Acid-Base Equilibria and Solubility Equilibria

Common Ion Effect

  • The common ion effect refers to the shift in equilibrium caused by the addition of a compound that shares an ion with the dissolved substance.
  • The presence of a common ion suppresses the ionization of a weak acid or a weak base.
  • Example: A mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).
    • CH3COONa (s) \rightleftharpoons Na^+ (aq) + CH3COO^- (aq)
    • CH3COOH (aq) \rightleftharpoons H^+ (aq) + CH3COO^- (aq)
    • CH_3COO^- is the common ion.

Henderson-Hasselbalch Equation

  • Consider a mixture of a salt NaA and a weak acid HA.
    • HA (aq) \rightleftharpoons H^+ (aq) + A^- (aq)
    • NaA (s) \rightleftharpoons Na^+ (aq) + A^- (aq)
  • K_a = \frac{[H^+][A^-]}{[HA]}
  • [H^+] = K_a \frac{[HA]}{[A^-]}
  • -log[H^+] = -log K_a - log \frac{[HA]}{[A^-]}
  • -log[H^+] = -log K_a + log \frac{[A^-]}{[HA]}
  • pH = pK_a + log \frac{[A^-]}{[HA]}
  • Where pKa = -log K_a
  • The Henderson-Hasselbalch equation is: pH = pK_a + log \frac{[conjugate \ base]}{[acid]}

Example: pH Calculation

  • What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?
    • HCOOH (aq) \rightleftharpoons H^+ (aq) + HCOO^- (aq)
    • Initial: [HCOOH] = 0.30 M, [HCOO-] = 0.52 M, [H+] = 0
    • Change: -x, +x, +x
    • Equilibrium: 0.30 - x, 0.52 + x, x
  • Common ion effect: 0.30 – x ≈ 0.30 and 0.52 + x ≈ 0.52
  • pH = pK_a + log \frac{[HCOO^-]}{[HCOOH]}
  • Given pK_a of HCOOH = 3.77
  • pH = 3.77 + log \frac{[0.52]}{[0.30]} = 4.01
  • This is a mixture of a weak acid and its conjugate base.

Buffer Solutions

  • A buffer solution consists of:
    1. A weak acid or a weak base, and
    2. The salt of the weak acid or weak base.
  • Both components must be present.
  • A buffer solution resists changes in pH upon the addition of small amounts of either acid or base.
  • Adding strong acid: H^+ (aq) + CH3COO^- (aq) \rightleftharpoons CH3COOH (aq)
  • Adding strong base: OH^- (aq) + CH3COOH (aq) \rightleftharpoons CH3COO^- (aq) + H_2O (l)
  • Consider an equal molar mixture of CH3COOH and CH3COONa

Identifying Buffer Systems

  • (a) KF/HF: KF is a salt of a weak acid (HF), so it's a buffer solution.
  • (b) KBr/HBr: HBr is a strong acid, so it's not a buffer solution.
  • (c) Na2CO3/NaHCO3: CO3^{2-} is a weak base, and HCO_3^- is its conjugate acid, so it's a buffer solution.

Example: Buffer Calculation with NaOH Addition

  • Calculate the pH of the 0.30 M NH3 / 0.36 M NH4Cl buffer system.
  • What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?
  • NH4^+ (aq) \rightleftharpoons H^+ (aq) + NH3 (aq)
  • pH = pKa + log \frac{[NH3]}{[NH_4^+]}
  • Given pK_a = 9.25
  • pH = 9.25 + log \frac{[0.30]}{[0.36]} = 9.17
  • NH4^+ (aq) + OH^- (aq) \rightleftharpoons H2O (l) + NH_3 (aq)
  • Initial moles: 0.029, 0.001, 0.024
  • Final moles: 0.028, 0, 0.025
  • Final volume = 80.0 mL + 20.0 mL = 100 mL
  • [NH_4^+] = \frac{0.028}{0.10}
  • [NH_3] = \frac{0.025}{0.10}
  • pH = 9.25 + log \frac{[0.25]}{[0.28]}

Titrations

  • Titration is a process where a solution of known concentration is gradually added to another solution of unknown concentration until the reaction between the two is complete.
  • Equivalence point: The point at which the 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) → H_2O (l) + NaCl (aq)
  • Net ionic equation: OH^- (aq) + H^+ (aq) \rightleftharpoons H_2O (l)

Weak Acid-Strong Base Titrations

  • Example: CH3COOH (aq) + NaOH (aq) \rightleftharpoons CH3COONa (aq) + H_2O (l)
  • Net ionic equation: CH3COOH (aq) + OH^- (aq) \rightleftharpoons CH3COO^- (aq) + H_2O (l)
  • Hydrolysis: CH3COO^- (aq) + H2O (l) \rightleftharpoons OH^- (aq) + CH_3COOH (aq)
  • At the equivalence point, pH > 7.

