Chemical Equilibrium, Solutions, Acids & Bases, and Buffers: Study Notes

Chemical Equilibrium

Introduction to Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. The system is dynamic, with ongoing reactions, but the observable properties remain unchanged.

• Equilibrium Constant (K): A ratio that quantifies the relative concentrations of products and reactants at equilibrium.

• Expression: For a general reaction $aA + bB \rightleftharpoons cC + dD$, the equilibrium constant is:

$K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

• Key Points:

  • K is unitless (or has units depending on the reaction stoichiometry).

  • The value of K indicates the extent to which a reaction favors products (K > 1) or reactants (K < 1).

  • K is constant at a given temperature.

• Example: For $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$, $K = \frac{[HI]^2}{[H_2][I_2]}$

Reaction Quotient (Q) and Direction of Shift

The reaction quotient Q is calculated using the same expression as K, but with initial (not equilibrium) concentrations. Comparing Q and K predicts the direction the reaction will shift to reach equilibrium.

• If Q < K: Reaction proceeds forward (toward products).

• If Q > K: Reaction proceeds in reverse (toward reactants).

• If Q = K: System is at equilibrium.

$Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

ICE Tables (Initial, Change, Equilibrium)

ICE tables are used to organize and solve equilibrium problems by tracking the changes in concentrations or pressures as the system moves toward equilibrium.

• Steps:

  1. List initial concentrations/pressures.

  2. Define changes using variables (e.g., x).

  3. Express equilibrium values in terms of x.

  4. Substitute into the K expression and solve for x.

• Example: For $A \rightleftharpoons 2B$, starting with 1.0 atm A and K = 0.5 atm:

  • Let change in A be -x, so B is +2x.

  • At equilibrium: [A] = 1.0 - x, [B] = 2x.

  • Set up: $K = \frac{[B]^2}{[A]} = \frac{(2x)^2}{1.0 - x} = 0.5$

  • Solve for x (may require quadratic equation).

Le Chatelier's Principle

When a system at equilibrium is disturbed (by changes in concentration, pressure, or temperature), it will shift to counteract the disturbance and re-establish equilibrium.

• Concentration: Adding reactant shifts equilibrium toward products; removing reactant shifts toward reactants.

• Pressure: Increasing pressure (by decreasing volume) shifts equilibrium toward the side with fewer moles of gas.

• Temperature: Increasing temperature favors the endothermic direction; decreasing temperature favors the exothermic direction.

Equilibrium Constants in Terms of Pressure (Kp)

For gaseous reactions, equilibrium can be expressed in terms of partial pressures:

$K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}$

• Relationship between $K_p$ and $K_c$ (concentration-based):

$K_p = K_c (RT)^{\Delta n}$

Where $\Delta n$ = (moles of gaseous products) - (moles of gaseous reactants), R = 0.08206 L·atm/(mol·K), T = temperature in Kelvin.

Van't Hoff Equation (Temperature Dependence of K)

The Van't Hoff equation relates the change in the equilibrium constant to temperature:

$\ln \left( \frac{K_2}{K_1} \right) = -\frac{\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)$

• Where $\Delta H^\circ$ is the standard enthalpy change, R is the gas constant, and T is in Kelvin.

Solutions and Solubility Equilibria

Types of Solutions

A solution is a homogeneous mixture of two or more substances. The solute is the minor component, and the solvent is the major component.

• Examples: Sugar in water, air (gas in gas), alloys (solid in solid).

Concentration Units

• Molarity (M): $M = \frac{\text{mol solute}}{\text{L solution}}$

• Molality (m): $m = \frac{\text{mol solute}}{\text{kg solvent}}$

• Mass Fraction: $\text{mass fraction} = \frac{\text{mass solute}}{\text{mass solution}}$

• Mole Fraction: $X = \frac{\text{mol solute}}{\text{total mol}}$

Solubility Product (Ksp)

The solubility product constant ($K_{sp}$) describes the equilibrium between a solid and its ions in solution.

