Equilibrium
6.1 – 6.5 Fundamentals of Equilibrium
6.1 Effect of Volume Change on Liquid-Vapour Equilibrium.
(a) Initial effect on vapour pressure:
When volume is suddenly increased, the density of vapour molecules decreases. According to Boyle’s Law ($P \propto 1/V$), the vapour pressure will decrease initially.
(b) Rates of evaporation and condensation:
The rate of evaporation depends on temperature, which is fixed, so it remains constant.
The rate of condensation depends on the number of vapour molecules per unit volume striking the liquid surface. Since vapour density decreases, the rate of condensation decreases initially.
(c) Final Equilibrium:
Since rate of evaporation > rate of condensation, more liquid evaporates to fill the space. Eventually, the density of vapour increases until the rate of condensation equals the rate of evaporation again. The final vapour pressure will be the same as the initial vapour pressure (since VP depends only on Temperature).
6.2 Calculate Kc for SO₂ Oxidation.
Given: $[SO_2] = 0.60M, [O_2] = 0.82M, [SO_3] = 1.90M$.
$K_c = \frac{(1.90)^2}{(0.60)^2 (0.82)}$
$K_c = \frac{3.61}{0.36 \times 0.82} = \frac{3.61}{0.2952} \approx \mathbf{12.23 \text{ L mol}^{-1}}$.
6.3 Calculate Kp for Iodine Dissociation.
Total Pressure $P_{total} = 10^5 \text{ Pa}$.
Volume of I atoms = 40% $\Rightarrow$ Mole fraction $\chi_I = 0.40$.
Mole fraction $\chi_{I_2} = 1 – 0.40 = 0.60$.
Partial Pressures:
$p_I = \chi_I \times P_{total} = 0.40 \times 10^5 \text{ Pa}$.
$p_{I_2} = \chi_{I_2} \times P_{total} = 0.60 \times 10^5 \text{ Pa}$.
$K_p = \frac{0.16 \times 10^{10}}{0.6 \times 10^5} = 0.267 \times 10^5 = \mathbf{2.67 \times 10^4 \text{ Pa}}$.
6.4 Write Kc Expressions.
(i) $K_c = \frac{[NO]^2 [Cl_2]}{[NOCl]^2}$
(ii) Ignore solids $Cu(NO_3)_2$ and $CuO$.
$K_c = [NO_2]^4 [O_2]$
(iii) Ignore liquid water (solvent).
$K_c = \frac{[CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H_5]}$
(iv) Ignore solid $Fe(OH)_3$.
$K_c = \frac{1}{[Fe^{3+}][OH^-]^3}$
(v) Ignore solid $I_2$.
$K_c = \frac{[IF_5]^2}{[F_2]^5}$
6.5 Calculate Kc from Kp.
Formula: $K_p = K_c(RT)^{\Delta n_g} \Rightarrow K_c = \frac{K_p}{(RT)^{\Delta n_g}}$.
$R = 0.0831 \text{ L bar K}^{-1} \text{mol}^{-1}$.
(i) $2NOCl \rightleftharpoons 2NO + Cl_2$:
$\Delta n_g = (2+1) – 2 = 1$. $T = 500 \text{ K}$. $K_p = 1.8 \times 10^{-2}$.
$K_c = \frac{1.8 \times 10^{-2}}{(0.0831 \times 500)^1} = \frac{1.8 \times 10^{-2}}{41.55} \approx \mathbf{4.33 \times 10^{-4}}$.
(ii) $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$:
$\Delta n_g = 1 – 0 = 1$. $T = 1073 \text{ K}$. $K_p = 167$.
$K_c = \frac{167}{(0.0831 \times 1073)} = \frac{167}{89.16} \approx \mathbf{1.87}$.
6.6 – 6.10 Reaction Quotients & Stoichiometry
6.6 Kc for Reverse Reaction.
Forward reaction: $NO + O_3 \rightleftharpoons NO_2 + O_2$.
$K_c(\text{forward}) = 6.3 \times 10^{14}$.
Reverse reaction: $NO_2 + O_2 \rightleftharpoons NO + O_3$.
$K_c(\text{reverse}) = \frac{1}{K_c(\text{forward})}$.
6.7 Why ignore Pure Liquids & Solids?
The molar concentration of a pure solid or liquid is directly proportional to its density ($C = \rho / M$). Since the density of pure substances remains constant at a fixed temperature, their active mass or concentration remains constant during the reaction. Therefore, they are incorporated into the equilibrium constant value rather than appearing as variables in the expression.
6.8 Composition of Equilibrium Mixture.
$K_c = 2.0 \times 10^{-37}$. This value is extremely small.
This implies the reaction hardly proceeds to the right. The amount of $N_2$ and $O_2$ reacted is negligible.
Equilibrium Concentrations:
$[N_2] \approx \frac{0.482}{10} = 0.0482 \text{ M}$.
$[O_2] \approx \frac{0.933}{10} = 0.0933 \text{ M}$.
Let eq. conc of $N_2O$ be $x$.
$K_c = \frac{[N_2O]^2}{[N_2]^2 [O_2]} \Rightarrow 2.0 \times 10^{-37} = \frac{x^2}{(0.0482)^2 (0.0933)}$.
$x^2 = 2.0 \times 10^{-37} \times 0.00232 \times 0.0933 = 4.33 \times 10^{-41}$.
$x = \sqrt{43.3 \times 10^{-42}} \approx \mathbf{6.6 \times 10^{-21} \text{ M}}$.
