Equilibrium

Q1 (10)

(a) Initial $\text{PCl}_5 = 2.0$ mol; dissociated $= 0.40\times2.0 = 0.80$ mol. (1) At equilibrium: $\text{PCl}_5 = 1.2$ mol, $\text{PCl}_3 = 0.80$ mol, $\text{Cl}_2 = 0.80$ mol. Concentrations ($V=4.0$ L): $[\text{PCl}_5]=0.30$, $[\text{PCl}_3]=0.20$, $[\text{Cl}_2]=0.20$ M. (1) $K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{0.20\times0.20}{0.30} = 0.133\ \text{M}$ (2)

(b) $\Delta n = (2-1) = 1$. $K_p = K_c(RT)^{\Delta n} = 0.133\times(0.0831\times500)^1$ (2) $= 0.133\times41.55 = 5.54\ \text{bar}$. (1)

(c) Doubling volume halves all concentrations: $[\text{PCl}_5]=0.15$, $[\text{PCl}_3]=[\text{Cl}_2]=0.10$. $Q_c = \dfrac{0.10\times0.10}{0.15} = 0.0667$. (1) $Q_c (0.0667) < K_c (0.133)$, so the reaction proceeds forward (more dissociation). (1) Consistent with Le Chatelier: decreased pressure favours the side with more moles of gas. (1)

Q2 (10)

(a) $[\text{H}^+] = 10^{-2.87} = 1.35\times10^{-3}\ \text{M}$. (1) $\alpha = \dfrac{[\text{H}^+]}{C} = \dfrac{1.35\times10^{-3}}{0.10} = 0.0135$ (1.35%). (1) $K_a = \frac{C\alpha^2}{1-\alpha}\approx C\alpha^2 = 0.10\times(0.0135)^2 = 1.82\times10^{-5}$ (2)

(b) $\text{p}K_a = -\log(1.82\times10^{-5}) = 4.74$. (2)

(c) New $C = 0.0010$ M. Ostwald: $\alpha=\sqrt{K_a/C}=\sqrt{1.82\times10^{-5}/0.0010}=\sqrt{0.0182}=0.135$ (13.5%). (2) $\alpha$ increases on dilution. New $[\text{H}^+]=C\alpha=0.0010\times0.135=1.35\times10^{-4}$, pH $= 3.87$; pH increases (solution less acidic) despite greater dissociation, because total $[\text{H}^+]$ falls. (2)

Q3 (10)

(a) Moles: $\text{CH}_3\text{COOH}=0.050\times0.20=0.010$; $\text{NaOH}=0.050\times0.10=0.0050$. (1) NaOH is limiting → forms $0.0050$ mol acetate, leaving $0.0050$ mol acid → acid/salt buffer. (1) Equal moles of acid and conjugate base: $\text{pH}=\text{p}K_a+\log\frac{[\text{A}^-]}{[\text{HA}]}=4.74+\log\frac{0.0050}{0.0050}=4.74$ (3)

(b) HCl added $=0.010\times0.10=0.0010$ mol. It converts acetate→acid: acetate $=0.0050-0.0010=0.0040$; acid $=0.0050+0.0010=0.0060$. (1) $\text{pH}=4.74+\log\frac{0.0040}{0.0060}=4.74+\log(0.667)=4.74-0.176=4.56$ (2)

(c) The buffer (weak acid + conjugate base) neutralises added $\text{H}^+$ by shifting the acetate/acid ratio only slightly, so pH changes little. (2)

Q4 (10)

(a) For AgCl: $[\text{Ag}^+]=\dfrac{K_{sp}}{[\text{Cl}^-]}=\dfrac{1.8\times10^{-10}}{0.010}=1.8\times10^{-8}\ \text{M}$. (2) For $\text{Ag}_2\text{CrO}_4$: $[\text{Ag}^+]=\sqrt{\dfrac{K_{sp}}{[\text{CrO}_4^{2-}]}}=\sqrt{\dfrac{1.1\times10^{-12}}{0.010}}=\sqrt{1.1\times10^{-10}}=1.05\times10^{-5}\ \text{M}$. (2)

