Equilibrium

Question 1

(a) Initial moles $n_0$; at dissociation $\alpha$: $\mathrm{PCl_5}=n_0(1-\alpha)$, $\mathrm{PCl_3}=n_0\alpha$, $\mathrm{Cl_2}=n_0\alpha$. Total $=n_0(1+\alpha)$. Concentrations = moles/$V$. $K_c=\frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}=\frac{(n_0\alpha/V)(n_0\alpha/V)}{n_0(1-\alpha)/V}=\frac{n_0\alpha^2}{V(1-\alpha)}$ (3 marks derivation)

Numeric: $n_0=1.0,\ \alpha=0.40,\ V=2.0$: $K_c=\frac{(1.0)(0.16)}{2.0(0.60)}=\frac{0.16}{1.20}=0.1333\ \text{mol L}^{-1}$ (2 marks) ✓ $K_c \approx 0.133$ M.

(b) For each gas $[i]=n_i/V=P_i/RT$ (ideal gas). Substituting partial pressures: $K_p=\frac{P_{PCl_3}P_{Cl_2}}{P_{PCl_5}}=\frac{([PCl_3]RT)([Cl_2]RT)}{[PCl_5]RT}=K_c(RT)^{(2-1)}=K_c(RT)^{\Delta n}$ (3 marks) $\Delta n=(1+1)-1=+1$. (1) $K_p=0.1333\times(0.0821\times500)^1=0.1333\times41.05=5.47\ \text{atm}$ (1 mark) ✓ $K_p\approx5.47$.

(c) Doubling $V$ at fixed $T$: $Q$ recomputed with old moles but new $V=4.0$: $Q=\frac{(0.4/4)(0.4/4)}{(0.6/4)}=\frac{(0.1)(0.1)}{0.15}=0.0667 < K_c=0.1333$ $Q<K_c\Rightarrow$ reaction shifts forward (more dissociation). (2 marks) New $\alpha$ from $K_c=\dfrac{n_0\alpha^2}{V(1-\alpha)}$ with $V=4.0$: $0.1333=\frac{\alpha^2}{4(1-\alpha)}\Rightarrow \alpha^2=0.5333(1-\alpha)\Rightarrow \alpha^2+0.5333\alpha-0.5333=0$ $\alpha=\frac{-0.5333+\sqrt{0.2844+2.1333}}{2}=\frac{-0.5333+1.5548}{2}=0.511$ (3 marks) ✓ $\alpha\approx0.51$ (increased, consistent with forward shift).

(d) (i) Reaction endothermic ($\Delta H>0$): raising $T$ shifts forward, $\alpha$ increases. (ii) Inert gas at constant volume: partial pressures/concentrations unchanged ⇒ no shift, $\alpha$ unchanged. (iii) Catalyst: speeds both directions equally, equilibrium position and $\alpha$ unchanged. (1 each = 3 marks)

Question 2

(a) For weak acid HA ⇌ H⁺ + A⁻, $K_a=\dfrac{C\alpha^2}{1-\alpha}$. If $\alpha\ll1$: $K_a\approx C\alpha^2\Rightarrow \alpha=\sqrt{K_a/C}$. (2) $\alpha=\sqrt{1.8\times10^{-5}/0.100}=\sqrt{1.8\times10^{-4}}=0.0134$ $[\mathrm{H^+}]=C\alpha=0.100\times0.0134=1.34\times10^{-3}$, $\text{pH}=2.87$. Valid since $\alpha=1.3\%\ll5\%$. (2 marks) ✓

(b) $K_a=\dfrac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}\Rightarrow[\mathrm{H^+}]=K_a\dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]}$. Taking $-\log$: $\text{pH}=\mathrm{p}K_a+\log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$ (2) $\mathrm{p}K_a=-\log(1.8\times10^{-5})=4.745$. Ratio $=0.100/0.100=1$, $\log1=0$: $\text{pH}=4.745$ (2 marks) ✓

(c) NaOH converts HA→A⁻: HA $=0.100-0.020=0.080$; A⁻ $=0.100+0.020=0.120$. $\text{pH}=4.745+\log(0.120/0.080)=4.745+\log1.5=4.745+0.176=4.921$ (3) $\Delta\text{pH}=+0.176$ for 0.020 mol base ⇒ small. Reason: added OH⁻ is consumed by the large reservoir of undissociated HA (common ion A⁻ already present suppresses shift); ratio changes only modestly, so $\log$ term barely moves — Le Chatelier buffering. (2 marks) ✓

(d) Exact treatment. Charge balance $[\mathrm{Na^+}]+[\mathrm{H^+}]=[\mathrm{A^-}]+[\mathrm{OH^-}]$, mass balance $[\mathrm{HA}]+[\mathrm{A^-}]=0.200$, $[\mathrm{Na^+}]=0.100$.

from sympy import symbols, nsolve, log
h = symbols('h', positive=True)
Ka, Kw = 1.8e-5, 1e-14
# A- = 0.100 + h - Kw/h  ; HA = 0.200 - A-
# Ka = h*A/HA
A = 0.100 + h - Kw/h
HA = 0.200 - A
sol = nsolve(Ka*HA - h*A, h, 1e-5)
pH = -log(sol,10)
print(float(pH))   # ~4.745

Exact pH $\approx4.745$ (differs from HH by <0.001) ⇒ Henderson–Hasselbalch fully justified here. (5 marks: eqns 2, code 2, conclusion 1) ✓

Question 3

(a) Onset when $Q=K_{sp}$.