6.9 Calculate Equilibrium Amount.
Initial: $NO=0.087, Br_2=0.0437, NOBr=0$.
Equilibrium: $NOBr = 0.0518$.
From stoichiometry, to form 2 mol NOBr, 2 mol NO and 1 mol Br2 are consumed.
Change:
NO consumed = $0.0518$ mol.
Br2 consumed = $0.0518 / 2 = 0.0259$ mol.
At Equilibrium:
$n_{NO} = 0.087 – 0.0518 = \mathbf{0.0352 \text{ mol}}$.
$n_{Br_2} = 0.0437 – 0.0259 = \mathbf{0.0178 \text{ mol}}$.
6.10 Kp to Kc conversion.
$K_p = 2.0 \times 10^{10} \text{ bar}^{-1}$. $T = 450 \text{ K}$.
$\Delta n_g = 2 – (2+1) = -1$.
$K_p = K_c (RT)^{\Delta n_g} \Rightarrow K_c = \frac{K_p}{(RT)^{-1}} = K_p (RT)$.
$R = 0.0831 \text{ L bar K}^{-1}$.
$K_c = 2.0 \times 10^{10} \times 37.395 \approx \mathbf{7.48 \times 10^{11} \text{ L mol}^{-1}}$.
6.11 – 6.15 Partial Pressure & Kc Calculations
6.11 Kp from Partial Pressures.
Initial $P_{HI} = 0.2$ atm.
Equilibrium $P_{HI} = 0.04$ atm.
Decrease in Pressure = $0.2 – 0.04 = 0.16$ atm.
From stoichiometry (2:1:1), decrease of 2x produces x of H2 and I2.
$2x = 0.16 \Rightarrow x = 0.08$.
$P_{H_2} = P_{I_2} = 0.08$ atm.
6.12 Reaction Quotient Check.
$[N_2] = 1.57/20 \text{ M}, [H_2] = 1.92/20 \text{ M}, [NH_3] = 8.13/20 \text{ M}$.
$K_c = 1.7 \times 10^2$.
$Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(8.13/20)^2}{(1.57/20)(1.92/20)^3}$.
Simplifying volume terms ($V^{-2}$ over $V^{-4}$ becomes $V^2$ in numerator):
$Q_c = \frac{(8.13)^2 \times (20)^2}{(1.57)(1.92)^3} = \frac{66.1 \times 400}{1.57 \times 7.078} = \frac{26440}{11.11} \approx 2380 \approx 2.38 \times 10^3$.
Result: $Q_c (2380) > K_c (170)$.
The mixture is not at equilibrium. Since $Q_c > K_c$, reaction will proceed in the reverse direction.
6.13 Equation from Expression.
Expression: $K_c = \frac{[NH_3]^4 [O_2]^5}{[NO]^4 [H_2O]^6}$.
Numerator = Products. Denominator = Reactants.
Coefficients correspond to powers.
Balanced Equation:
$\mathbf{4NO(g) + 6H_2O(g) \rightleftharpoons 4NH_3(g) + 5O_2(g)}$
6.14 Kc Calculation (Water Gas Shift).
Initial: 1 mol H2O, 1 mol CO in 10L.
Reacted: 40% of water = 0.4 mol.
Equilibrium Moles:
$H_2O = 1 – 0.4 = 0.6$.
$CO = 1 – 0.4 = 0.6$.
$H_2 = 0.4$. $CO_2 = 0.4$.
Since $\Delta n_g = 0$, volume cancels out.
6.15 Calculate Conc from Reverse Equilibrium.
Given reaction $H_2 + I_2 \rightleftharpoons 2HI$ has $K_c = 54.8$.
We start with HI, so we consider: $2HI \rightleftharpoons H_2 + I_2$.
$K’_c = 1/54.8$.
Equilibrium $[HI] = 0.5 \text{ M}$. Let $[H_2] = [I_2] = x$.
$x^2 = \frac{0.25}{54.8} = 0.00456$
$x = \sqrt{0.00456} \approx \mathbf{0.0675 \text{ M}}$.
$[H_2] = [I_2] = 0.0675 \text{ M}$.
6.16 – 6.22 Advanced Equilibrium Problems
6.16 ICl Decomposition.
$K_c = 0.14$. Initial $[ICl] = 0.78$.
Change: $-2x$ for ICl, $+x$ for I2, Cl2.
Eq: $[ICl] = 0.78 – 2x$, $[I_2] = x$, $[Cl_2] = x$.
$0.14 = \frac{x \cdot x}{(0.78 – 2x)^2}$. Take square root:
$\sqrt{0.14} = \frac{x}{0.78 – 2x} \Rightarrow 0.374 = \frac{x}{0.78 – 2x}$.
$x = 0.374(0.78 – 2x) = 0.292 – 0.748x$.
$1.748x = 0.292 \Rightarrow x = 0.167$.
Concentrations:
$[I_2] = [Cl_2] = \mathbf{0.167 \text{ M}}$.
$[ICl] = 0.78 – 2(0.167) = \mathbf{0.446 \text{ M}}$.
6.17 Ethane Decomposition (Kp).
$K_p = 0.04$. Initial Pressure $P = 4.0$ atm.
Eq Pressures: $C_2H_6 = 4.0 – p$, $C_2H_4 = p$, $H_2 = p$.
$K_p = \frac{p \cdot p}{4.0 – p} = 0.04$.
$p^2 = 0.16 – 0.04p \Rightarrow p^2 + 0.04p – 0.16 = 0$.
