Rutherford - Equilibrium and Acid_Bases

• Solution Equilibrium: Saturated solutions in closed systems are in equilibrium.

  • Adding solute to a saturated solution leads to dissolution balanced by recrystallization.

• Phase Equilibrium: In a closed system, a substance at its boiling and melting point is in equilibrium.

  • Example: H2O(l) ↔ H2O(g)

Chemical Equilibrium

• Definition: Chemical equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction.

• Reversible Reactions: Only reversible reactions can reach equilibrium.

• Conditions Preventing Equilibrium: Equilibrium cannot be reached if:

  • A reaction produces a precipitate.

  • Water is produced in the reaction.

  • The reaction occurs in a non-closed system with a gas.

The Concept of Equilibrium

• Dynamic Equilibrium: Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate resulting in a dynamic state where both reactions are continuously occurring.

  • As a system approaches equilibrium, both forward and reverse reactions occur.

  • At equilibrium, both reactions proceed at the same rate.

System at Equilibrium

• Constant Amounts: Once equilibrium is achieved, the amounts of each reactant and product remain constant.

• Equilibrium Condition: The forward reaction rate is equal to the reverse reaction rate (Answer c).

• Correct Description: The concentrations of products and reactants are constant (Answer b).

Depicting Equilibrium

• Double Arrow: In a system at equilibrium, both forward and reverse reactions occur, represented by a double arrow in the equation.

  • Example: N2O4(g) ↔ 2 NO2(g)

The Equilibrium Constant

• Forward Reaction: N2O4(g) → 2 NO2(g)

• Rate Law: Rate = kf[N2O4]

• Reverse Reaction: 2 NO2(g) → N2O4(g)

• Rate Law: Rate = kr[NO2]^2

• Equilibrium Condition: At equilibrium, the forward rate equals the reverse rate (Ratef = Rater).

• kf[N2O4] = kr[NO2]^2

• Rewriting the Equation:

  • kf/kr = [NO2]^2/[N2O4]

• Equilibrium Constant Definition

  • The ratio of rate constants is constant at a specific temperature and is denoted as Keq.

  • Keq = kf/kr = [NO2]^2/[N2O4]

• Generalized Equilibrium Expression

  • For the reaction: aA + bB ↔ cC + dD

  • The equilibrium expression is: Kc = [C]^c[D]^d/[A]^a[B]^b

Equilibrium Expression vs. Rate Law

• Focus: Unlike rate laws that are based on the rate-determining step, the equilibrium expression does not deal with mechanisms.

• Stoichiometry: Equilibrium expression relates to the stoichiometry of the balanced equation.

Concentrations of Solids and Liquids

• Constant Concentrations: The concentrations of solids and liquids are essentially constant.

  • These concentrations are obtained by dividing the density of the substance by its molar mass, both of which are constants at constant temperature.

• Exclusion from Equilibrium Expression: Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression.

• Example Equations

  • Ti(s) + 2Cl2(g) ↔ TiCl4(l)

  • PbCl2(s) ↔ Pb^2+(aq) + 2 Cl^-(aq)

  • Kc = [Pb^2+][Cl^-]^2

• More on Solid and Gasses

  • As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.

  • CaCO3(s) ↔ CO2(g) + CaO(s)

  • Kc = [CO2]

Equilibrium Expression Examples

• N2(g) + 3 H2(g) ↔ 2 NH3(g)

• Cu(s) + 2 Ag^+(aq) ↔ Cu^2+(aq) + 2 Ag(s)

• Kc Expression Examples

  • N2(g) + 3 H2(g) ↔ 2 NH3(g)

  • Kc = [NH3]^2/[N2][H2]^3

  • Cu(s) + 2 Ag^+(aq) ↔ Cu^2+(aq) + 2 Ag(s)

  • Kc = [Cu^2+]/[Ag^+]^2

Equilibrium from Either Direction

• Reaching Equilibrium: Equilibrium can be reached from either direction.

• Proportions: Whether starting with N2 and H2 or NH3, the same proportions of all three substances will be present at equilibrium.

Value of K

• K >> 1: If K is much greater than 1, the reaction is product-favored, and the product predominates at equilibrium.

  • The equilibrium lies to the right.

• K << 1: If K is much less than 1, the reaction is reactant-favored, and the reactant predominates at equilibrium.

  • The equilibrium lies to the left.

Manipulating Equilibrium Constants

• Reverse Reaction: The equilibrium constant of a reaction in the reverse direction is the reciprocal of the equilibrium constant of the forward reaction.