Example: Titration Calculation

  • Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid by sodium hydroxide after the addition of:
    • (a) 10.0 mL of 0.100 M NaOH
    • (b) 25.0 mL of 0.100 M NaOH
    • (c) 35.0 mL of 0.100 M NaOH
  • The reaction: CH3COOH(aq) + NaOH(aq) \rightleftharpoons CH3COONa(aq) + H_2O(l)

Solution Strategy

  • 1 mol CH_3COOH reacts with 1 mol NaOH.
  • Calculate the number of moles of base reacting with the acid at each stage.
  • The pH of the solution is determined by the excess acid or base left over.
  • At the equivalence point, the neutralization is complete, and the pH depends on the hydrolysis of the salt formed (CH_3COONa).

Part A: 10.0 mL of 0.100 M NaOH

  • Moles of NaOH = 0.100 \frac{mol}{L} * 10.0 \ mL * \frac{1 \ L}{1000 \ mL} = 1.00 * 10^{-3} \ mol
  • Moles of CH_3COOH = 0.100 \frac{mol}{L} * 25.0 \ mL * \frac{1 \ L}{1000 \ mL} = 2.50 * 10^{-3} \ mol
  • Changes in moles:
    • CH3COOH (aq) + NaOH (aq) \rightarrow CH3COONa(aq) + H_2O(l)
    • Initial: 2.50 * 10^{-3}, 1.00 * 10^{-3}, 0
    • Change: -1.00 * 10^{-3}, -1.00 * 10^{-3}, +1.00 * 10^{-3}
    • Final: 1.50 * 10^{-3}, 0, 1.00 * 10^{-3}
  • We have a buffer system of CH3COOH and CH3COO^-
  • Ka = \frac{[H^+][CH3COO^-]}{[CH_3COOH]}
  • [H^+] = Ka * \frac{[CH3COOH]}{[CH_3COO^-]} = 1.8 * 10^{-5} * \frac{1.50 * 10^{-3}}{1.00 * 10^{-3}} = 2.7 * 10^{-5} M
  • pH = -log(2.7 * 10^{-5}) = 4.57

Part B: Equivalence Point (25.0 mL of 0.100 M NaOH)

  • Moles of NaOH = 0.100 \frac{mol}{L} * 25.0 \ mL * \frac{1 \ L}{1000 \ mL} = 2.50 * 10^{-3} \ mol
  • Changes in moles:
    • CH3COOH (aq) + NaOH (aq) \rightarrow CH3COONa(aq) + H_2O(l)
    • Initial: 2.50 * 10^{-3}, 2.50 * 10^{-3}, 0
    • Change: -2.50 * 10^{-3}, -2.50 * 10^{-3}, +2.50 * 10^{-3}
    • Final: 0, 0, 2.50 * 10^{-3}
  • Concentration of CH_3COONa = \frac{2.50 * 10^{-3} \ mol}{50.0 \ mL} * \frac{1000 \ mL}{1 \ L} = 0.0500 M
  • Hydrolysis of CH_3COO^-

Part C: After Equivalence Point (35.0 mL of 0.100 M NaOH)

  • Moles of NaOH = 0.100 \frac{mol}{L} * 35.0 \ mL * \frac{1 \ L}{1000 \ mL} = 3.50 * 10^{-3} \ mol
  • Changes in moles:
    • CH3COOH (aq) + NaOH (aq) \rightarrow CH3COONa(aq) + H_2O(l)
    • Initial: 2.50 * 10^{-3}, 3.50 * 10^{-3}, 0
    • Change: -2.50 * 10^{-3}, -2.50 * 10^{-3}, +2.50 * 10^{-3}
    • Final: 0, 1.00 * 10^{-3}, 2.50 * 10^{-3}
  • Excess OH^- governs the pH.
  • [OH^-] = \frac{1.00 * 10^{-3} \ mol}{60.0 \ mL} * \frac{1000 \ mL}{1 \ L} = 0.0167 M
  • pOH = -log[0.0167] = 1.78
  • pH = 14.00 - 1.78 = 12.22

Strong Acid-Weak Base Titrations

  • Example: HCl (aq) + NH3 (aq) → NH4Cl (aq)
  • NH4^+ (aq) + H2O (l) \rightleftharpoons NH_3 (aq) + H^+ (aq)
  • At the equivalence point, pH < 7.