Solving quadratic: $p = \frac{-0.04 + \sqrt{0.0016 + 0.64}}{2} \approx \frac{0.76}{2} = 0.38$.
Eq Pressure of $C_2H_6 = 4.0 – 0.38 = \mathbf{3.62 \text{ atm}}$.
6.18 Esterification Calculations.
(i) Qc: $Q_c = \frac{[Ester][H_2O]}{[Acid][Alcohol]}$. (Liquid phase reaction, water is product, not solvent).
(ii) Kc Calculation:
Start: Acid=1.0, EtOH=0.18. Eq Ester=0.171.
Reaction is 1:1:1:1. So water formed = 0.171.
Eq Acid = $1.0 – 0.171 = 0.829$. Eq EtOH = $0.18 – 0.171 = 0.009$.
$K_c = \frac{0.171 \times 0.171}{0.829 \times 0.009} \approx \mathbf{3.92}$.
(iii) Check Equilibrium:
Start: Acid=1.0, EtOH=0.5. Ester=0.214.
Water=0.214. Acid=$1-0.214=0.786$. EtOH=$0.5-0.214=0.286$.
$Q_c = \frac{0.214 \times 0.214}{0.786 \times 0.286} = \frac{0.0458}{0.2248} \approx 0.204$.
$Q_c (0.204) \neq K_c (3.92)$. Not reached equilibrium.
6.19 PCl₅ Equilibrium Concentrations.
$PCl_5 \rightleftharpoons PCl_3 + Cl_2$. $K_c = 8.3 \times 10^{-3}$.
Given $[PCl_5]_{eq} = 0.05 \text{ M}$.
Let $[PCl_3] = [Cl_2] = x$.
$8.3 \times 10^{-3} = \frac{x \cdot x}{0.05}$.
$x^2 = 8.3 \times 10^{-3} \times 5 \times 10^{-2} = 41.5 \times 10^{-5} = 4.15 \times 10^{-4}$.
$x = \sqrt{4.15} \times 10^{-2} \approx 2.04 \times 10^{-2}$.
Conc of $PCl_3, Cl_2 = \mathbf{0.02 \text{ M}}$.
6.20 Iron Oxide Reduction Pressures.
$K_p = 0.265 = \frac{P_{CO_2}}{P_{CO}}$.
Initial: $P_{CO}=1.4, P_{CO_2}=0.80$.
$Q_p = 0.8/1.4 = 0.57 > K_p$. Reverse reaction.
Let pressure change be $p$. $CO$ increases, $CO_2$ decreases.
Eq: $P_{CO} = 1.4 + p$, $P_{CO_2} = 0.8 – p$.
$0.265 = \frac{0.8 – p}{1.4 + p} \Rightarrow 0.371 + 0.265p = 0.8 – p$.
$1.265p = 0.429 \Rightarrow p = 0.339$ atm.
Results:
$P_{CO} = 1.4 + 0.339 = \mathbf{1.739 \text{ atm}}$.
$P_{CO_2} = 0.8 – 0.339 = \mathbf{0.461 \text{ atm}}$.
6.21 Direction of Reaction.
$K_c = 0.061$.
Given: $[N_2]=3.0, [H_2]=2.0, [NH_3]=0.5$.
$Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.5)^2}{(3.0)(2.0)^3} = \frac{0.25}{3 \times 8} = \frac{0.25}{24} \approx 0.0104$.
Since $Q_c (0.0104) < K_c (0.061)$, the reaction will proceed in the forward direction to produce more ammonia.
6.22 BrCl Equilibrium Concentration.
$K_c = 32$. Initial $[BrCl] = 3.3 \times 10^{-3} \text{ M}$.
At eq: $[BrCl] = 3.3 \times 10^{-3} – 2x$, $[Br_2]=x, [Cl_2]=x$.
$32 = \frac{x^2}{(3.3 \times 10^{-3} – 2x)^2}$. Take square root.
$5.656 = \frac{x}{0.0033 – 2x}$.
$x = 5.656(0.0033) – 11.312x \Rightarrow 12.312x = 0.01866$.
$x \approx 1.5 \times 10^{-3}$.
$[BrCl]_{eq} = 3.3 \times 10^{-3} – 2(1.5 \times 10^{-3}) = 3.3 \times 10^{-3} – 3.0 \times 10^{-3} = \mathbf{3.0 \times 10^{-4} \text{ M}}$.
Equilibrium
6.23 – 6.24 Kc & Gibbs Energy
6.23 Calculate Kc for C(s) + CO₂ ⇌ 2CO.
Total mass of gases = 100g (Assume).
Mass CO = 90.55g. Mass $CO_2$ = $100 – 90.55 = 9.45g$.
Moles CO = $90.55/28 = 3.234$. Moles $CO_2$ = $9.45/44 = 0.215$.
Total moles $n = 3.449$.
$P_{total} = 1$ atm.
$p_{CO} = \frac{3.234}{3.449} \times 1 = 0.938$ atm.
$p_{CO_2} = \frac{0.215}{3.449} \times 1 = 0.062$ atm.
$K_c = \frac{K_p}{(RT)^{\Delta n_g}} = \frac{14.19}{(0.0821 \times 1127)^1} = \frac{14.19}{92.53} \approx \mathbf{0.153}$.
6.24 ∆G° and K for NO₂ Formation.