  • Example: N2O4(g) ↔ 2 NO2(g)

  • Kc = [NO2]^2/[N2O4] = 4.72

    • 2 NO2(g) ↔ N2O4(g)

    • Kc = [N2O4]/[NO2]^2 = 1/4.72 = 0.212

Equilibrium Calculations

• Example Problem: A closed 1.0 L system at equilibrium contains 5 mol A, 2 mol B, and 3 mol of A2B3 at 448°C. Calculate Kc at 448°C for the reaction:

  • 2A(g) + 3B(l) ↔ A2B3(g)

  • Kc = [A2B3]/[A]^2 = 3/5^2 = 0.12

• Example Problem: A closed system initially containing 1.000 x 10^-3 M H2 and 2.000 x 10^-3 M I2 at 448°C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10^-3 M. Calculate Kc at 448°C for the reaction:

  • H2(g) + I2(g) ↔ 2 HI(g)

Calculating Equilibrium Concentrations

• Initial Conditions:

  • [H2] = 1.000 x 10^-3 M

  • [I2] = 2.000 x 10^-3 M

  • [HI] = 0

• Change: [HI] increases by 1.87 x 10^-3 M

• Stoichiometry: The stoichiometry tells us that [H2] and [I2] decrease by half as much.

• Equilibrium Concentrations:

  • [H2] = 6.5 x 10^-5 M

  • [I2] = 1.065 x 10^-3 M

  • [HI] = 1.87 x 10^-3 M

• Equilibrium Constant:

  • Kc = [HI]^2/[H2][I2] = (1.87 x 10^-3)^2/(6.5 x 10^-5)(1.065 x 10^-3) = 51

The Reaction Quotient (Q)

• Calculation: To calculate Q, substitute the initial concentrations of reactants and products into the equilibrium expression.

• Description: Q gives the same ratio as the equilibrium expression but for a system not at equilibrium.

• Meaning:

  • If Q = K, the system is at equilibrium.

  • If Q > K, there is too much product, and the equilibrium shifts to the left, forming reactants.

  • If Q < K, there is too much reactant, and the equilibrium shifts to the right, forming products.

Equilibrium with Q

• Q = K: System is at equilibrium.

• Q > K: Reaction forms reactants; equilibrium shifts left.

• Q < K: Reaction forms products; equilibrium shifts right.

Equilibrium Calculations with Q

• Example: For the reaction 2A(g) + 3B(l) ↔ A2B3(g), a closed 1.0 L system at equilibrium contains 5 mol A and 2 mol B and 3 mol of A2B3 at 448°C.

  • The Kc at 448°C is 0.12 (calculated earlier).

• Problem: A student has a closed 1.0 L system not at equilibrium, containing 2 mol of A, 1 mol of B, and 5 mol of A2B3. Calculate Q and describe how the system will adjust.

  • Q = [A2B3]/[A]^2 = 5/2^2 = 1.25

• Conclusion: Q > Kc, so the system favors the reverse reaction (concentration of products decreases and reactants increase until equilibrium is established).

Le Châtelier's Principle

• Definition: If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position to counteract the effect of the disturbance.

• Stresses: Concentration, temperature, and pressure disturb equilibrium.

  • "Add away, remove replace."

  • N2(g) + 3 H2(g) ↔ 2 NH3(g) ΔH = -91.8 kJ

• Concentration: Increasing concentration will speed up the reaction going in the opposite direction (and vice versa).

• Temperature: Increasing temperature will always speed up the endothermic reaction (and vice versa).

• Pressure: Increasing pressure increases the concentration of gas particles, which will speed up the reaction going away from the side with more moles of gas (and vice versa).

• Example Explanation: Increasing the concentration of H2(g) increases the concentration of NH3.

  • Increase in frequency of collision between N2 and H2 will cause the equilibrium to shift right (or favor the forward reaction).

Q vs. K and Equilibrium Shifts

• Shifts: If equilibrium shifts right (forward reaction favored), all product amounts increase, and all reactants decrease until equilibrium is reestablished.

Catalysts Effect

• Rate Increase: Catalysts increase the rate of both the forward and reverse reactions.

• Equilibrium: Equilibrium is achieved faster, but the equilibrium composition remains unaltered.

The Haber Process

• Significance: The transformation of nitrogen and hydrogen into ammonia (NH3) is of tremendous significance in agriculture, where ammonia-based fertilizers are critically important.

