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)