(a) Calculate $\Delta G^\ominus$:
$\Delta G^\ominus = \Delta_f G^\ominus(NO_2) – [\Delta_f G^\ominus(NO) + \frac{1}{2}\Delta_f G^\ominus(O_2)]$
$\Delta G^\ominus = 52.0 – [87.0 + 0] = \mathbf{-35.0 \text{ kJ/mol}}$.
(b) Calculate K:
$\Delta G^\ominus = -2.303 RT \log K$
$-35000 = -2.303 \times 8.314 \times 298 \times \log K$
$\log K = \frac{35000}{5705.8} \approx 6.134$.
$K = \text{antilog}(6.134) \approx \mathbf{1.36 \times 10^6}$.
6.25 – 6.30 Le Chatelier’s Principle
6.25 Effect of Pressure Decrease (Volume Increase).
Decreasing pressure favors the direction with more moles of gas.
(a) $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$:
$\Delta n_g = 2 – 1 = +1$. Forward reaction favored. Products increase.
(b) $CaO(s) + CO_2(g) \rightleftharpoons CaCO_3(s)$:
Reactants have gas, products don’t. Backward reaction favored (more gas). Products decrease.
(c) $3Fe(s) + 4H_2O(g) \rightleftharpoons Fe_3O_4(s) + 4H_2(g)$:
$\Delta n_g = 4 – 4 = 0$. No change in moles of gas. Products remain same.
6.26 Effect of Increasing Pressure.
Pressure increase favors direction with fewer gas moles.
- (i) $1 \to 2$ gas moles. Backward. Affected.
- (ii) $3 \to 3$ gas moles. No Change. Not affected.
- (iii) $1 \to 2$ gas moles. Backward. Affected.
- (iv) $3 \to 1$ gas moles. Forward. Affected.
- (v) $1 \to 0$ gas moles (solid product). Forward. Affected.
- (vi) $9 \to 10$ gas moles. Backward. Affected.
6.27 Equilibrium Pressures (HBr Reaction).
$K_p = 1.6 \times 10^5$.
Reverse reaction (decomposition of HBr): $2HBr \rightleftharpoons H_2 + Br_2$.
$K’_p = 1 / (1.6 \times 10^5) = 6.25 \times 10^{-6}$.
Initial $P_{HBr} = 10.0$ bar. Let decrease be $p$.
Eq: $P_{HBr} \approx 10.0$ (since K is small), $P_{H_2} = p/2, P_{Br_2} = p/2$.
$p^2 = 2500 \times 10^{-6} = 2.5 \times 10^{-3}$
$p = 0.05$ bar.
$P_{H_2} = P_{Br_2} = 0.025$ bar. $P_{HBr} = 10.0$ bar.
6.28 Kp Expression and Factors.
(a) Kp Expression: $K_p = \frac{p_{CO} \times (p_{H_2})^3}{p_{CH_4} \times p_{H_2O}}$
(b) Effects:
(i) Pressure Increase: No change in Kp. Equilibrium shifts Backward (fewer moles).
(ii) Temperature Increase: Kp Increases (Endothermic). Equilibrium shifts Forward.
(iii) Catalyst: No change in Kp or composition. Only rate increases.
6.29 Concentration Effects on Methanol Synthesis.
(a) Add H₂: Forward shift (consume reactant).
(b) Add CH₃OH: Backward shift (consume product).
(c) Remove CO: Backward shift (form reactant).
(d) Remove CH₃OH: Forward shift (form product).
6.30 PCl₅ Decomposition Factors.
(a) Kc Expression: $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}$
(b) Reverse Kc: $K’_c = 1/K_c = 1/(8.3 \times 10^{-3}) = \mathbf{120.48}$.
(c) Effect on Kc:
(i) Add PCl5: No change in Kc.
(ii) Pressure Increase: No change in Kc.
(iii) Temp Increase: Increase (Endothermic).
6.31 – 6.34 Calculations
6.31 Water Gas Shift Calculations.
$K_p = 10.1$. Initial $P_{CO}=4.0, P_{H_2O}=4.0$.
At Eq: $P_{CO}=4-x, P_{H_2O}=4-x, P_{CO_2}=x, P_{H_2}=x$.
$\sqrt{10.1} = 3.178 = \frac{x}{4-x}$
$x = 3.178(4-x) = 12.712 – 3.178x$
$4.178x = 12.712 \Rightarrow x = \mathbf{3.04 \text{ bar}}$.
Partial pressure of $H_2 = 3.04$ bar.
6.32 Predict Extent of Reaction.
Rule: High Kc (>10³) $\to$ Products favored. Low Kc (<10⁻³) $\to$ Reactants favored.
(a) Kc = 5×10⁻³⁹: Very small. Reactants dominate. Negligible products.
(b) Kc = 3.7×10⁸: Very large. Products dominate. Reaction nearly complete.
(c) Kc = 1.8: Intermediate. Appreciable concentrations of both reactants and products.
6.33 Ozone Concentration.
$K_c = 2.0 \times 10^{-50}$. $[O_2] = 1.6 \times 10^{-2}$.
$K_c = \frac{[O_3]^2}{[O_2]^3}$.
$[O_3]^2 = 2.0 \times 10^{-50} \times 4.096 \times 10^{-6}$
$[O_3]^2 = 8.192 \times 10^{-56}$
$[O_3] = \sqrt{8.192} \times 10^{-28} = \mathbf{2.86 \times 10^{-28} \text{ M}}$.
6.34 CH₄ Concentration Calculation.
$K_c = 3.90$. Volume = 1L (Molarity = Moles).
Given: $[CO]=0.30, [H_2]=0.10, [H_2O]=0.02$. Let $[CH_4] = x$.