Example: pH Calculation at Equivalence Point

  • Calculate the pH at the equivalence point when 25.0 mL of 0.100 M NH_3 is titrated by a 0.100 M HCl solution.
  • The reaction: NH3(aq) + HCl(aq) \rightarrow NH4Cl(aq)
  • 1 mol NH_3 reacts with 1 mol HCl.
  • At the equivalence point, the major species are NH4Cl (dissociated into NH4^+ and Cl^- ions) and H_2O.
  • First, determine the concentration of NH_4Cl formed.
  • Then, calculate the pH due to the NH_4^+ ion hydrolysis.
  • The Cl^- ion does not react with water.

Solution

  • Moles of NH_3 = 0.100 \frac{mol}{L} * 25.0 \ mL * \frac{1 \ L}{1000 \ mL} = 2.50 * 10^{-3} \ mol
  • At the equivalence point, moles of HCl added = moles of NH_3.
  • Changes in moles:
    • NH3(aq) + HCl(aq) \rightarrow NH4Cl(aq)
    • Initial: 2.50 * 10^{-3}, 2.50 * 10^{-3}, 0
    • Change: -2.50 * 10^{-3}, -2.50 * 10^{-3}, +2.50 * 10^{-3}
    • Final: 0, 0, 2.50 * 10^{-3}
  • [NH_4Cl] = \frac{2.50 * 10^{-3} \ mol}{50.0 \ mL} * \frac{1000 \ mL}{1 \ L} = 0.0500 M
  • Hydrolysis of NH_4^+ ions:
    • NH4^+ + H2O \rightleftharpoons NH_3 + H^+
    • Initial (M): 0.0500, 0, 0
    • Change (M): -x, +x, +x
    • Equilibrium (M): (0.0500-x), x, x
    • Ka = \frac{[NH3][H^+]}{[NH_4^+]} \approx 5.6 * 10^{-10} = \frac{x^2}{0.0500}
    • x = \sqrt{5.6 * 10^{-10} * 0.0500} = 5.3 * 10^{-6} M
    • pH = -log(5.3 * 10^{-6}) = 5.28

Example: pH at Equivalence Point

  • Exactly 100 mL of 0.10 M HNO_2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point?
  • HNO2 (aq) + OH^- (aq) \rightleftharpoons NO2^- (aq) + H_2O (l)
  • Initial moles: 0.01, 0.01, 0.0
  • Final moles: 0.0, 0.0, 0.01
  • [NO_2^-] = \frac{0.01}{0.200} = 0.05 M
  • NO2^- (aq) + H2O (l) \rightleftharpoons OH^- (aq) + HNO_2 (aq)
  • Initial (M): 0.05, 0, 0
  • Change (M): -x, +x, +x
  • Equilibrium (M): 0.05 - x, x, x
  • Kb = \frac{[OH^-][HNO2]}{[NO_2^-]} = \frac{x^2}{0.05-x} = 2.2 * 10^{-11}
  • x ≈ 1.05 * 10^{-6} = [OH^-]
  • pOH = 5.98
  • pH = 14 – pOH = 8.02

Acid-Base Indicators

  • Indicators are weak acids that have different colors in their acid (HIn) and conjugate base (In-) forms.
  • HIn (aq) \rightleftharpoons H^+ (aq) + In^- (aq)
  • When [HIn] / [In^-] >= 10, the color of the acid (HIn) predominates.
  • When [HIn] / [In^-] <= 0.1, the color of the conjugate base (In-) predominates.

Solubility Equilibria

  • AgCl (s) \rightleftharpoons Ag^+ (aq) + Cl^- (aq)
  • K_{sp} = [Ag^+][Cl^-] (solubility product constant)
  • MgF_2 (s) \rightleftharpoons Mg^{2+} (aq) + 2F^- (aq)
  • K_{sp} = [Mg^{2+}][F^-]^2
  • Ag2CO3 (s) \rightleftharpoons 2Ag^+ (aq) + CO_3^{2-} (aq)
  • K{sp} = [Ag^+]^2[CO3^{2-}]
  • Ca3(PO4)2 (s) \rightleftharpoons 3Ca^{2+} (aq) + 2PO4^{3-} (aq)
  • K{sp} = [Ca^{2+}]^3[PO4^{3-}]^2

Dissolution of an ionic solid in aqueous solution:

  • Q = K_{sp} Saturated solution
  • Q < K_{sp} Unsaturated solution No precipitate
  • Q > K_{sp} Supersaturated solution Precipitate will form

Molar Solubility

  • Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution.
  • Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution.

