IB DP Chemistry Structure 2.2 The covalent model HL Paper 2- Exam Style Questions - New Syllabus
▶️ Answer/Explanation
Markscheme (with detailed working)
(a)
(i) Lewis structure of H2S (any equivalent with dots/crosses/lines accepted):
Central S with two bonding pairs (to H) and two lone pairs.
(ii) Bent / non-linear / angular / v-shaped (VSEPR: 4 electron domains ⇒ tetrahedral EDG; 2 bonding + 2 lone pairs ⇒ bent shape).
(b)
(i) Conjugate base: HS−. [1]
(ii) Given c(H2S) = 0.10 mol dm−3, pH = 4.0 ⇒ \([H^+] = 10^{-pH} = 10^{-4} = 1.0\times 10^{-4}\) mol dm−3.
If H2S were a strong monoprotic acid at 0.10 mol dm−3, then \([H^+] \approx 0.10\) mol dm−3 ⇒ pH ≈ 1.
Observed \([H^+]=10^{-4}\ll 0.10\), so H2S is a weak acid at this concentration. [1]
If H2S were a strong monoprotic acid at 0.10 mol dm−3, then \([H^+] \approx 0.10\) mol dm−3 ⇒ pH ≈ 1.
Observed \([H^+]=10^{-4}\ll 0.10\), so H2S is a weak acid at this concentration. [1]
(iii) At 298 K, \(K_w = [H^+][OH^-] = 1.00\times 10^{-14}\). Using \([H^+] = 1.0\times 10^{-4}\):
\([OH^-] = \dfrac{K_w}{[H^+]} = \dfrac{1.00\times 10^{-14}}{1.0\times 10^{-4}} = \mathbf{1.0\times 10^{-10}\ \text{mol dm}^{-3}}\). [1]
\([OH^-] = \dfrac{K_w}{[H^+]} = \dfrac{1.00\times 10^{-14}}{1.0\times 10^{-4}} = \mathbf{1.0\times 10^{-10}\ \text{mol dm}^{-3}}\). [1]
(c) [3]
Gas volume decreases from 550 cm3 to 525 cm3 when H2S is removed by NaOH.
Volume of H2S absorbed: \(V_{\text{H}_2\text{S}} = 550 – 525 = 25\ \text{cm}^3\).
Under the same T and P, Avogadro’s law ⇒ volume fraction = mole fraction. Hence \[ \text{mol \% H}_2\text{S} = \frac{25}{550}\times 100 = \mathbf{4.55\%\ \approx\ 4.5\%}. \]
Assumption (state one): gases behave ideally / only H2S reacts with NaOH (N2 inert) / reaction goes to completion / measurements at constant T and P.
(Total for Q2 = 7)