• For $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$:

$K_{sp} = [Pb^{2+}][I^-]^2$

• Example: If $K_{sp} = 1.7 \times 10^{-5}$, and $[Pb^{2+}] = x$, $[I^-] = 2x$, then $K_{sp} = x(2x)^2 = 4x^3$

Common Ion Effect

The solubility of a salt decreases when one of its constituent ions is already present in the solution. This is called the common ion effect.

• Example: Solubility of $PbCl_2$ in 0.100 M NaCl is less than in pure water due to the presence of Cl-.

Acids and Bases

Definitions

• Brønsted-Lowry Acid: Proton (H+) donor.

• Brønsted-Lowry Base: Proton (H+) acceptor.

• Lewis Acid: Electron pair acceptor.

• Lewis Base: Electron pair donor.

Conjugate Acid-Base Pairs

Acids and bases exist in pairs that differ by one proton.

• Example: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$

• NH3 is the base, NH4+ is its conjugate acid.

Acid and Base Strength

• Strong acids/bases: Completely ionize in water (e.g., HCl, NaOH).

• Weak acids/bases: Partially ionize (e.g., CH3COOH, NH3).

• Acid Dissociation Constant (Ka): $K_a = \frac{[H^+][A^-]}{[HA]}$

• Base Dissociation Constant (Kb): $K_b = \frac{[BH^+][OH^-]}{[B]}$

• Relationship: $K_a \times K_b = K_w = 1.0 \times 10^{-14}$ at 25°C

• pKa and pKb: $pK_a = -\log_{10} K_a$, $pK_b = -\log_{10} K_b$

• pH and pOH: $pH = -\log_{10} [H^+]$, $pOH = -\log_{10} [OH^-]$, $pH + pOH = 14$

Polyprotic Acids

Polyprotic acids can donate more than one proton, each with its own dissociation constant.

• Example: Phosphoric acid ($H_3PO_4$) has three pKa values: 2.12, 7.25, 12.58.

Calculating pH

• Strong Acid: $[H^+] =$ initial concentration of acid.

• Weak Acid: Use ICE table and $K_a$ to solve for $[H^+]$.

• Example: For 0.10 M acetic acid ($K_a = 1.74 \times 10^{-5}$):

  • Set up ICE table, solve $K_a = \frac{x^2}{0.10 - x}$ for x (where x = $[H^+]$).

  • Calculate pH: $pH = -\log_{10} x$

Buffers

Buffer Solutions

Buffers are solutions that resist changes in pH upon addition of small amounts of acid or base. They are typically made from a weak acid and its conjugate base, or a weak base and its conjugate acid.

• Henderson-Hasselbalch Equation:

$pH = pK_a + \log_{10} \left( \frac{[A^-]}{[HA]} \right)$

• Example: For a buffer with 0.100 M acetic acid and 0.0850 M sodium acetate ($pK_a = 4.76$):

  • $pH = 4.76 + \log_{10} \left( \frac{0.0850}{0.100} \right) = 4.76 + (-0.07) = 4.69$

• Buffer Capacity: The amount of acid or base a buffer can neutralize before pH changes significantly.

Polyprotic Buffer Systems

Polyprotic acids (like carbonic acid or phosphoric acid) can form buffer systems at each dissociation step.

• Example: Carbonate system: $H_2CO_3 / HCO_3^-$ and $HCO_3^- / CO_3^{2-}$

Tables

Sample Table: Comparison of Acid Strengths

Acid	Formula	pKa	Strength
Hydrochloric acid	HCl	<0	Strong
Acetic acid	CH3COOH	4.76	Weak
Phosphoric acid (1st dissociation)	H3PO4	2.12	Weak (polyprotic)
Carbonic acid	H2CO3	6.4	Weak

Sample Table: Solubility Product Constants (Ksp)

Salt	Ksp	Solubility (mol/L)
PbI2	1.7 × 10-5	1.62 × 10-2
Mg(OH)2	5.6 × 10-12	7.4 × 10-5

Additional info:

• Quadratic equations are often needed to solve equilibrium problems when changes are not negligible.

• For buffer calculations, the ratio of conjugate base to acid determines the pH relative to pKa.

• Polyprotic acids have multiple buffer regions, each corresponding to a different pKa.

• Solubility is affected by the presence of common ions and by pH (for salts containing basic or acidic ions).