Example: Solubility of Silver Chloride

  • What is the solubility of silver chloride in g/L?
  • AgCl (s) \rightleftharpoons Ag^+ (aq) + Cl^- (aq)
  • K_{sp} = [Ag^+][Cl^-]
  • Initial (M): 0, 0
  • Change (M): +s, +s
  • Equilibrium (M): s, s
  • K_{sp} = s^2
  • s = \sqrt{K_{sp}} = \sqrt{1.6 * 10^{-10}} = 1.3 * 10^{-5} M
  • [Ag^+] = 1.3 * 10^{-5} M
  • [Cl^-] = 1.3 * 10^{-5} M
  • Solubility of AgCl = 1.3 * 10^{-5} \frac{mol \ AgCl}{1 \ L \ soln} * \frac{143.35 \ g \ AgCl}{1 \ mol \ AgCl} = 1.9 * 10^{-3} \frac{g}{L}

Relationship Between Ksp and Molar Solubility (s)

  • AgCl: K{sp} = [Ag^+][Cl^-] = s^2; s = \sqrt{K{sp}}
  • BaSO4: K{sp} = [Ba^{2+}][SO4^{2-}] = s^2; s = \sqrt{K{sp}}
  • Ag2CO3: K{sp} = [Ag^+]^2[CO3^{2-}] = (2s)^2(s) = 4s^3; s = \sqrt[3]{\frac{K_{sp}}{4}}
  • PbF2: K{sp} = [Pb^{2+}][F^-]^2 = (s)(2s)^2 = 4s^3; s = \sqrt[3]{\frac{K_{sp}}{4}}
  • Al(OH)3: K{sp} = [Al^{3+}][OH^-]^3 = (s)(3s)^3 = 27s^4; s = \sqrt[4]{\frac{K_{sp}}{27}}
  • Ca3(PO4)2: K{sp} = [Ca^{2+}]^3[PO4^{3-}]^2 = (3s)^3(2s)^2 = 108s^5; s = \sqrt[5]{\frac{K{sp}}{108}}

Example: Precipitation Prediction

  • If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl_2, will a precipitate form?
  • Ions present: Na^+, OH^-, Ca^{2+}, Cl^-
  • Possible precipitate: Ca(OH)_2
  • Is Q > K{sp} for Ca(OH)2?
  • [Ca^{2+}]_0 = 0.100 M
  • [OH^-]_0 = 0.200 \frac{mol}{L} * 0.002 \ L = 4.0 * 10^{-4} M
  • K_{sp} = [Ca^{2+}][OH^-]^2 = 8.0 * 10^{-6}
  • Q = [Ca^{2+}]0[OH^-]0^2 = 0.10 * (4.0 * 10^{-4})^2 = 1.6 * 10^{-8}
  • Q < K_{sp}, so no precipitate will form.

Selective Precipitation

  • What concentration of Ag^+ is required to precipitate ONLY AgBr in a solution that contains both Br^- and Cl^- at a concentration of 0.02 M?
  • AgCl (s) \rightleftharpoons Ag^+ (aq) + Cl^- (aq) K_{sp} = [Ag^+][Cl^-] = 1.6 * 10^{-10}
  • AgBr (s) \rightleftharpoons Ag^+ (aq) + Br^- (aq) K_{sp} = [Ag^+][Br^-] = 7.7 * 10^{-13}
  • [Ag^+] = \frac{K_{sp}}{[Br^-]} = \frac{7.7 * 10^{-13}}{0.020} = 3.9 * 10^{-11} M
  • [Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.6 * 10^{-10}}{0.020} = 8.0 * 10^{-9} M
  • 3.9 * 10^{-11} M < [Ag^+] < 8.0 * 10^{-9} M

Common Ion Effect and Solubility

  • The presence of a common ion decreases the solubility of the salt.
  • What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr?
  • AgBr (s) \rightleftharpoons Ag^+ (aq) + Br^- (aq) K_{sp} = 7.7 * 10^{-13}
  • In pure water: s^2 = K_{sp} => s = 8.8 * 10^{-7}
  • In 0.0010 M NaBr: NaBr (s) \rightleftharpoons Na^+ (aq) + Br^- (aq) -> [Br^-] = 0.0010 M
  • AgBr (s) \rightleftharpoons Ag^+ (aq) + Br^- (aq)
  • [Ag^+] = s, [Br^-] = 0.0010 + s \approx 0.0010
  • K_{sp} = 0.0010 * s => s = 7.7 * 10^{-10}