(b) AgCl requires the lower $[\text{Ag}^+]$ ($1.8\times10^{-8}$), so AgCl precipitates first. (1)

(c) $\text{Ag}_2\text{CrO}_4$ begins at $[\text{Ag}^+]=1.05\times10^{-5}$. At that point: $[\text{Cl}^-]=\frac{K_{sp}(\text{AgCl})}{[\text{Ag}^+]}=\frac{1.8\times10^{-10}}{1.05\times10^{-5}}=1.71\times10^{-5}\ \text{M}$ (3) Fraction of Cl⁻ remaining $=1.71\times10^{-5}/0.010=1.7\times10^{-3}$ (0.17%). Since >99.8% of Cl⁻ is removed before chromate starts, selective precipitation is very effective. (2)

Q5 (10)

(a) (1 each)

• $\text{NH}_4\text{Cl}$: salt of SA(HCl)/WB(NH₃) → $\text{NH}_4^+$ hydrolyses → acidic.
• $\text{CH}_3\text{COONa}$: salt of WA/SB → acetate hydrolyses → basic.
• $\text{CH}_3\text{COON H}_4$: WA/WB, $K_a\approx K_b$ → ≈ neutral (slightly acidic).

(b) $K_a(\text{NH}_4^+)=K_w/K_b=10^{-14}/1.8\times10^{-5}=5.56\times10^{-10}$. (1) $[\text{H}^+]=\sqrt{K_a\cdot C}=\sqrt{5.56\times10^{-10}\times0.10}=\sqrt{5.56\times10^{-11}}=7.45\times10^{-6}$. (2) $\text{pH}=-\log(7.45\times10^{-6})=5.13$. (1)

(c) Conjugate base of $\text{H}_2\text{PO}_4^-$ is $\text{HPO}_4^{2-}$ (lose H⁺). (1) Conjugate acid of $\text{CO}_3^{2-}$ is $\text{HCO}_3^-$ (gain H⁺). (1) $\text{H}_2\text{PO}_4^-$ can donate or accept a proton → amphiprotic (amphoteric Brønsted species). (1)

[
  {"claim":"Q1a Kc = 0.133 M","code":"Kc=(0.20*0.20)/0.30; result = abs(Kc-0.1333)<1e-3"},
  {"claim":"Q1b Kp = 5.54 bar","code":"Kc=Rational(2,15); Kp=float(Kc)*(0.0831*500); result = abs(Kp-5.54)<0.05"},
  {"claim":"Q1c Qc=0.0667 < Kc=0.133 -> forward","code":"Qc=(0.10*0.10)/0.15; Kc=0.1333; result = Qc<Kc"},
  {"claim":"Q2a Ka = 1.82e-5","code":"import math; H=10**-2.87; a=H/0.10; Ka=0.10*a**2; result = abs(Ka-1.82e-5)<0.1e-5"},
  {"claim":"Q2c alpha after 100x dilution = 0.135","code":"import math; a=math.sqrt(1.82e-5/0.0010); result = abs(a-0.135)<0.005"},
  {"claim":"Q3b buffer pH = 4.56","code":"import math; pH=4.74+math.log10(0.0040/0.0060); result = abs(pH-4.56)<0.02"},
  {"claim":"Q4c Cl- remaining = 1.71e-5 M","code":"import math; Ag=math.sqrt(1.1e-12/0.010); Cl=1.8e-10/Ag; result = abs(Cl-1.71e-5)<0.1e-5"},
  {"claim":"Q5b NH4Cl pH = 5.13","code":"import math; Ka=1e-14/1.8e-5; H=math.sqrt(Ka*0.10); pH=-math.log10(H); result = abs(pH-5.13)<0.03"}
]