Level 3 — ProductionEquilibrium

Equilibrium

60 marksprintable — key stays hidden on paper

Level 3 Paper (Production: Derivations & Explain-out-loud)

Time: 45 minutes Total Marks: 60

Answer all questions. Show every derivation step from first principles. State assumptions.

Q1. [10 marks] Starting from the law of mass action for the general gas-phase reaction $aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)$ derive the relationship $K_p = K_c (RT)^{\Delta n}$ from scratch. Clearly define $\Delta n$ and state the assumption(s) required. Then compute $K_p$ at 500 K given $K_c = 4.0\times10^{-2}$ for the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ (Use $R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}$ .)

Q2. [10 marks] (a) Derive Ostwald's dilution law for a weak monobasic acid $HA$ of initial concentration $C$ and degree of dissociation $\alpha$ , giving the exact expression for $K_a$ and the approximate form $\alpha = \sqrt{K_a/C}$ . State the approximation used and when it fails. [6] (b) A weak acid has $K_a = 1.8\times10^{-5}$ . Using the approximation, find $\alpha$ and the pH of a $0.10\ \text{M}$ solution. [4]

Q3. [12 marks] (a) Derive the Henderson–Hasselbalch equation $\text{pH} = pK_a + \log\dfrac{[\text{A}^-]}{[\text{HA}]}$ starting from the acid dissociation equilibrium. [5] (b) A buffer is made by mixing $0.20\ \text{mol}$ acetic acid and $0.30\ \text{mol}$ sodium acetate in $1.0\ \text{L}$ . $K_a(\text{CH}_3\text{COOH}) = 1.8\times10^{-5}$ . Calculate the pH. [3] (c) Explain out loud (in words + equations) how this buffer resists pH change when a small amount of strong base is added. Recalculate the pH after adding $0.05\ \text{mol}$ NaOH. [4]

Q4. [10 marks] (a) Define the reaction quotient $Q$ . State the three cases relating $Q$ to $K$ and the direction the reaction shifts in each. [4] (b) For $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ , $K_c = 50$ at a given temperature. A vessel contains $[SO_2]=0.10$ , $[O_2]=0.20$ , $[SO_3]=0.30\ \text{M}$ . Calculate $Q$ and predict the direction of net reaction. [4] (c) Using Le Chatelier's principle, state and justify the effect on the equilibrium yield of $SO_3$ of (i) increasing pressure, (ii) adding a catalyst. [2]

Q5. [10 marks] (a) Derive the expression for the pH of an aqueous solution of a salt of a weak acid and strong base (WA/SB), of concentration $C$ , in terms of $K_w$ , $K_a$ and $C$ . Show that $\text{pH} = 7 + \tfrac12 pK_a + \tfrac12\log C$ . [6] (b) Calculate the pH of $0.10\ \text{M}$ sodium acetate ( $K_a = 1.8\times10^{-5}$ ). [4]

Q6. [8 marks] (a) Write the $K_{sp}$ expression for $Ag_2CrO_4$ and derive its molar solubility $s$ in pure water in terms of $K_{sp}$ . [3] (b) $K_{sp}(Ag_2CrO_4) = 1.1\times10^{-12}$ . Find $s$ in pure water, then find $s$ in $0.10\ \text{M}$ AgNO $_3$ . Explain the common-ion effect using the numbers. [5]

Answer keyMark scheme & solutions

Q1 [10]

Derivation [6]: For an ideal gas, $PV=nRT \Rightarrow P_i = \dfrac{n_i}{V}RT = [i]RT$ (1). So each partial pressure equals concentration × RT.

$K_p = \dfrac{P_C^c P_D^d}{P_A^a P_B^b}$ (1). Substitute $P_i=[i]RT$ : $K_p = \frac{([C]RT)^c([D]RT)^d}{([A]RT)^a([B]RT)^b} = \frac{[C]^c[D]^d}{[A]^a[B]^b}(RT)^{(c+d)-(a+b)}$ (2) $= K_c (RT)^{\Delta n}$ , where $\Delta n = (c+d)-(a+b)$ = moles gaseous product − moles gaseous reactant (1). Assumption: ideal gas behaviour; only gaseous species counted in $\Delta n$ (1).

Numerical [4]: $\Delta n = 2-(1+3) = -2$ (1). $K_p = K_c(RT)^{\Delta n} = 4.0\times10^{-2}\times(0.0821\times500)^{-2}$ (1). $RT = 41.05$ ; $(41.05)^{-2}=5.935\times10^{-4}$ (1). $K_p = 4.0\times10^{-2}\times5.935\times10^{-4} = 2.37\times10^{-5}\ \text{atm}^{-2}$ (1).

Q2 [10]

(a) [6] $HA \rightleftharpoons H^+ + A^-$ . Initial $C$ , 0, 0; equilibrium $C(1-\alpha),\ C\alpha,\ C\alpha$ (2). $K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$ (2) — exact form. For weak acid $\alpha \ll 1 \Rightarrow 1-\alpha \approx 1$ (1): $K_a \approx C\alpha^2 \Rightarrow \alpha = \sqrt{K_a/C}$ (1). Fails at high dilution / larger $K_a$ where $\alpha$ not small.