• $\mathrm{MgF_2}$: $[\mathrm{F^-}]=\sqrt{K_{sp}/[\mathrm{Mg^{2+}}]}=\sqrt{6.4\times10^{-9}/0.010}=\sqrt{6.4\times10^{-7}}=8.0\times10^{-4}$ M.
• $\mathrm{CaF_2}$: $[\mathrm{F^-}]=\sqrt{3.9\times10^{-11}/0.010}=\sqrt{3.9\times10^{-9}}=6.2\times10^{-5}$ M. $\mathrm{CaF_2}$ needs lower $[\mathrm{F^-}]$ ⇒ CaF₂ precipitates first. (4 marks) ✓

(b) Mg²⁺ starts precipitating when $[\mathrm{F^-}]=8.0\times10^{-4}$. Residual Ca²⁺ then: $[\mathrm{Ca^{2+}}]=\frac{K_{sp}(\mathrm{CaF_2})}{[\mathrm{F^-}]^2}=\frac{3.9\times10^{-11}}{(8.0\times10^{-4})^2}=\frac{3.9\times10^{-11}}{6.4\times10^{-7}}=6.1\times10^{-5}\ \text{M}$ (3) Fraction remaining $=6.1\times10^{-5}/0.010=0.61\%$; >99% Ca removed before Mg starts ⇒ selective separation feasible. (1 mark) ✓

(c) F⁻ + H₂O ⇌ HF + OH⁻, $K_b=K_w/K_a=10^{-14}/6.8\times10^{-4}=1.47\times10^{-11}$. $[\mathrm{OH^-}]=\sqrt{K_b\,C}=\sqrt{1.47\times10^{-11}\times0.050}=\sqrt{7.35\times10^{-13}}=8.57\times10^{-7}$. $\text{pOH}=6.07\Rightarrow \text{pH}=14-6.07=7.93$. (4 marks) ✓ (slightly basic — WA/SB salt).

(d)

• $\mathrm{NH_4Cl}$: SA/WB salt → acidic (NH₄⁺ hydrolyses).
• $\mathrm{CH_3COONa}$: WA/SB → basic.
• $\mathrm{NaCl}$: SA/SB → neutral.
• $\mathrm{NH_4CH_3COO}$: WA/WB, $K_a(\mathrm{NH_4^+})=5.6\times10^{-10}\approx K_a(\mathrm{CH_3COOH})=1.8\times10^{-5}$? Since $K_a(\text{acid})=1.8\times10^{-5} > K_a(\text{cation})=5.6\times10^{-10}$, but standard comparison is $K_a(\mathrm{NH_4^+})$ vs $K_b(\mathrm{Ac^-})$: both ≈$5.6\times10^{-10}$ ⇒ ≈ neutral (very slightly acidic/near 7). (2 marks) ✓

[
{"claim":"Q1a Kc = 0.1333","code":"Kc=(1.0*0.40**2)/(2.0*(1-0.40)); result=abs(Kc-0.13333)<1e-3"},
{"claim":"Q1b Kp=Kc*(RT)^1 = 5.47","code":"Kc=0.13333; Kp=Kc*(0.0821*500)**1; result=abs(Kp-5.473)<0.05"},
{"claim":"Q1c new alpha ~0.511 after doubling V","code":"from sympy import symbols,solve; a=symbols('a',positive=True); s=solve(0.13333-a**2/(4*(1-a)),a); val=[float(x) for x in s if 0<float(x)<1][0]; result=abs(val-0.511)<0.01"},
{"claim":"Q2a pure acetic acid pH=2.87","code":"import math; alpha=math.sqrt(1.8e-5/0.100); H=0.100*alpha; pH=-math.log10(H); result=abs(pH-2.87)<0.02"},
{"claim":"Q2c buffer pH after NaOH = 4.921","code":"import math; pKa=-math.log10(1.8e-5); pH=pKa+math.log10(0.120/0.080); result=abs(pH-4.921)<0.01"},
{"claim":"Q3b residual Ca2+ = 6.1e-5","code":"Ca=3.9e-11/(8.0e-4)**2; result=abs(Ca-6.09e-5)<2e-6"},
{"claim":"Q3c NaF pH = 7.93","code":"import math; Kb=1e-14/6.8e-4; OH=math.sqrt(Kb*0.050);