Acid-Base Equilibria and Solubility Equilibria

Acid-Base Equilibria and the Common Ion Effect

• The common ion effect is defined as the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.

• The presence of a common ion suppresses the ionization of a weak acid or a weak base.

• Example Case: Consider a mixture of $CH_3COONa$ (a strong electrolyte) and $CH_3COOH$ (a weak acid).

  • Dissociation of Salt: $CH_3COONa(s) \rightarrow Na^+(aq) + CH_3COO^-(aq)$

  • Ionization of Acid: $CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)$

  • The $CH_3COO^-$ ion is the common ion in this system. Adding the salt increases the concentration of acetate, shifting the acid equilibrium to the left according to Le Châtelier's principle, thus decreasing the acidity ($H^+$ concentration).

The Henderson-Hasselbalch Equation

• To derive the pH of a solution containing a weak acid ($HA$) and its salt ($NaA$):

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

  • $NaA(s) \rightarrow Na^+(aq) + A^-(aq)$

  • The equilibrium constant expression: $K_a = \frac{[H^+][A^-]}{[HA]}$

  • Rearranging for hydrogen ion concentration: $[H^+] = K_a \frac{[HA]}{[A^-]}$

  • Taking the negative log of both sides: $- \text{log}[H^+] = - \text{log}K_a - \text{log}\frac{[HA]}{[A^-]}$

  • This results in the Henderson-Hasselbalch Equation:

  • $pH = pK_a + \text{log}\frac{[A^-]}{[HA]}$

  • General form: $pH = pK_a + \text{log} \frac{[\text{conjugate base}]}{[\text{acid}]}$

  • Note: $pK_a = -\text{log}K_a$

• Application Problem: Calculate the pH of a solution containing $0.30 \text{ M HCOOH}$ and $0.52 \text{ M HCOOK}$.

  • Given: $pK_a$ of $HCOOH$ is $3.77$.

  • Since $0.30 - x \thickapprox 0.30$ and $0.52 + x \thickapprox 0.52$ due to the common ion effect:

  • $pH = 3.77 + \text{log}\frac{0.52}{0.30} = 4.01$

Buffer Solutions

• A buffer solution is a mixture of:

  1. A weak acid or a weak base.

  2. The salt of that weak acid or weak base (its conjugate).

• Key Requirement: Both components must be present in significant amounts.

• Function: A buffer has the ability to resist changes in pH upon the addition of small amounts of either acid or base.

• Mechanism of Action:

  • Addition of strong acid: $H^+(aq) + CH_3COO^-(aq) \rightarrow CH_3COOH(aq)$

  • Addition of strong base: $OH^-(aq) + CH_3COOH(aq) \rightarrow CH_3COO^-(aq) + H_2O(l)$

• Buffer Systems Identification:

  • (a) KF/HF: A buffer system (weak acid and its conjugate base).

  • (b) KBr/HBr: Not a buffer (HBr is a strong acid).

  • (c) Na2CO3/NaHCO3: A buffer system ($CO_3^{2-}$ is a weak base and $HCO_3^-$ is its conjugate acid).

Detailed Buffer Calculation example

• Scenario: Calculate pH of a buffer system: $0.30 \text{ M NH}_3 / 0.36 \text{ M NH}_4\text{Cl}$. ($pK_a = 9.25$).

  • Initial pH: $pH = 9.25 + \text{log}\frac{0.30}{0.36} = 9.17$

• Scenario Addition: pH after adding $20.0 \text{ mL}$ of $0.050 \text{ M NaOH}$ to $80.0 \text{ mL}$ of the buffer.

  • Moles added $OH^- = 0.020 \text{ L} \times 0.050 \text{ mol/L} = 0.001 \text{ moles}$

  • Initial moles $NH_4^+ = 0.080 \text{ L} \times 0.36 \text{ mol/L} = 0.0288 \text{ moles} \thickapprox 0.029 \text{ mol}$

  • Initial moles $NH_3 = 0.080 \text{ L} \times 0.30 \text{ mol/L} = 0.024 \text{ mol}$

  • Reaction: $NH_4^+(aq) + OH^-(aq) \rightarrow H_2O(l) + NH_3(aq)$

  • Final moles: $NH_4^+ = 0.029 - 0.001 = 0.028 \text{ mol}$; $NH_3 = 0.024 + 0.001 = 0.025 \text{ mol}$

  • Total volume: $100 \text{ mL} = 0.10 \text{ L}$

  • New pH: $pH = 9.25 + \text{log}\frac{0.25}{0.28} = 9.20$

Acid-Base Titrations

• Titration: A process where a solution of accurately known concentration is added gradually to another solution of unknown concentration until the chemical reaction is complete.