(b) [4] $\alpha=\sqrt{1.8\times10^{-5}/0.10}=\sqrt{1.8\times10^{-4}}=1.342\times10^{-2}$ (2). $[H^+]=C\alpha = 0.10\times1.342\times10^{-2}=1.342\times10^{-3}$ (1). $\text{pH}=-\log(1.342\times10^{-3})=2.87$ (1).

Q3 [12]

(a) [5] $HA\rightleftharpoons H^+ + A^-$ , $K_a=\dfrac{[H^+][A^-]}{[HA]}$ (1). $[H^+]=K_a\dfrac{[HA]}{[A^-]}$ (1). Take $-\log$ : $-\log[H^+] = -\log K_a - \log\dfrac{[HA]}{[A^-]}$ (2). $\text{pH}=pK_a+\log\dfrac{[A^-]}{[HA]}$ (1).

(b) [3] $pK_a=-\log(1.8\times10^{-5})=4.745$ (1). $\text{pH}=4.745+\log(0.30/0.20)=4.745+\log1.5=4.745+0.176$ (1) $=4.92$ (1).

(c) [4] Added OH⁻ reacts with HA: $HA+OH^-\to A^-+H_2O$ , converting acid to conjugate base; ratio changes slightly, pH nearly stable (2). New: HA = $0.20-0.05=0.15$ , A⁻ = $0.30+0.05=0.35$ (1). $\text{pH}=4.745+\log(0.35/0.15)=4.745+0.368=5.11$ (1).

Q4 [10]

(a) [4] $Q$ = same mass-action ratio as $K$ but with current (non-equilibrium) concentrations (1). $Q<K$ : forward shift (1); $Q=K$ : at equilibrium (1); $Q>K$ : reverse shift (1).

(b) [4] $Q=\dfrac{[SO_3]^2}{[SO_2]^2[O_2]}=\dfrac{0.30^2}{0.10^2\times0.20}=\dfrac{0.09}{0.002}=45$ (3). $Q=45<K=50$ ⇒ shifts forward (right) (1).

(c) [2] (i) $\Delta n_g = 2-3=-1$ ; increasing pressure shifts toward fewer gas moles → more $SO_3$ (1). (ii) Catalyst speeds both directions equally → no change in equilibrium position/yield, only faster attainment (1).

Q5 [10]

(a) [6] Salt $A^-$ hydrolyses: $A^- + H_2O \rightleftharpoons HA + OH^-$ , $K_h=\dfrac{K_w}{K_a}$ (2). Let $x=[OH^-]$ : $K_h=\dfrac{x^2}{C-x}\approx\dfrac{x^2}{C}$ (1) ⇒ $x=\sqrt{K_hC}=\sqrt{\dfrac{K_wC}{K_a}}$ (1). $pOH=-\log x = \tfrac12(pK_w - pK_a - \log C)$ (1). $\text{pH}=14-pOH = 7+\tfrac12 pK_a+\tfrac12\log C$ (1).

(b) [4] $pK_a=4.745$ (1). $\text{pH}=7+\tfrac12(4.745)+\tfrac12\log(0.10)$ (1) $=7+2.372-0.5$ (1) $=8.87$ (1).

Q6 [8]

(a) [3] $Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}$ ; $K_{sp}=[Ag^+]^2[CrO_4^{2-}]$ (1). If solubility $s$ : $[Ag^+]=2s,[CrO_4^{2-}]=s$ (1) ⇒ $K_{sp}=(2s)^2 s=4s^3 \Rightarrow s=(K_{sp}/4)^{1/3}$ (1).

(b) [5] Pure water: $s=(1.1\times10^{-12}/4)^{1/3}=(2.75\times10^{-13})^{1/3}=6.50\times10^{-5}\ \text{M}$ (2). In $0.10$ M AgNO₃: $[Ag^+]\approx0.10$ ; $K_{sp}=(0.10)^2 s \Rightarrow s=\dfrac{1.1\times10^{-12}}{0.010}=1.1\times10^{-10}\ \text{M}$ (2). Common-ion Ag⁺ suppresses dissolution — solubility falls ~6 orders of magnitude (1).

[
{"claim":"Q1 Kp = 2.37e-5","code":"Kc=Rational(4,100); R=0.0821; T=500; dn=-2; Kp=Kc*(R*T)**dn; result=abs(float(Kp)-2.37e-5)<1e-7"},
{"claim":"Q2 pH of 0.10M weak acid Ka=1.8e-5 is 2.87","code":"Ka=1.8e-5; C=0.10; alpha=(Ka/C)**0.5; H=C*alpha; import sympy; pH=-sympy.log(H,10); result=abs(float(pH)-2.87)<0.02"},
{"claim":"Q3b buffer pH 4.92","code":"import sympy; pKa=-sympy.log(1.8e-5,10); pH=pKa+sympy.log(Rational(30,20),10); result=abs(float(pH)-4.92)<0.02"},
{"claim":"Q5b sodium acetate pH 8.87","code":"import sympy; pKa=-sympy.log(1.8e-5,10); pH=7+pKa/2+sympy.log(0.10,10)/2; result=abs(float(pH)-8.87)<0.02"},
{"claim":"Q6 solubility Ag2CrO4 pure water 6.50e-5","code":"s=(1.1e-12/4)**(1/3); result=abs(s-6.50e-5)<1e-6"}
]