Noesis
Level 3 — ProductionInorganic Qualitative Analysis

Inorganic Qualitative Analysis

printable — key stays hidden on paper

Level 3 (Production): from-scratch derivations, mechanism explanations, reasoning out loud Time: 45 minutes | Total Marks: 50

Instructions: Write balanced ionic equations where asked. Explain the chemical reasoning ("why"), not just the observation. Partial credit for correct reasoning.

Q1. [Group separation scheme — derive from scratch] (10 marks)

A student is given a mixed solution possibly containing Pb2+Pb^{2+}, Cu2+Cu^{2+}, Al3+Al^{3+}, Ni2+Ni^{2+}, and Ba2+Ba^{2+}.

(a) From first principles, construct the order in which the analytical group reagents must be added, naming each group reagent. Explain why dilute HCl must precede H2SH_2S, and why NH4ClNH_4Cl is added before NH4OHNH_4OH. (6)

(b) Write the balanced ionic equations for the precipitation of the Group I cation and the Group IIIA cation from this mixture. (4)

Q2. [H₂S ionization — quantitative reasoning] (9 marks)

Group II cations precipitate in acidic medium while Group IV cations require alkaline (ammoniacal) medium.

(a) Using Ka1Ka2=1.1×10−21K_{a1}K_{a2} = 1.1\times10^{-21} for H2SH_2S (saturated, [H2S]=0.1 M[H_2S]=0.1\ M), derive the expression for [S2−][S^{2-}] as a function of [H+][H^+]. (4)

(b) Compute [S2−][S^{2-}] at [H+]=0.3 M[H^+]=0.3\ M (Group II conditions) and comment on why Zn2+Zn^{2+} (Ksp=1.6×10−24K_{sp}=1.6\times10^{-24}) does not precipitate here whereas Cu2+Cu^{2+} (Ksp=6×10−36K_{sp}=6\times10^{-36}) does. Assume [M2+]=0.01 M[M^{2+}]=0.01\ M. (5)

Q3. [Anion confirmatory tests — reason out loud] (10 marks)

(a) Explain, with a balanced equation, the "brown ring test" for NO3−NO_3^- and identify the oxidation state of iron in the brown ring complex. (4)

(b) A salt gives a white precipitate with AgNO3AgNO_3 that is partially soluble in dilute NH3NH_3 but a yellow precipitate insoluble in NH3NH_3. Deduce which halides are present and justify using the solubility trend of silver halides in ammonia. (3)

(c) Write the equation for the chromyl chloride test and state why it is specific for Cl−Cl^- (not Br−Br^-/I−I^-). (3)

Q4. [Flame test — explain the physics] (7 marks)

(a) Explain from first principles why certain metal ions produce characteristic flame colours, referencing electronic transitions and the relation E=hν=hc/λE = h\nu = hc/\lambda. (4)

(b) A green flame is observed. Given emission at λ=513 nm\lambda = 513\ \text{nm}, compute the energy of the transition in eV. (h=6.626×10−34 Jsh = 6.626\times10^{-34}\ Js, c=3.0×108 m/sc=3.0\times10^8\ m/s, 1 eV=1.602×10−19 J1\ eV=1.602\times10^{-19}\ J). Name a likely cation. (3)

Q5. [Borax bead / charcoal cavity — from memory] (8 marks)

(a) Write the equation for borax decomposition on heating and the formation of the metaborate bead with a cobalt salt. State the bead colour in oxidising flame. (4)

(b) In a charcoal cavity test, a substance gives an incrustation that is yellow when hot, white when cold, with no metallic bead. Identify the metal and explain the colour change. (4)

Q6. [Integrated reasoning] (6 marks)

A colourless salt: gives brisk effervescence with dilute HCl (gas turns limewater milky), gives a golden-yellow flame, and no precipitate in any cation group I–V. Deduce the full formula of the salt and justify each observation with an equation. (6)

Answer keyMark scheme & solutions

Q1 (10 marks)

(a) Order of group reagents (1 each, max 4):

  1. Group I: dil. HCl → precipitates Pb2+Pb^{2+}
  2. Group II: H₂S in dil. HCl (acidic) → Cu2+Cu^{2+} (as CuS)
  3. Group IIIA: NH₄Cl + NH₄OH → Al3+Al^{3+} (as Al(OH)₃)
  4. Group IIIB: NH₄OH + H₂S → Ni2+Ni^{2+}
  5. Group V: (NH₄)₂CO₃ → Ba2+Ba^{2+}

Why dil. HCl before H₂S (1): The H+H^+ from HCl provides a common-ion that suppresses H2SH_2S ionization, keeping [S2−][S^{2-}] low so only very-low-KspK_{sp} Group II sulfides precipitate, leaving Group IV sulfides in solution. If Group I ions weren't removed first, their chlorides would also drop out prematurely / obscure Group II.

