a solute that provides an ion already present in the equilibrium. This effect suppresses the ionization of a weak acid or base.

• Example: Acetic acid/sodium acetate system:

$\mathrm{CH_3COOH~(aq) + H_2O~(l) \rightleftharpoons CH_3COO^-~(aq) + H_3O^+~(aq)}$

Adding sodium acetate ($\mathrm{NaCH_3COO}$) increases $\mathrm{CH_3COO^-}$, shifting equilibrium left and decreasing $\mathrm{H_3O^+}$, thus increasing pH.

• Buffering: The solution resists changes in pH due to the presence of both a weak acid and its conjugate base.

Buffers

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Buffers are typically prepared by mixing a weak acid with its conjugate base or a weak base with its conjugate acid.

• Preparation: Mix equal molar amounts of a weak acid and its salt (conjugate base), or a weak base and its salt (conjugate acid).

• Buffer Solution Example A: $1~\mathrm{M}~\mathrm{NH_3}$ and $1~\mathrm{M}~\mathrm{NH_4Cl}$ in water.

• Buffer Solution Example B: $1~\mathrm{M}$ acetic acid and sodium acetate in water.

• Neutralization: Buffers neutralize added acids or bases, shifting equilibrium to minimize pH change.

The pH of a Buffer

The pH of a buffer can be calculated using the concentrations of the acid and its conjugate base (or base and its conjugate acid) and the appropriate equilibrium constant.

• Example: Mixing $60~\mathrm{mL}$ of $0.100~\mathrm{M}~\mathrm{NH_3}$ with $40~\mathrm{mL}$ of $0.100~\mathrm{M}~\mathrm{NH_4Cl}$ gives $[\mathrm{NH_3}] = 0.060~\mathrm{M}$ and $[\mathrm{NH_4^+}] = 0.040~\mathrm{M}$ after dilution.

• Set up an ICE table and solve for $[\mathrm{OH^-}]$ using $K_b$:

$\mathrm{NH_3~(aq) + H_2O~(l) \rightleftharpoons NH_4^+~(aq) + OH^-~(aq)}$

$K_b = \frac{[\mathrm{NH_4^+}][\mathrm{OH^-}]}{[\mathrm{NH_3}]}$

$[\mathrm{OH^-}] = 2.7 \times 10^{-5}$

$pOH = -\log(2.7 \times 10^{-5}) = 4.57$

$pH = 14 - 4.57 = 9.43$

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation provides a convenient way to calculate the pH of a buffer solution.

• For an acid buffer:

$pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right)$

• For a base buffer:

$pOH = pK_b + \log \left( \frac{[HB^+]}{[B]} \right)$

• General form:

$pH = pK_a + \log \left( \frac{\text{base}}{\text{acid}} \right)$

• Relationship: $pK_a + pK_b = 14$

• When $[\text{acid}] = [\text{base}]$, $pH = pK_a$

Derivation of the Henderson-Hasselbalch Equation

Starting from the acid dissociation constant:

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

Taking logarithms and rearranging gives:

$pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right)$

Acid-Base Titration

An acid-base titration curve plots the pH of a solution as a function of the volume of added acid or base. The equivalence point is when stoichiometric amounts of acid and base have reacted.

• Strong Acid/Strong Base: $\mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{H_2O} + \mathrm{NaCl}$

• At the equivalence point, $pH = 7$

• All neutralization reactions go to completion.

Titration of a Weak Acid by a Strong Base

• At the end point, the solution contains the conjugate base of the weak acid.

• pH at the end point is greater than 7.

Titration of a Weak Base by a Strong Acid

• At the end point, the solution contains the conjugate acid of the weak base.

• pH at the end point is less than 7.

Example: Titration Calculations

Calculate the pH at various points during the titration of $50~\mathrm{mL}$ of $0.10~\mathrm{M}$ acetic acid with $0.10~\mathrm{M}$ NaOH:

• (a) 0 mL NaOH (before titration): $pH = 2.88$

• (b) 10 mL NaOH: $pH = 4.17$

• (c) 25 mL NaOH (half-equivalence): $pH = 4.77$

• (d) 50 mL NaOH (equivalence): $pH = 8.74$

• (e) 60 mL NaOH (after equivalence): $pH = 12.00$

At the half-equivalence point, $pH = pK_a$.

Equivalents and Normality

Equivalent (eq): In acid-base reactions, one equivalent is the amount of acid that yields 1 mole of $H^+$ or base that reacts with 1 mole of $H^+$.

• Equivalent mass (eq mass): The mass of one equivalent, calculated as:

$\text{eq mass} = \frac{\text{molar mass}}{\#~H^+~\text{released or neutralized}}$

• Normality (N): $N = \frac{\text{equivalents of solute}}{\text{liter of solution}}$

Example: $0.10~\mathrm{mol}$ $\mathrm{H_2SO_4}$ in $0.10~\mathrm{L}$ solution: $N = 2.0~\mathrm{N}$

Redox Equivalents

In redox reactions, one equivalent is the amount of oxidizing or reducing agent that uses or provides one mole of electrons.

