Inorganic Qualitative Analysis

Q1. [10]

(a) White ppt with dil. HCl → Group I cation (Pb²⁺, Ag⁺, Hg₂²⁺). Insoluble in hot water rules out complete PbCl₂ dissolution (PbCl₂ is soluble in hot water), partial solubility hints at PbCl₂ mixed behaviour → points to Pb²⁺ (PbCl₂ is only slightly soluble in hot water). (1 mark ppt = Group I; 1 mark Pb²⁺ reasoning)

(b) Canary-yellow ppt with K₂CrO₄: $\text{Pb}^{2+} + \text{CrO}_4^{2-} \rightarrow \text{PbCrO}_4\downarrow \text{ (yellow)}$ Confirms Pb²⁺. (equation 1, identification 1)

(c) Colourless gas turning lime-water milky, milkiness disappearing in excess → CO₂ from carbonate: $\text{CO}_3^{2-} + 2\text{H}^+ \rightarrow \text{H}_2\text{O} + \text{CO}_2\uparrow$ $\text{CO}_2 + \text{Ca(OH)}_2 \rightarrow \text{CaCO}_3\downarrow + \text{H}_2\text{O};\quad \text{CaCO}_3 + \text{CO}_2 + \text{H}_2\text{O}\rightarrow \text{Ca(HCO}_3)_2$ Anion = CO₃²⁻. (gas ID 1, 2 equations 2)

(d) Pb²⁺ is in Group I; group reagent = dilute HCl (precipitates as PbCl₂). (group 1, reagent 1, note Pb also in Group II with H₂S as PbS 1)

So X = PbCO₃ (basic lead carbonate). (implicit)

Q2. [12]

(a) Group I negative → no Ag⁺/Pb²⁺/Hg₂²⁺. Group II black ppt → Cu²⁺ (CuS black) or Pb²⁺/Hg²⁺/etc; black + subsequent chemistry → Cu²⁺. Group III reddish-brown gelatinous ppt = Fe(OH)₃, so cation = Fe³⁺. Two cations: Cu²⁺ and Fe³⁺. (Cu²⁺ 2, Fe³⁺ 2)

(b) NH₄Cl supplies common ion NH₄⁺, suppressing ionisation of NH₄OH (Le Chatelier), lowering $[\text{OH}^-]$ so that only Group III hydroxides (high charge, low $K_{sp}$) precipitate while Group IV/V cations like Mg²⁺ stay in solution. (concept 2)

Numerical: ionic product $Q = [\text{Mg}^{2+}][\text{OH}^-]^2 = (0.01)(1.0\times10^{-5})^2 = 0.01\times10^{-10} = 1.0\times10^{-12}$. Since $Q = 1.0\times10^{-12} < K_{sp}=1.8\times10^{-11}$, Mg(OH)₂ does not precipitate — Mg²⁺ stays in solution as intended. (calc 2, conclusion 1)

(c) Cu²⁺ confirmatory (with K₄[Fe(CN)₆] → chocolate brown): $2\text{Cu}^{2+} + [\text{Fe(CN)}_6]^{4-} \rightarrow \text{Cu}_2[\text{Fe(CN)}_6]\downarrow \text{ (reddish-brown)}$ Or with NH₃ deep blue $[\text{Cu(NH}_3)_4]^{2+}$. (equation 3)

Q3. [10]

(a) Table [6, 1 per correct cell]:

(b) $\text{Cl}_2 + 2\text{I}^- \rightarrow 2\text{Cl}^- + \text{I}_2 \text{ (violet in CCl}_4)$ (2)

(c) $K_{sp}$ decreases AgCl > AgBr > AgI. Ammonia complexation $[\text{Ag(NH}_3)_2]^+$ can dissolve AgCl fully and AgBr only in concentrated NH₃ (intermediate $K_{sp}$), but not AgI ($K_{sp}$ too small). (2)

Q4. [10]

(a) Blue borax bead (both flames) + red-brown malleable metallic bead in charcoal cavity + bluish-green flame → Copper (Cu). Cu bead is red-brown/malleable; Cu²⁺ borax bead blue; Cu flame bluish-green. All three consistent. (metal 1, three justifications 3)

(b) Borax on heating gives sodium metaborate + boric anhydride: $\text{Na}_2\text{B}_4\text{O}_7 \rightarrow 2\text{NaBO}_2 + \text{B}_2\text{O}_3$ $\text{B}_2\text{O}_3$ combines with metal oxide to form coloured metaborate e.g. $\text{Cu(BO}_2)_2$ (blue), giving the bead its colour. (equation/species 2, reason 1)

(c) Yellow hot/colourless cold oxidising bead + grey bead with white incrustation in charcoal cavity → volatile white oxide → Zinc (Zn) (ZnO yellow hot, white cold; incrustation ZnO). (Zn 3)

Q5. [8]

(a) No ppt with AgNO₃ (not halide), BaCl₂ (not sulphate/carbonate), no gas with dil. acid (not carbonate/sulphite), but brown ring → NO₃⁻ (nitrate). $\text{NO}_3^- + 4\text{H}^+ + 3\text{Fe}^{2+} \rightarrow 3\text{Fe}^{3+} + \text{NO} + 2\text{H}_2\text{O}$ $[\text{Fe(H}_2\text{O})_6]^{2+} + \text{NO} \rightarrow [\text{Fe(H}_2\text{O})_5\text{NO}]^{2+} + \text{H}_2\text{O} \text{ (brown ring)}$ (ID 1, equations 3)

(b) Both give gas milky with lime-water (CO₂ / SO₂). Distinguish: pass gas through acidified K₂Cr₂O₇ (orange→green) or KMnO₄ (pink decolourised) — only SO₂ (reducing) reacts; carbonate/CO₂ does not. $\text{SO}_2 + \text{Cr}_2\text{O}_7^{2-} + \ldots \rightarrow \text{Cr}^{3+}(\text{green}) + \text{SO}_4^{2-}$ $3\text{SO}_2 + \text{Cr}_2\text{O}_7^{2-} + 2\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{SO}_4^{2-} + \text{H}_2\text{O}$ CO₂ gives no change. (test 2, equation 2)

[
  {"claim":"Q2b: Q = [Mg2+][OH-]^2 = 1.0e-12","code":"Q=Rational(1,100)*(Rational(1,10)**5)**2; result = (float(Q)==1.0e-12)"},
  {"claim":"Q2b: Q < Ksp so no precipitate","code":"Q=Rational(1,100)*(Rational(1,10)**5)**2; Ksp=Rational(18,10)*Rational(1,10)**11; result = (Q < Ksp)"},
  {"claim":"Q5a brown ring: electron balance 3 Fe2+ -> 3 Fe3+ with 1 NO3->NO (3 e-)","code":"result = (3*1 == 3)"},
  {"claim":"Q5b dichromate: Cr goes +6 to +3 (3 e- each, 2 Cr = 6), 3 SO2 give 6 e-","code":"result = (2*3 == 3*2)"}
]