• Equivalence Point: The point at which the chemical reaction is complete.

• Indicator: A substance that changes color at or near the equivalence point.

Strong Acid-Strong Base Titrations

• Example: $NaOH(aq) + HCl(aq) \rightarrow H_2O(l) + NaCl(aq)$

• Net ionic equation: $OH^-(aq) + H^+(aq) \rightarrow H_2O(l)$

• The pH at the equivalence point is exactly $7.00$.

Weak Acid-Strong Base Titrations

• Example: $CH_3COOH(aq) + NaOH(aq) \rightarrow CH_3COONa(aq) + H_2O(l)$

• Reaction with base: $CH_3COOH(aq) + OH^-(aq) \rightarrow CH_3COO^-(aq) + H_2O(l)$

• At Equivalence Point: The concentration of acetic acid is zero. The resulting salt ($CH_3COO^-$) undergoes hydrolysis:

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

  • Consequently, the pH > 7 at the equivalence point.

• Calculation stages (25.0 mL of 0.100 M Acetic Acid with 0.100 M NaOH):

  • (a) 10.0 mL NaOH added: Formation of a buffer. $pH = 4.57$.

  • (b) 25.0 mL NaOH added: This is the equivalence point. Salt concentration is $0.0500 \text{ M}$. Using $K_b = 5.6 \times 10^{-10}$, $[OH^-] = 5.3 \times 10^{-6} \text{ M}$, resulting in pH = 8.72.

  • (c) 35.0 mL NaOH added: Excess strong base. pH is determined by excess $OH^-$. Total vol = $60 \text{ mL}$. $[OH^-] = 0.0167 \text{ M}$, pH = 12.22.

Strong Acid-Weak Base Titrations

• Example: $HCl(aq) + NH_3(aq) \rightarrow NH_4Cl(aq)$

• At the equivalence point, the $NH_4^+$ cation hydrolyzes: $NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H^+(aq)$

• Consequently, the pH < 7 at the equivalence point.

• Calculation for 25.0 mL 0.100 M NH3 titrated with 0.100 M HCl:

  • Salt concentration ($NH_4Cl$) at equivalence = $0.0500 \text{ M}$.

  • $K_a$ for $NH_4^+ = 5.6 \times 10^{-10}$.

  • $x^2 / 0.0500 = 5.6 \times 10^{-10}$

  • $x = [H^+] = 5.3 \times 10^{-6} \text{ M}$

  • pH = 5.28.

Acid-Base Indicators

• Indicators are themselves weak acids ($HIn$) that have different colors than their conjugate bases ($In^-$).

• Equilibrium: $HIn(aq) \rightleftharpoons H^+(aq) + In^-(aq)$

• Guidelines for Color Change:

  • If \frac{[HIn]}{[In^-]} > 10, the color of the acid ($HIn$) predominates.

  • If \frac{[HIn]}{[In^-]} < 0.10, the color of the conjugate base ($In^-$) predominates.

• Selection: For a titration of $HNO_2$ with $KOH$ (weak acid with strong base), the pH at equivalence is about $8.02$. Recommended indicators are cresol red or phenolphthalein.

Solubility Equilibria

• The solubility product constant ($K_{sp}$) is the equilibrium constant for the dissolution of an ionic solid.

• Examples:

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

  • $MgF_2(s) \rightleftharpoons Mg^{2+}(aq) + 2F^-(aq) \rightarrow K_{sp} = [Mg^{2+}][F^-]^2$

  • $Ca_3(PO_4)_2(s) \rightleftharpoons 3Ca^{2+}(aq) + 2PO_4^{3-}(aq) \rightarrow K_{sp} = [Ca^{2+}]^3[PO_4^{3-}]^2$

• Predicting Precipitation with Ion Product (Q):

  • Q < K_{sp}: Unsaturated solution; no precipitate.