• Shifting Equilibrium: If H2 is added to the system, N2 will be consumed, and the two reagents will form more NH3.

• Removal of Ammonia: This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid.

Solubility Products

• Equilibrium in Saturated Solution: Consider the equilibrium that exists in a saturated solution of BaSO4 in water.

  • BaSO4(s) ↔ Ba^2+(aq) + SO4^2−(aq)

• Solubility Product Constant (Ksp):

  • The equilibrium constant expression for this equilibrium is

  • Ksp = [Ba^2+][SO4^2−]

• Distinction: Ksp is not the same as solubility.

• Solubility Expression: Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution or in mol/L (M).

• Determine Ksp:

  • Determine the Ksp value for AgCl at 25°C if it has a molar solubility of s = 1.33 x 10^-5 moles per liter.

  • AgCl(s) → Ag^+(aq) + Cl^−(aq)

    • Ksp = [Ag^+][Cl^-]

    • Ksp = [1.33 x 10^-5][1.33 x 10^-5]

    • Ksp = 1.77 x 10^-10

  • Determine the molar solubility of Fe(OH)2 in pure water. The Ksp for Fe(OH)2 is 4.87 x 10^-17.

    • Fe(OH)2(s) ↔ Fe^2+(aq) + 2OH^−(aq)

    • Ksp = [Fe^2+][OH−]^2

    • 4.87 x 10^-17 = [x][2x]^2

    • x = 2.3 x 10^-6 M

Factors Affecting Solubility

• The Common-Ion Effect: If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left, and the solubility of the salt will decrease.

  • BaSO4(s) ↔ Ba^2+(aq) + SO4^2−(aq)

• Common-Ion Effect Example: What is the molar solubility of AgCl in seawater that has an [NaCl] of 0.10 M?

  • The Ksp for AgCl at 25°C is 1.77 x 10^-10.

  • AgCl(s) → Ag^+(aq) + Cl^−(aq)

  • Ksp = [Ag^+][Cl^-]

  • 1.77 x 10^-10 = [x][0.10 + x]

    • (x will be so much smaller than 0.1 so is negligible)

  • 1.77 x 10^-10 = [x][0.10]

  • x = 1.77 x 10^-9

• Will a Precipitate Form?: The system will proceed based on the comparison of Q and Ksp.

  • If Q = Ksp, the system is at equilibrium, and the solution is saturated.

  • If Q < Ksp, more solid will dissolve until Q = Ksp.

  • If Q > Ksp, the salt will precipitate until Q = Ksp.

Acids and Bases: Arrhenius Definition

• Arrhenius Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions (H+).

  • Examples: HCl(aq), H2SO4(aq)

  • Special Cases: Ends in -COOH (organic acid).

  • Non-Acids: CH4 or CH3OH

• Arrhenius Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions.

  • Examples: NaOH(aq), Ca(OH)2(aq)

  • Special Cases: NH3(aq)

  • Non-Bases: CH3OH (Carbon Chain ending in -OH)

Hydrogen Ion and Hydronium Ion

• Proton in Water: A proton (H+) will not remain just a proton in water.

• Hydrogen Atom: A hydrogen atom consists only of a proton and an electron.

• Hydronium Ion Formation: The proton is attracted to a water molecule to form a hydronium ion (H3O+).

Acids and Bases: Brønsted–Lowry Definition

• Brønsted–Lowry Acid: Proton donor.

• Brønsted–Lowry Base: Proton acceptor.

  • Example: NH3 + H2O → NH4+ + OH−

• Conjugate Acid-Base: After an acid reacts, the remaining ion can accept a proton and is therefore a base.

  • Example: HNO3 + H2O → H3O+ + NO3−

Conjugate Acids and Bases

• Origin: From the Latin word conjugare, meaning "to join together."

• Reactions: Reactions between acids and bases always yield their conjugate bases and acids.

Acids Dissolving in Water

• Proton Abstraction: Water acts as a Brønsted–Lowry base and abstracts a proton (H+) from the acid.

• Products: As a result, the conjugate base of the acid and a hydronium ion are formed.

Amphiprotic Nature

• Definition: A substance that can act as either an acid or a base is amphiprotic.

  • Examples: HCO3−, HSO4−, H2O

• Water's Behavior: Depending on its environment, water can behave either as an acid or a base.

  • NH3 + H2O → NH4+ + OH−

  • H2SO4 + H2O → H3O+ + HSO4−

Identifying Acids, Bases, and Salts

• Compound Identification: Classify compounds as acid, base, salt, or alcohol.

