IX. Chemical Equilibrium
Key focus of this chapter: equilibrium constant
This chapter reviews equilibrium constants, Le Châtelier’s principle, acid–base equilibria, pH, and solubility products.
A. Reversible and irreversible reactions
An irreversible reaction is treated as proceeding essentially to completion under the stated conditions. Reversibility can depend on the conditions and on whether products are removed.
Reversible reaction |  |
Irreversible reaction |  |
B. Equilibrium state
: a state of balance in a reversible reaction, which is reached when the forward and reverse reactions have the same reaction rate and the reactant and product concentrations are no longer changing.
Fig. 1 Equilibrium state of reaction rate and concentration over time
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- Both forward and reverse reactions continue at equal rates.
- Equilibrium does not require equal reactant and product concentrations; each concentration remains constant under unchanged conditions.
C. The equilibrium constant, K
1. Keq
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- Dependent on temperature but independent of reactant and product concentrations and the presence of a catalyst
- Pure solids and pure liquids are omitted from the equilibrium-constant expression because their activities are constant.
The thermodynamic equilibrium constant is defined using activities. Concentration and partial-pressure expressions are the common dilute-solution and ideal-gas forms used here.
2. Kc and Kp
For conventional ideal-gas concentration and partial-pressure expressions, Kp = Kc(RT)Δn, where Δn is the sum of gaseous product coefficients minus the sum of gaseous reactant coefficients. Use consistent pressure and concentration units; do not count pure solids or liquids.
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Fig. 1 Factors that affect Kc and Kp, reaction rate, and equilibrium
D. Le Chatelier’s principle
: an equilibrium system responds to a disturbance by shifting in a direction that partly counteracts the disturbance.
Pressure shifts here refer to changing volume at constant temperature, and only gaseous stoichiometric coefficients are counted. Adding an inert gas at constant volume does not change reacting-gas partial pressures or shift an ideal-gas equilibrium.
Changes of conditions | Direction of reactions |
Concentration |  |
Pressure (volume) & Mole |  |
Temperature & ΔH |  |
Question: Which of the following shifts the equilibrium in the forward direction in the reaction shown?
A. Adding WO3
B. Adding H2
C. Removing W
D. Removing H2O
E. Both B and D
ANS: E Adding WO3 or removing W does not affect the direction of equilibrium, provided each pure solid phase remains present. Removing an entire solid phase can invalidate that assumption. Adding H2 or removing H2O lowers Q below K, favoring the net forward reaction.
Question: Which change shifts the equilibrium toward the reactants?
A. Increasing pressure by decreasing volume
B. Decreasing pressure
C. Increasing volume
D. Removing Y
E. A and C
ANS: A When the volume is decreased at constant temperature, the equilibrium moves from the side with more moles of gas, the products (2Y + 2Z) to the side with fewer moles of gas, the reactant (3X).
Question: Which of the following shifts the equilibrium in the forward direction in the reaction shown?
A. Adding X
B. Removing Y
C. Decreasing pressure by increasing volume
D. Increasing temperature
E. All are correct
ANS: E When the reaction is endothermic (ΔH > 0), the direction of the equilibrium is shifted forward by an increase in temperature.
E. Acid and base equilibrium
1. Acid and base theory
Acid | Base | |
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2. Acid and base strength
a. Acid strength
Acid and base strength describe equilibrium ionization in a specified solvent and are compared using Ka or Kb at a fixed temperature. Strength differs from concentration: the fraction ionized also depends on concentration. H2SO4 is strong for its first ionization; later ionizations require separate treatment. Solubility and base strength are different properties, especially for sparingly soluble metal hydroxides.
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b. Acids and bases in a solution
- Proton transfer from an acid to a base produces their conjugate base and conjugate acid
Proton transfer generally favors the weaker acid–base pair. At the equivalence point, a salt from a strong acid and weak base is acidic, a salt from a weak acid and strong base is basic, and a strong-acid/strong-base salt is neutral when its ions do not hydrolyze. Excess acid or base must be accounted for separately; weak-acid/weak-base salts depend on their conjugate ion equilibria.
Strong acids | • Essentially complete ionization of the relevant acidic proton in dilute aqueous solution |
Weak acids | • Partial ionization in water; an equilibrium remains between acid and conjugate base |
Strong acids and strong bases |  |
Strong acids and weak bases Weak acids and strong bases Weak acids and weak bases |  |
| with halogens | 
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Common strong oxoacids | 
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| Group 1 hydroxides | 
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Group 2 hydroxides (see solubility note) | 
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3. Acid–base equilibrium constants
a. Ka (acid-dissociation constant) and Kb (base-dissociation constant)
: equilibrium constants for acid and base ionization.
- The equilibrium constants Ka and Kb depend on temperature, but are independent of the concentration of acids and bases
- Stronger acids and bases have higher values of Ka and Kb
b. Kw (ion-product constant for water)
: the equilibrium constant for water autoionization
The relation KaKb = Kw applies to a conjugate acid–base pair at the same temperature. Kw varies with temperature; Kw = 1.0 × 10−14 is the usual dilute-solution value at 25 °C. Formal HCl/Cl− expressions illustrate the algebra: HCl is essentially fully ionized in water and Cl− has negligible basicity.
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Equilibrium constants | Equilibrium Equations |
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Question: Calculate the value of Ka when 0.002M of HOBr is 0.1% dissociated in a solution.
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Question: Ka of HNO2 is 4.5 × 10-4. At 25 °C, calculate Kb for the conjugate base NO2−.
Use Ka(HNO2)Kb(NO2−) = Kw. The calculation gives 2.22… × 10−11, reported as 2.2 × 10−11 to two significant figures.
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F. pH
pH = −log10aH+; for dilute solutions, pH ≈ −log10[H3O+]. Also, pH + pOH = pKw. Neutrality means [H3O+] = [OH−], so neutral pH is pKw/2 and equals 7 at 25 °C.
For a buffer containing a weak acid HA and its conjugate base A−, pH ≈ pKa + log10([A−]/[HA]). Use concentrations after any stoichiometric neutralization.
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Question: Calculate the pH of a solution containing [H+] = 0.000035 M.
Calculator result: pH = −log10(3.5 × 10−5) = 4.46. The figure shows a rough estimate of about 4.5.
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Question: 0.0050 M CH3COOH is present in an aqueous solution. When the Ka of CH3COOH is 1.8×10-5, calculate the pH of the solution.
Assume 25 °C and ideal dilute behavior. The estimate x ≈ √(KaC) gives about 6% ionization, exceeding the usual 5% check. Use the quadratic equation: Ka = x2/(C−x). It gives [H3O+] ≈ 2.91 × 10−4 M and pH ≈ 3.54.
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G. Solubility product constant, Ksp
: the equilibrium constant for dissolution of a sparingly soluble ionic solid in a saturated solution; the pure solid is omitted from the expression
In the diagram below, compare salts with the same dissolution stoichiometry at the same temperature, in the same pure solvent, with ideal dilute behavior and no side reactions. The conductivity arrow additionally assumes fixed ionic mobilities; conductivity cannot be predicted from Ksp alone. The freezing-point trend also assumes the same solvent and ideal colligative behavior.
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Question: What is the expression for Ksp of Ca3(PO4)2 in a solution?
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Question (hypothetical exercise): Calculate Ksp for a saturated solution of BaF2 when the concentration of Ba2+ is 2.0 × 10-5 M.
Assume pure water and no other source of Ba2+ or F−; use [F−] = 2[Ba2+].
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