$3.90 = \frac{0.02x}{0.30 \times 0.001} = \frac{0.02x}{0.0003}$
$3.90 = 66.67x$
$x = \frac{3.90}{66.67} \approx \mathbf{0.0585 \text{ M}}$.
6.35 – 6.40 Acid-Base Concepts
6.35 Conjugate Acid-Base Pairs.
Conjugate Acid: Add $H^+$. Conjugate Base: Remove $H^+$.
- $HNO_2$: Base ($NO_2^-$).
- $CN^-$: Acid ($HCN$).
- $HClO_4$: Base ($ClO_4^-$).
- $F^-$: Acid ($HF$).
- $OH^-$: Acid ($H_2O$), Base ($O^{2-}$).
- $CO_3^{2-}$: Acid ($HCO_3^-$).
- $S^{2-}$: Acid ($HS^-$).
6.36 Identify Lewis Acids.
Lewis Acid: Electron pair acceptor.
- $BF_3$: Incomplete octet (B has 6e). Lewis Acid.
- $H^+$: Empty orbital. Lewis Acid.
- $NH_4^+$: Saturated, positive charge but no empty orbital to accept pair easily? Actually acts as Bronsted acid. However, in this list, BF3 and H+ are clear Lewis acids.
6.37 Conjugate Bases of Acids.
Remove $H^+$.
- $HF \to \mathbf{F^-}$
- $H_2SO_4 \to \mathbf{HSO_4^-}$
- $HCO_3^- \to \mathbf{CO_3^{2-}}$
6.38 Conjugate Acids of Bases.
Add $H^+$.
- $NH_2^- \to \mathbf{NH_3}$
- $NH_3 \to \mathbf{NH_4^+}$
- $HCOO^- \to \mathbf{HCOOH}$
6.39 Amphoteric Species (Conjugates).
| Species | Conj Acid (+H) | Conj Base (-H) |
|---|---|---|
| $H_2O$ | $H_3O^+$ | $OH^-$ |
| $HCO_3^-$ | $H_2CO_3$ | $CO_3^{2-}$ |
| $HSO_4^-$ | $H_2SO_4$ | $SO_4^{2-}$ |
| $NH_3$ | $NH_4^+$ | $NH_2^-$ |
6.40 Classify Lewis Acids/Bases.
- (a) $OH^-$: Has lone pairs. Electron donor. Lewis Base.
- (b) $F^-$: Has lone pairs. Electron donor. Lewis Base.
- (c) $H^+$: Empty 1s orbital. Electron acceptor. Lewis Acid.
- (d) $BCl_3$: Incomplete octet (6e). Electron acceptor. Lewis Acid.
Equilibrium
6.41 – 6.45 pH and Ionization Constants
6.41 pH of Soft Drink.
Given: $[H^+] = 3.8 \times 10^{-3} \text{ M}$.
$= -(\log 3.8 – 3)$
$= 3 – 0.58 = \mathbf{2.42}$.
6.42 Concentration of Hydrogen Ion in Vinegar.
Given: pH $= 3.76$.
$\log[H^+] = -3.76 = \bar{4}.24$
$[H^+] = \text{antilog}(-3.76) = 10^{-3.76}$
$[H^+] = \mathbf{1.74 \times 10^{-4} \text{ M}}$.
6.43 Conjugate Base Ionization Constants.
Formula: $K_a \times K_b = K_w = 1.0 \times 10^{-14}$.
- $F^-$ (from HF):
$K_b = 10^{-14} / (6.8 \times 10^{-4}) = \mathbf{1.47 \times 10^{-11}}$. - $HCOO^-$ (from HCOOH):
$K_b = 10^{-14} / (1.8 \times 10^{-4}) = \mathbf{5.56 \times 10^{-11}}$. - $CN^-$ (from HCN):
$K_b = 10^{-14} / (4.8 \times 10^{-9}) = \mathbf{2.08 \times 10^{-6}}$.
6.44 Phenol Ionization and Common Ion Effect.
$K_a = 1.0 \times 10^{-10}$. $C = 0.05 \text{ M}$.
(i) Pure Phenol:
$\alpha = \sqrt{K_a/C} = \sqrt{10^{-10}/0.05} = \sqrt{20 \times 10^{-10}} \approx 4.47 \times 10^{-5}$.
$[PhO^-] = C\alpha = 0.05 \times 4.47 \times 10^{-5} = \mathbf{2.2 \times 10^{-6} \text{ M}}$.
(ii) With 0.01M Phenolate (Common Ion):
Let degree of ionization be $\alpha’$. $[PhO^-] \approx 0.01 \text{ M}$.
$K_a = \frac{[PhO^-][H^+]}{[PhOH]} \approx \frac{(0.01)(C\alpha’)}{C}$.
$10^{-10} = 0.01 \alpha’$.
$\alpha’ = 10^{-8}$.
6.45 H₂S Ionization (With/Without HCl).
(i) Pure 0.1M $H_2S$:
$[H^+] = [HS^-] = \sqrt{K_{a1} C} = \sqrt{9.1 \times 10^{-8} \times 0.1} = \mathbf{9.5 \times 10^{-5} \text{ M}}$.
(ii) With 0.1M HCl:
$[H^+] \approx 0.1 \text{ M}$.
$K_{a1} = \frac{(0.1)[HS^-]}{0.1} \Rightarrow [HS^-] = K_{a1} = \mathbf{9.1 \times 10^{-8} \text{ M}}$.