pH and Solubility

  • The presence of a common ion decreases the solubility.
  • Insoluble bases dissolve in acidic solutions.
  • Insoluble acids dissolve in basic solutions.
  • Mg(OH)2 (s) \rightleftharpoons Mg^{2+} (aq) + 2OH^- (aq) K{sp} = [Mg^{2+}][OH^-]^2 = 1.2 * 10^{-11}
  • K_{sp} = (s)(2s)^2 = 4s^3
  • 4s^3 = 1.2 * 10^{-11}
  • s = 1.4 * 10^{-4} M
  • [OH^-] = 2s = 2.8 * 10^{-4} M
  • pOH = 3.55
  • pH = 10.45

Complex Ion Equilibria and Solubility

  • A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions.
  • Co^{2+} (aq) + 4Cl^- (aq) \rightleftharpoons CoCl_4^{2-} (aq)
  • Kf = \frac{[CoCl4^{2-}]}{[Co^{2+}][Cl^-]^4} (formation constant or stability constant)
  • Co(H2O)6^{2+} + 4Cl^- \rightleftharpoons CoCl_4^{2-}

Example: Complex Ion Formation

  • A 0.20-mole quantity of CuSO4 is added to a liter of 1.20 M NH3 solution. What is the concentration of Cu^{2+} ions at equilibrium?

Solution

  • Cu^{2+}(aq) + 4NH3(aq) \rightleftharpoons Cu(NH3)_4^{2+}(aq)
  • K_f = 5.0 * 10^{13}
  • Since (Kf) is very large, the reaction lies mostly to the right. At equilibrium, the concentration of Cu^{2+} will be very small.
  • The amount of NH_3 consumed in forming the complex ion is 4 x 0.20 mol, or 0.80 mol.
  • [NH_3] at equilibrium = (1.20 - 0.80) mol/L soln = 0.40 M
  • [Cu(NH3)4^{2+}] = 0.20 M
  • Kf = \frac{[Cu(NH3)4^{2+}]}{[Cu^{2+}][NH3]^4} = \frac{0.20}{x (0.40)^4} = 5.0 * 10^{13}
  • Solving for x:
    • [Cu^{2+}] = 1.6 * 10^{-13} M

Example: Molar Solubility with Complex Ion Formation

  • Calculate the molar solubility of AgCl in a 1.0 M NH_3 solution.

Strategy

  • AgCl is only slightly soluble in water: AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)
  • The Ag^+ ions form a complex ion with NH3: Ag^+(aq) + 2NH3(aq) \rightleftharpoons Ag(NH3)2^+(aq)

Solution

  • The equilibrium reactions are

    • AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq) \qquad K_{sp} = [Ag^+][Cl^-] = 1.6 * 10^{-10}

    • Ag^+(aq) + 2NH3(aq) \rightleftharpoons Ag(NH3)2^+(aq) \qquad Kf = \frac{[Ag(NH3)2^+]}{[Ag^+][NH_3]^2} = 1.5 * 10^{7}

  • Overall: AgCl(s) + 2NH3(aq) \rightleftharpoons Ag(NH3)_2^+(aq) + Cl^-(aq)

    • K = K{sp} * Kf = (1.6 * 10^{-10}) * (1.5 * 10^7) = 2.4 * 10^{-3}

  • Let s be the molar solubility of AgCl (mol/L):

    • AgCl(s) + 2NH3(aq) \rightleftharpoons Ag(NH3)_2^+(aq) + Cl^-(aq)

    • Initial: 1.0, 0.0, 0.0

    • Change: -2s, +s, +s

    • Equilibrium: (1.0 – 2s), s, s

    • K = \frac{[Ag(NH3)2^+][Cl^-]}{[NH_3]^2} = \frac{s^2}{(1.0 - 2s)^2} = 2.4 * 10^{-3}

    • \frac{s}{(1.0 - 2s)} = \sqrt{2.4 * 10^{-3}} = 0.049

  • Solve for s:

    • s = 0.045 M

Qualitative Analysis of Cations

  • Group 1: Ag+, Hg_2^{2+}, Pb2+ (precipitated by HCl)

  • Group 2: Bismuth, Cadmium, Copper, Mercury, Tin (precipitated by H2S in acidic solutions)

  • Group 3: Aluminum, Cobalt, Chromium, Iron, Manganese, Nickel, Zinc (precipitated by H2S in basic solutions)

  • Group 4: Barium, Calcium, Strontium (precipitated by Na2CO3)

  • Group 5: Sodium, Potassium, Ammonium (no precipitating reagent)