  • NaBr

  • HC2H3O2

  • NaOH

  • CH3OH

  • CH3COOH

  • H2S

  • Ca(OH)2

  • NH3

• Conjugate Acid/Base Pair: Identify a conjugate acid/base pair from the reaction:

  • HCO3− + HCl → H2CO3 + Cl−

• Solved

  • NaBr (s)

  • HC2H3O2 (ac)

  • NaOH (b)

  • CH3OH (al)

  • CH3COOH (ac)

  • H2S (ac)

  • Ca(OH)2 (b)

  • NH3 (b)

• Solved

  • HCO3− is a base, and H2CO3 is its conjugate acid.

  • HCl is an acid, and Cl− is its conjugate base.

Naming Acids and Bases

• Binary Acids: Composed of hydrogen and one other element.

  • Begin with Hydro-, followed by the element name modified to end with –ic, then acid.

  • HCl(aq) - Hydrochloric Acid

    • H2S(aq) - Hydrosulfuric Acid

• Polyatomic Acids: Use the polyatomic ion name.

  • If the ion ends in –ate, change it to –ic.

  • If the ion ends in –ite, change it to -ous.

  • H2SO4 - Sulfuric Acid

    • HNO2 - Nitrous Acid

• Naming Bases: Use the metal name followed by Hydroxide.

  • NaOH - Sodium Hydroxide

  • Fe(OH)2 - Iron (II) Hydroxide

Properties of Acids and Bases

• Acids: Sour taste, conduct electricity (electrolyte). Strong acids are good conductors; weak acids conduct slightly. Acids change the color of acid-base indicators and turn litmus paper RED.

• Bases: Bitter taste, slippery/soapy feeling, conduct electricity (electrolyte). Strong bases are good conductors; weak bases conduct slightly. Bases change the color of acid-base indicators and turn litmus paper BLUE.

Acid and Base Strength

• Strong Acids: Completely dissociated in water. Their conjugate bases are extremely weak.

• Weak Acids: Only dissociate partially in water. Their conjugate bases are weak bases.

• Negligible Acidity: Substances with negligible acidity do not dissociate in water. Their conjugate bases are exceedingly strong.

• Acidic Solution: An acidic solution is one where [H+] ions is greater than the [OH−].

• Alkaline Solution: An alkaline solution is one where [OH−] is greater, this is considered a basic solution.

• **[H+] is inversely proportional to [OH−].

Strong Acids

• Seven Strong Acids: HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.

• Strong Electrolytes: Exist totally as ions in aqueous solution.

• Monoprotic: [H3O+] = [acid] for monoprotic strong acids.

Strong Bases

• Strong Bases: Soluble hydroxides are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+).

  • NH4OH is weak base.

• Complete Dissociation: These substances dissociate completely in aqueous solution.

Strong, Weak, or Non-Electrolyte

• NaBr (strong)

• HNO3 (strong)

• NaOH (strong)

• CH3OH (non)

• CH3COOH (weak)

• H2S (weak)

• Al(OH)3 (weak)

• NH3 (weak)

Acid-Base Reaction Equilibrium

• Equilibrium Favor: In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.

  • Example: HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq)

  • H2O is a much stronger base than Cl−, so the equilibrium lies so far to the right K is not measured (K >> 1).

Autoionization of Water

• Amphoteric Nature: Water is amphoteric.

• Autoionization: In pure water, a few molecules act as bases, and a few act as acids.

  • H2O(l) + H2O(l) ↔ H3O+(aq) + OH−(aq)

Ion-Product Constant

• Equilibrium Expression: The equilibrium expression for the autoionization of water is:

  • Kc = [H3O+][OH−]

• Ion-Product Constant for Water (Kw): At 25°C, Kw = 1.0 x 10−14

pH Definition

• Definition: pH is defined as the negative base-10 logarithm of the hydrogen (hydronium) ion concentration.

  • pH = −log [H3O+] or pH= -log[H+]

pH in Pure Water

• Pure Water: In pure water, Kw = [H3O+][OH−] = 1.0 x 10−14.

• Concentrations: Because in pure water [H3O+] = [OH−],

  • [H3O+] = (1.0 x 10−14)^{1/2} = 1.0 x 10−7

pH Scale

• Therefore, in pure water,

  • pH = −log(1.0 x 10−7) = 7.00

• Acids: An acid has a higher [H3O+] than pure water, so its pH is <7.