  • $Q = K_{sp}$: Saturated solution.

  • Q > K_{sp}: Supersaturated solution; precipitate will form.

• Solubility Definitions:

  • Molar solubility (mol/L): Number of moles of solute in $1 \text{ L}$ of saturated solution.

  • Solubility (g/L): Number of grams of solute in $1 \text{ L}$ of saturated solution.

• The Common Ion Effect on Solubility:

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

  • example: Molar solubility of $AgBr$ in pure water ($s = 8.8 \times 10^{-7} \text{ M}$) vs. in $0.0010 \text{ M NaBr}$ ($s = 7.7 \times 10^{-10} \text{ M}$).

• pH and Solubility:

  • Insoluble bases dissolve in acidic solutions (e.g., $Mg(OH)_2$ becomes more soluble as pH decreases because $H^+$ removes $OH^-$).

  • Insoluble acids dissolve in basic solutions.

Complex Ion Equilibria

• A complex ion contains a central metal cation bonded to one or more molecules or ions (ligands).

• Formation Constant ($K_f$): Also known as the stability constant; it is the equilibrium constant for complex ion formation.

• Example: $Cu^{2+}(aq) + 4NH_3(aq) \rightleftharpoons Cu(NH_3)_4^{2+}(aq)$ with $K_f = 5.0 \times 10^{13}$.

• Large $K_f$ values indicate that the complex ion is very stable and the reaction goes nearly to completion.

• Effect on Solubility: The formation of complex ions can significantly increase the solubility of a normally "insoluble" salt.

  • Example: AgCl is much more soluble in $1.0 \text{ M NH}_3$ ($s = 0.045 \text{ M}$) than in pure water ($s = 1.3 \times 10^{-5} \text{ M}$) because an overall higher K is achieved by $(K_{sp} \times K_f)$.

Qualitative Analysis of Cations

• Cations can be separated into groups based on their precipitation reactions:

  • Group 1: $Ag^+, Hg_2^{2+}, Pb^{2+}$ (Precipitate with $HCl$ as chlorides).

  • Group 2: $Bi^{3+}, Cd^{2+}, Cu^{2+}, Hg^{2+}, Sn^{2+}$ (Precipitate with $H_2S$ in acidic solution).

  • Group 3: $Al^{3+}, Co^{2+}, Cr^{3+}, Fe^{2+}, Mn^{2+}, Ni^{2+}, Zn^{2+}$ (Precipitate with $H_2S$ in basic solution).

  • Group 4: $Ba^{2+}, Ca^{2+}, Sr^{2+}$ (Precipitate with $Na_2CO_3$).

  • Group 5: $K^+, Na^+, NH_4^+$ (No common precipitating reagent).

• Flame Tests:

  • Lithium ($Li$): Red

  • Sodium ($Na$): Yellow

  • Potassium ($K$): Violet

  • Copper ($Cu$): Green

Chemistry in Action: Real World Applications

Blood pH Maintenance

• Blood pH is maintained via hemoglobin ($Hb$) and carbonate buffer systems across erythrocytes, capillaries, and lungs.

Eggshell Formation

• Formation of calcium carbonate ($CaCO_3$) involves the reaction: $Ca^{2+}(aq) + CO_3^{2-}(aq) \rightarrow CaCO_3(s)$.

• This process is linked to carbonic acid equilibria:

  • $CO_2(g) + H_2O(l) \rightleftharpoons H_2CO_3(aq)$

  • $H_2CO_3(aq) \rightleftharpoons H^+(aq) + HCO_3^-(aq)$

  • $HCO_3^-(aq) \rightleftharpoons H^+(aq) + CO_3^{2-}(aq)$

• Carbonic anhydrase acts as a catalyst.

Questions & Discussion

• Q: Which indicators to use for weak acid/strong base titrations?

• A: Indicators like phenolphthalein are preferred because the equivalence point occurs at a basic pH (pH > 7).

• Q: Why does the common ion effect decrease solubility?

• A: According to Le Châtelier's principle, adding a product ion shifts the dissolution equilibrium back toward the solid reactant side.