(iii) $[S^{2-}]$ (Second dissociation):
$HS^- \rightleftharpoons H^+ + S^{2-}$ ($K_{a2} = 1.2 \times 10^{-13}$)
Since $K_{a2}$ is very small, $[S^{2-}] \approx K_{a2}$ in pure solution (if $[H^+] \approx [HS^-]$).
Actually, $K_{a2} = \frac{[H^+][S^{2-}]}{[HS^-]}$.
Case 1 (Pure): $[H^+] \approx [HS^-] \Rightarrow [S^{2-}] = K_{a2} = \mathbf{1.2 \times 10^{-13} \text{ M}}$.
Case 2 (With HCl): $[H^+]=0.1, [HS^-]=9.1 \times 10^{-8}$.
$[S^{2-}] = \frac{K_{a2}[HS^-]}{[H^+]} = \frac{1.2 \times 10^{-13} \times 9.1 \times 10^{-8}}{0.1} = \mathbf{1.09 \times 10^{-19} \text{ M}}$.
6.46 – 6.50 Weak Acids and Bases
6.46 Acetic Acid Dissociation.
$K_a = 1.74 \times 10^{-5}$. $C = 0.05 \text{ M}$.
Degree of Dissociation ($\alpha$):
$\alpha = \sqrt{K_a/C} = \sqrt{1.74 \times 10^{-5} / 0.05} = \sqrt{34.8 \times 10^{-5}} = \sqrt{3.48 \times 10^{-4}}$
$\alpha = \mathbf{0.0186}$.
Acetate Ion Concentration:
$[CH_3COO^-] = C\alpha = 0.05 \times 0.0186 = \mathbf{9.3 \times 10^{-4} \text{ M}}$.
pH:
$[H^+] = 9.3 \times 10^{-4} \text{ M}$.
$pH = -\log(9.3 \times 10^{-4}) = 4 – 0.97 = \mathbf{3.03}$.
6.47 Organic Acid pKa Calculation.
$C = 0.01 \text{ M}$. pH $= 4.15$.
$[H^+] = \text{antilog}(-4.15) = 7.08 \times 10^{-5} \text{ M}$.
Concentration of Anion $[A^-] = [H^+] = \mathbf{7.08 \times 10^{-5} \text{ M}}$.
$K_a = \frac{50.1 \times 10^{-10}}{10^{-2}} = \mathbf{5.01 \times 10^{-8}}$.
$pK_a = -\log(5.01 \times 10^{-8}) = 8 – 0.70 = \mathbf{7.3}$.
6.48 pH of Strong Acids/Bases.
(a) 0.003 M HCl: $[H^+] = 3 \times 10^{-3}$. pH $= 3 – \log 3 = 3 – 0.48 = \mathbf{2.52}$.
(b) 0.005 M NaOH: $[OH^-] = 5 \times 10^{-3}$. pOH $= 3 – \log 5 = 2.3$. pH $= 14 – 2.3 = \mathbf{11.7}$.
(c) 0.002 M HBr: $[H^+] = 2 \times 10^{-3}$. pH $= 3 – \log 2 = 2.7$.
(d) 0.002 M KOH: $[OH^-] = 2 \times 10^{-3}$. pOH $= 2.7$. pH $= \mathbf{11.3}$.
6.49 Solution pH Calculations.
(a) TlOH: Molar Mass $\approx 221$. 2g/2L = 1g/L.
$M = 1/221 \approx 4.52 \times 10^{-3}$. pOH $\approx 2.34$. pH $\approx \mathbf{11.66}$.
(b) Ca(OH)₂: Molar Mass = 74. 0.3g/0.5L = 0.6g/L.
$M = 0.6/74 = 8.1 \times 10^{-3}$. $[OH^-] = 2 \times M = 1.62 \times 10^{-2}$.
pOH $= 1.79$. pH $= \mathbf{12.21}$.
(c) NaOH: Molar Mass = 40. 0.3g/0.2L = 1.5g/L.
$M = 1.5/40 = 0.0375$. pOH $= 1.43$. pH $= \mathbf{12.57}$.
(d) Diluted HCl: $M_1V_1 = M_2V_2$.
$13.6 \times 1 = M_2 \times 1000$. $M_2 = 0.0136$.
pH $= -\log(0.0136) = \mathbf{1.87}$.
6.50 Bromoacetic Acid pH & pKa.
$C = 0.1 \text{ M}$. $\alpha = 0.132$.
$[H^+] = C\alpha = 0.1 \times 0.132 = 0.0132 \text{ M}$.
pH $= -\log(0.0132) = \mathbf{1.88}$.
$K_a = \frac{C\alpha^2}{1-\alpha} = \frac{0.1 \times (0.132)^2}{1 – 0.132} = \frac{0.00174}{0.868} = 2.0 \times 10^{-3}$.
$pK_a = -\log(2 \times 10^{-3}) = \mathbf{2.70}$.
6.51 – 6.60 Advanced pH Calculations
6.51 Codeine pKb.
pH $= 9.95$. pOH $= 14 – 9.95 = 4.05$.
$[OH^-] = \text{antilog}(-4.05) = 8.91 \times 10^{-5}$.
$C = 0.005$. $\alpha = [OH^-]/C = 0.0178$.
$K_b = C\alpha^2 = 0.005 \times (0.0178)^2 = \mathbf{1.58 \times 10^{-6}}$.
$pK_b = \mathbf{5.8}$.