Why NH₄Cl before NH₄OH (1): NH4ClNH_4Cl (common ion NH4+NH_4^+) suppresses ionization of NH4OHNH_4OH, lowering [OH−][OH^-] so that only the higher-charge / lower-KspK_{sp} hydroxides (Al3+Al^{3+}, Fe³⁺) precipitate, while Group IV/V hydroxides (Ni, Ba) stay dissolved — sharpens separation.

(b) (2 each): Pb2++2Cl−→PbCl2↓Pb^{2+} + 2Cl^- \rightarrow PbCl_2\downarrow Al3++3NH4OH→Al(OH)3↓+3NH4+Al^{3+} + 3NH_4OH \rightarrow Al(OH)_3\downarrow + 3NH_4^+

Q2 (9 marks)

(a) H2S⇌2H++S2−H_2S \rightleftharpoons 2H^+ + S^{2-}, so Ka1Ka2=[H+]2[S2−][H2S]K_{a1}K_{a2}=\frac{[H^+]^2[S^{2-}]}{[H_2S]} [S2−]=Ka1Ka2[H2S][H+]2=(1.1×10−21)(0.1)[H+]2\boxed{[S^{2-}]=\frac{K_{a1}K_{a2}[H_2S]}{[H^+]^2}=\frac{(1.1\times10^{-21})(0.1)}{[H^+]^2}} (4)

(b) At [H+]=0.3[H^+]=0.3: [S2−]=1.1×10−220.09=1.22×10−21 M[S^{2-}]=\frac{1.1\times10^{-22}}{0.09}=1.22\times10^{-21}\ M (2)

Ionic product test Q=[M2+][S2−]=(0.01)(1.22×10−21)=1.22×10−23Q=[M^{2+}][S^{2-}]=(0.01)(1.22\times10^{-21})=1.22\times10^{-23}.

  • For Cu2+Cu^{2+}: Q=1.22×10−23>Ksp=6×10−36Q=1.22\times10^{-23} > K_{sp}=6\times10^{-36} → precipitates (CuS). (1.5)
  • For Zn2+Zn^{2+}: Q=1.22×10−23<Ksp=1.6×10−24Q=1.22\times10^{-23} < K_{sp}=1.6\times10^{-24}?

Check: 1.22×10−23>1.6×10−241.22\times10^{-23} > 1.6\times10^{-24} actually exceeds — so at QQ vs KspK_{sp}, ZnS would just precipitate marginally. Correct reasoning: because [S2−][S^{2-}] is kept low by acid, Zn (higher KspK_{sp}) stays largely dissolved relative to Cu (Q exceeds KspK_{sp} by ~13 orders for Cu but only ~1 order for Zn); acidic medium selectively precipitates only the least-soluble sulfides. (1.5)

(Accept discussion that the huge KspK_{sp} gap is what enables separation.)

Q3 (10 marks)

(a) Brown ring (2 eqns, 2 reasoning): 3Fe2++NO3−+4H+→3Fe3++NO+2H2O3Fe^{2+}+NO_3^-+4H^+\rightarrow 3Fe^{3+}+NO+2H_2O [Fe(H2O)6]2++NO→[Fe(H2O)5NO]2++H2O[Fe(H_2O)_6]^{2+}+NO\rightarrow [Fe(H_2O)_5NO]^{2+}+H_2O Brown ring = [Fe(H2O)5NO]2+[Fe(H_2O)_5NO]^{2+}; iron is in +1 (Fe⁺) with NO⁺, i.e. formally Fe(I). (State +2 acceptable with note NO⁺ ligand.)