• Bases: A base has a lower [H3O+] than pure water, so its pH is >7.

pH Scale

• H+ ions: Measures the [H+] ions in solution.

• Scale: Scale typically ranges from 0 to 14.

• pH Levels: A pH of 0 is strongly acidic, 7 is neutral, 14 is basic.

• Logarithmic Scale: Is a logarithmic scale, a change in one pH unit is a tenfold change in [H+] ions.

Other “p” Scales

• Negative Logarithm: The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions).

  • Examples include:

    • pOH = −log [OH−]

    • pKw = −log Kw

Relationship Between pH and pOH

• Because [H3O+][OH−] = Kw = 1.0 x 10−14, we know that

  • −log [H3O+] + −log [OH−] = −log Kw = 14.00

  • pH + pOH = pKw = 14.00

Measuring pH

• Litmus Paper: Used for less accurate measurements.

  • “Red” paper turns blue above ~pH = 8.

  • “Blue” paper turns red below ~pH = 5.

• Indicator

• pH Meter: Uses a pH meter for more accurate measurements, which measures the voltage in the solution.

Calculating pH for strong acids and bases

• Strong Acids

• What is the [H+], pH, pOH, and [OH−] of a 0.00035 M HNO3 solution?

  • [H+] = 0.00035

  • pH = -log(0.00035) = 3.46

  • pOH = 14 - 3.46 = 10.54

  • [OH−] = 10^{-10.56} = 2.86 x 10^{-11}

• Strong Bases

• What is the [OH−], pOH, pH, and [H+] of a 0.0025 M KOH solution?

  • [OH−] = 0.0025

  • pOH = -log(0.0025) = 2.60

  • pH = 14 - 2.60 = 11.40

  • [H+] = 10^{-11.40} = 4.0 x 10^{-12}

• pH for strong acids and bases

• What is the pOH, [H+], [OH−],[HCl] for an HCl solution with a pH of 1.05?

  • pOH = 14 - 1.05 = 12.95

  • [H+] = 10^{-1.05} = 0.089

  • [OH−] = 10^{-12.95} = 1.12 x 10^{-13}

  • [HCl] = 0.089 M

• pH for strong bases

• What is the pOH, [OH−], [H+],[NaOH] for an NaOH solution with a pH of 12.25?

  • pOH = 14 - 12.25 = 1.75

  • [H+] = 10^{-12.25} = 5.6 x 10^{-13}

  • [OH−] = 10^{-1.75} = 0.018

  • [NaOH] = 0.018 M

Dissociation Constants

• Generalized Acid Dissociation:

  • HA(aq) + H2O(l) ↔ A−(aq) + H3O+(aq)

• Equilibrium Expression: The equilibrium expression would be:

  • Kc = [H3O+][A−]/[HA]

• Acid-Dissociation Constant (Ka)

  • For strong acids/bases, Ka is not calculated because [HA] is so small.

• Calculating pH:

  • Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25°C.

  • HC2H3O2(aq) + H2O(l) ↔ H3O+(aq) + C2H3O2−(aq)

  • Ka for acetic acid at 25°C is 1.8 x 10^{-5}.

• ICE Calculation for pH of acids

  • pH = -log [H3O+]

  • pH = -log(2.3 x 10^{-3})

• Calculating pH

  • pH = 2.64

Polyprotic Acids

• Definition: Have more than one acidic proton.

• pH Dependence: If the difference between the Ka for the first dissociation and subsequent Ka values is 10^3 or more, the pH generally depends only on the first dissociation.

pH of acids

• Calculate the pH of 0.1 mol/L hypochlorous acid(HClO). Ka = 3.5 x 10^{-8}.

  • Ka = [H3O+][ClO−]/[HClO]

  • ICE Calculation for pH of acids

  • HClO(aq) + H2O(l) ↔ H3O+(aq) + C2H3O2−(aq)

  • Calculating pH: x = 0.000059 | pH = -log(0.000059) = 4.23

  • 3.5 x 10^{-8} = [x][x]/[0.1]

Weak Bases

• Reaction with Water: Bases react with water to produce hydroxide ion.

  • NH3(aq) + H2O(l) ↔ NH4+(aq) + OH−(aq)

• Equilibrium Constant Expression: The equilibrium constant expression for this reaction is:

  • Kb = [HB][OH−]/[B−]

• Calculating pOH: Do the same thing for Kb, but will first solve for p