6.52 Aniline Solution Calculations.
$C = 0.001 \text{ M}$. $K_b = 4.27 \times 10^{-10}$ (Table).
$\alpha = \sqrt{K_b/C} = \sqrt{4.27 \times 10^{-7}} \approx 6.53 \times 10^{-4}$.
$[OH^-] = C\alpha = 6.53 \times 10^{-7}$.
pOH $= 6.18$. pH $= \mathbf{7.82}$.
$K_a(\text{conj}) = K_w/K_b = 10^{-14}/4.27 \times 10^{-10} = \mathbf{2.34 \times 10^{-5}}$.
6.53 Acetic Acid with HCl (Common Ion).
$pK_a = 4.74 \Rightarrow K_a = 1.82 \times 10^{-5}$. $C = 0.05$.
$\alpha_{pure} = \sqrt{K_a/C} = \mathbf{0.019}$.
With HCl: $[H^+]_{total} \approx [HCl]$.
$K_a = \frac{[H^+][Ac^-]}{[HAc]} \approx \frac{[HCl](C\alpha)}{C} = [HCl]\alpha$.
(a) 0.01M HCl: $\alpha = 1.82 \times 10^{-5} / 0.01 = \mathbf{0.0018}$.
(b) 0.1M HCl: $\alpha = 1.82 \times 10^{-5} / 0.1 = \mathbf{0.00018}$.
6.54 Dimethylamine ionization.
$K_b = 5.4 \times 10^{-4}$. $C = 0.02$.
$\alpha = \sqrt{K_b/C} = \sqrt{0.027} = \mathbf{0.164}$.
With 0.1M NaOH: $[OH^-] \approx 0.1$.
$K_b \approx [OH^-]\alpha’ \Rightarrow \alpha’ = K_b/0.1 = 5.4 \times 10^{-3}$.
Percentage $= \mathbf{0.54\%}$.
6.55 Hydrogen Ion Concentration from pH.
(a) pH 6.83: $\text{antilog}(-6.83) = \mathbf{1.48 \times 10^{-7} \text{ M}}$.
(b) pH 1.2: $\mathbf{6.3 \times 10^{-2} \text{ M}}$.
(c) pH 7.38: $\mathbf{4.17 \times 10^{-8} \text{ M}}$.
(d) pH 6.4: $\mathbf{3.98 \times 10^{-7} \text{ M}}$.
6.56 pH Calculations for Foods.
Milk (6.8): $1.58 \times 10^{-7}$.
Black Coffee (5.0): $1.0 \times 10^{-5}$.
Tomato Juice (4.2): $6.31 \times 10^{-5}$.
Lemon Juice (2.2): $6.31 \times 10^{-3}$.
Egg White (7.8): $1.58 \times 10^{-8}$.
6.57 KOH Solution Calculations.
0.561 g KOH in 200 mL. Molar Mass = 56.1.
Moles $= 0.01$. $M = 0.01/0.2 = 0.05 \text{ M}$.
$[K^+] = [OH^-] = \mathbf{0.05 \text{ M}}$.
$[H^+] = 10^{-14}/0.05 = \mathbf{2 \times 10^{-13} \text{ M}}$.
pH $= 13 – 0.3 = \mathbf{12.7}$.
6.58 Sr(OH)₂ Solubility and pH.
Solubility $19.23 \text{ g/L}$. Molar Mass Sr(OH)₂ $\approx 121.6$.
Molarity $= 19.23/121.6 \approx 0.158 \text{ M}$.
$[Sr^{2+}] = \mathbf{0.158 \text{ M}}$.
$[OH^-] = 2 \times 0.158 = \mathbf{0.316 \text{ M}}$.
pOH $= -\log(0.316) = 0.5$. pH $= \mathbf{13.5}$.
6.59 Propanoic Acid Dissociation.
$K_a = 1.32 \times 10^{-5}$. $C = 0.05$.
$\alpha = \sqrt{1.32 \times 10^{-5}/0.05} = \mathbf{0.016}$.
pH $= 3 – \log 1.6 = \mathbf{3.09}$.
With 0.01 M HCl: $\alpha = K_a/0.01 = \mathbf{1.32 \times 10^{-3}}$.
6.60 Cyanic Acid Ka Calculation.
pH $= 2.34 \Rightarrow [H^+] = 4.57 \times 10^{-3} \text{ M}$.
$\alpha = [H^+]/C = 4.57 \times 10^{-3}/0.1 = \mathbf{0.0457}$.
$K_a = C\alpha^2 = 0.1 \times (0.0457)^2 \approx \mathbf{2.09 \times 10^{-4}}$.
6.61 – 6.66 Hydrolysis & Mixtures
6.61 Sodium Nitrite Hydrolysis.
Salt of Weak Acid ($K_a = 4.5 \times 10^{-4}$) and Strong Base. Basic Solution.
$h = \sqrt{K_w / (K_a C)} = \sqrt{10^{-14} / (4.5 \times 10^{-4} \times 0.04)}$
$h = \sqrt{10^{-14} / 1.8 \times 10^{-5}} \approx \mathbf{2.36 \times 10^{-5}}$.
$[OH^-] = Ch = 0.04 \times 2.36 \times 10^{-5} = 9.4 \times 10^{-7}$.
pOH $= 6.03$. pH $= \mathbf{7.97}$.
6.62 Pyridinium Hydrochloride Kb.
Salt of Weak Base (Pyridine) + Strong Acid (HCl). Acidic.
pH $= 3.44$. $[H^+] = 3.63 \times 10^{-4}$.