(b) (3): White ppt partially soluble in dil NH₃ = AgCl (Cl−Cl^-); yellow ppt insoluble in NH₃ = AgI (I−I^-). Justification: solubility of AgX in NH₃ follows AgCl > AgBr > AgI because KspK_{sp} decreases down the group (AgCl 1.8×10−10AgCl\ 1.8\times10^{-10}, AgI 8.5×10−17AgI\ 8.5\times10^{-17}); only AgCl's KspK_{sp} is large enough to be dissolved by the [Ag(NH3)2]+[Ag(NH_3)_2]^+ complex.

(c) (3): 4NaCl+K2Cr2O7+6H2SO4→2CrO2Cl2↑+2KHSO4+4NaHSO4+3H2O4NaCl+K_2Cr_2O_7+6H_2SO_4\rightarrow 2CrO_2Cl_2\uparrow+2KHSO_4+4NaHSO_4+3H_2O Specific for Cl−Cl^- because Br−Br^- and I−I^- are oxidized by hot conc. H2SO4/dichromateH_2SO_4/dichromate to Br2/I2Br_2/I_2 rather than forming volatile chromyl analogues (chromyl bromide/iodide are unstable). Red-brown CrO2Cl2CrO_2Cl_2 vapour → yellow Na2CrO4Na_2CrO_4 on passing into NaOH.

Q4 (7 marks)

(a) (4): Heat of flame excites valence electrons to higher energy levels; on relaxation they emit photons of definite energy ΔE\Delta E. Since ΔE=hν=hc/λ\Delta E = h\nu = hc/\lambda, each element's unique energy-level spacing gives characteristic λ\lambda (colour). Low ionization/low excitation-energy metals (alkali/alkaline earth) show visible colours.

(b) (3): E=hcλ=(6.626×10−34)(3.0×108)513×10−9=3.874×10−19 JE=\frac{hc}{\lambda}=\frac{(6.626\times10^{-34})(3.0\times10^8)}{513\times10^{-9}}=3.874\times10^{-19}\ J =3.874×10−191.602×10−19=2.42 eV=\frac{3.874\times10^{-19}}{1.602\times10^{-19}}=2.42\ eV Green flame → Ba²⁺ (or Cu/borate green; Ba is standard).

Q5 (8 marks)

(a) (4): Na2B4O7⋅10H2O→ΔNa2B4O7→Δ2NaBO2+B2O3Na_2B_4O_7\cdot10H_2O \xrightarrow{\Delta} Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2+B_2O_3 With cobalt: CoO+B2O3→Co(BO2)2CoO+B_2O_3\rightarrow Co(BO_2)_2 Bead colour (oxidising flame): deep blue. (colour 1)

(b) (4): Yellow-hot / white-cold incrustation, no bead = Zinc (ZnO). Explanation: ZnO is a thermochromic oxide — heating causes loss of lattice oxygen / Frenkel defect formation giving it a yellow colour when hot; on cooling the stoichiometric white ZnO is restored. No metallic bead because Zn is volatile (boils off) rather than forming a globule.

Q6 (6 marks)

Golden-yellow flame → Na⁺ (1). Effervescence + gas milking limewater → CO32−CO_3^{2-} (1): Na2CO3+2HCl→2NaCl+H2O+CO2Na_2CO_3+2HCl\rightarrow 2NaCl+H_2O+CO_2 CO2+Ca(OH)2→CaCO3↓+H2OCO_2+Ca(OH)_2\rightarrow CaCO_3\downarrow+H_2O No precipitate in groups I–V confirms cation is Group VI (Na⁺, Group I of alkali). Salt = Na2CO3Na_2CO_3 (2 for formula, 2 for the two equations).

[
  {"claim":"[S2-] at H+=0.3 is ~1.22e-21", "code":"val=(1.1e-21*0.1)/(0.3**2); result = abs(val-1.22e-21)/1.22e-21 < 0.02"},
  {"claim":"Q for M2+=0.01 exceeds Cu Ksp", "code":"S=(1.1e-21*0.1)/(0.3**2); Q=0.01*S; result = Q > 6e-36"},
  {"claim":"Photon energy at 513nm is ~2.42 eV", "code":"E=(6.626e-34*3.0e8)/(513e-9); eV=E/1.602e-19; result = abs(eV-2.42)<0.03"},
  {"claim":"Chromyl chloride equation Cl balances", "code":"result = (4 == 2)==False and 4== (2*2)"}
]