• Example: $\mathrm{MnO_4^-} + 8H^+ + 5e^- \rightarrow \mathrm{Mn^{2+}} + 4H_2O$

• 1 mole $\mathrm{MnO_4^-}$ = 5 eq

Solubility Equilibrium

Solubility product constant ($K_{sp}$): The equilibrium constant for the solubility of a slightly soluble salt.

• Example: $\mathrm{CaC_2O_4~(s) \rightleftharpoons Ca^{2+}~(aq) + C_2O_4^{2-}~(aq)}$

• $K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{C_2O_4^{2-}}]$

Writing $K_{sp}$ expressions:

• $\mathrm{AgCl~(s) \rightleftharpoons Ag^+~(aq) + Cl^-~(aq)}$ $K_{sp} = [\mathrm{Ag^+}][\mathrm{Cl^-}]$

• $\mathrm{PbI_2~(s) \rightleftharpoons Pb^{2+}~(aq) + 2I^-~(aq)}$ $K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{I^-}]^2$

Calculating $K_{sp}$ from Solubility

• Molar Solubility (S): Number of moles of salt that dissolve per liter of water.

• Example: If the molar solubility of $\mathrm{CaC_2O_4}$ is $4.8 \times 10^{-5}~\mathrm{M}$, then:

$K_{sp} = S^2 = (4.8 \times 10^{-5})^2 = 2.3 \times 10^{-9}$

• For $\mathrm{PbI_2}$, $K_{sp} = 4S^3$ (since $[\mathrm{Pb^{2+}}] = S$, $[\mathrm{I^-}] = 2S$)

Calculating Solubility from $K_{sp}$

• Example: For $\mathrm{CaF_2}$, $K_{sp} = 3.4 \times 10^{-4}$

$K_{sp} = 4S^3 \implies S = \sqrt[3]{K_{sp}/4}$

$S = 2.0 \times 10^{-2}~\mathrm{M}$

Solubility and the Common-Ion Effect

• The presence of a common ion decreases the solubility of a salt.

• Example: Molar solubility of $\mathrm{CaC_2O_4}$ in $0.15~\mathrm{M}~\mathrm{CaCl_2}$ is:

$K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{C_2O_4^{2-}}]$

$S = \frac{K_{sp}}{[\mathrm{Ca^{2+}}]}$

Precipitation Calculations

The ion product ($Q_c$) is the product of ion concentrations in a solution, each raised to the power of its coefficient in the balanced equation.

• If $Q_c < K_{sp}$: No precipitation; solution is unsaturated.

• If $Q_c > K_{sp}$: Precipitation occurs; solution is supersaturated.

• If $Q_c = K_{sp}$: Solution is just saturated; no more salt dissolves.

Qualitative Effect of pH

• If the anion is the conjugate base of a weak acid, it reacts with $H^+$, affecting solubility.

• Salts like $\mathrm{Ca(CN)_2}$ are more soluble in acid than in base.

• Salts like $\mathrm{Ca(OH)_2}$ are more soluble in acid than in base due to reaction with $H^+$.

Separation of Metal Ions by Sulfide Precipitation

Many metal sulfides are insoluble in water but dissolve in acidic solution. The change in solubility with pH can be used to separate metal ions.

• Example: $\mathrm{H_2S~(g) \rightleftharpoons H^+~(aq) + HS^-~(aq)}$

• Solubility depends on $[H^+]$; lower pH increases solubility of metal sulfides.

Complex-Ion Equilibrium

Complex ions are formed from a metal ion with Lewis bases (ligands) attached by coordinate covalent bonds. The formation constant ($K_f$) describes the equilibrium for complex ion formation.

• General reaction: $\mathrm{M^{2+}~(aq) + 6~L~(aq) \rightleftharpoons ML_6^{2+}~(aq)}$

• $K_f = \frac{[ML_6^{2+}]}{[M^{2+}][L]^6}$

• The dissociation constant ($K_d$) is the reciprocal of $K_f$:

$K_d = \frac{[M^{2+}][L]^6}{[ML_6^{2+}]} = \frac{1}{K_f}$

Summary Table: Titration End-Point pH

Titration Type	pH at End-Point
Strong Acid / Strong Base	7
Weak Acid / Strong Base	> 7
Weak Base / Strong Acid	< 7

Key Equations

• $K_a = \frac{[H^+][A^-]}{[HA]}$

• $pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right)$

• $K_{sp} = [\text{cation}]^m [\text{anion}]^n$

• $Q_c = [\text{cation}]^m [\text{anion}]^n$ (using experimental concentrations)

• $K_f = \frac{[ML_n]}{[M][L]^n}$

• $K_d = \frac{1}{K_f}$

• $\text{Normality}~(N) = \frac{\text{equivalents}}{\text{liter}}$