$[H^+] = \sqrt{K_w C / K_b}$.
$K_b = K_w C / [H^+]^2 = 10^{-14} \times 0.02 / (3.63 \times 10^{-4})^2$
$K_b = 2 \times 10^{-16} / 1.32 \times 10^{-7} \approx \mathbf{1.5 \times 10^{-9}}$.
6.63 Nature of Salt Solutions.
- NaCl: Neutral (SA+SB).
- KBr: Neutral (SA+SB).
- NaCN: Basic (WA+SB).
- NH₄NO₃: Acidic (SA+WB).
- NaNO₂: Basic (WA+SB).
- KF: Basic (WA+SB).
6.64 Buffer Solution pH.
Acid (Chloroacetic) + Salt (Sodium salt). Buffer.
$pK_a = -\log(1.35 \times 10^{-3}) = 2.87$.
pH $= pK_a + \log([\text{Salt}]/[\text{Acid}])$.
Since concentrations are equal (0.1M), $\log(1) = 0$.
pH $= \mathbf{2.87}$.
6.65 pH of Neutral Water at 310 K.
$K_w = 2.7 \times 10^{-14}$.
$[H^+] = \sqrt{K_w} = \sqrt{2.7 \times 10^{-14}} = 1.64 \times 10^{-7}$.
pH $= -\log(1.64 \times 10^{-7}) = 7 – 0.21 = \mathbf{6.79}$.
6.66 pH of Mixtures.
(a) Ca(OH)₂ (2 mEq) + HCl (2.5 mEq). Excess Acid 0.5 mEq in 35 mL.
$[H^+] = 0.5/35 = 0.0143$. pH $= \mathbf{1.85}$.
(b) H₂SO₄ (0.2 mEq) + Ca(OH)₂ (0.2 mEq). Neutral. pH $= \mathbf{7.0}$.
(c) H₂SO₄ (2 mEq) + KOH (1 mEq). Excess Acid 1 mEq in 20 mL.
$[H^+] = 1/20 = 0.05$. pH $= \mathbf{1.3}$.
6.67 – 6.73 Solubility Product (Ksp)
6.67 Solubility from Ksp Table.
$Ag_2CrO_4$ ($4s^3$): $s = (K_{sp}/4)^{1/3}$.
$BaCrO_4$ ($s^2$): $s = \sqrt{K_{sp}}$.
$Fe(OH)_3$ ($27s^4$): $s = (K_{sp}/27)^{1/4}$.
Use table values to compute $s$.
6.68 Ratio of Molarities (Saturated Solutions).
$Ag_2CrO_4$: $s_1 = (1.1 \times 10^{-12} / 4)^{1/3} \approx 6.5 \times 10^{-5}$.
$AgBr$: $s_2 = \sqrt{5.0 \times 10^{-13}} \approx 7.07 \times 10^{-7}$.
Ratio $s_1/s_2 \approx 91.9$.
6.69 Precipitation of Copper Iodate.
Mixing equal volumes halves concentrations.
$[Cu^{2+}] = 0.001 \text{ M}$. $[IO_3^-] = 0.001 \text{ M}$.
$Q_{sp} = [Cu^{2+}][IO_3^-]^2 = (0.001)(0.001)^2 = 10^{-9}$.
$K_{sp} = 7.4 \times 10^{-8}$.
Since $Q_{sp} < K_{sp}$, Precipitation will NOT occur.
6.70 Solubility in Buffer vs Pure Water.
Pure water: $s = \sqrt{K_{sp}} = 1.58 \times 10^{-6.5}$.
Buffer pH 3.19: $[H^+] = 6.46 \times 10^{-4}$. $K_a = 6.46 \times 10^{-5}$.
Anion $Bz^-$ reacts with $H^+$ to form $HBz$.
Ratio of solubilities involves $1 + [H^+]/K_a$.
Factor $= \sqrt{1 + 10} \approx 3.3$.
Silver benzoate is approx 3.3 times more soluble.
6.71 Max Concentration to avoid Precipitation.
$FeS$ ($K_{sp} = 6.3 \times 10^{-18}$).
Mixing equal volumes $\to$ conc becomes $C/2$.
$Q = (C/2)(C/2) = C^2/4 \le K_{sp}$.
$C^2 = 4 \times 6.3 \times 10^{-18} = 25.2 \times 10^{-18}$.
$C = \mathbf{5.02 \times 10^{-9} \text{ M}}$.
6.72 Minimum Water for CaSO₄.
$s = \sqrt{9.1 \times 10^{-6}} = 3.02 \times 10^{-3} \text{ mol/L}$.
Molar Mass $CaSO_4 = 136$ g/mol.
Solubility in g/L $= 3.02 \times 10^{-3} \times 136 = 0.411$ g/L.
To dissolve 1g: $1 / 0.411 = \mathbf{2.43 \text{ Litres}}$.
6.73 Sulfide Precipitation.
$[S^{2-}]$ calculation depends on dilution. Final Vol = 15mL.
$[S^{2-}]$ provided? Question says conc is $10^{-19}$. Assume constant.
Metal ion conc after dilution $= 0.04 \times 5 / 15 = 0.0133 \text{ M}$.
$Q_{sp} = 0.0133 \times 10^{-19} = 1.33 \times 10^{-21}$.
Compare with Ksp values of sulfides (not given here, but generally):
ZnS, CdS have very low Ksp ($< 10^{-21}$) $\to$ Precipitate.
FeS, MnS have higher Ksp $\to$ No precipitate.