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Higher Paper Practice, Question 1

Higher Paper Practice, Question 1. Empirical formula is Na 2 S 2 O 3.

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Higher Paper Practice, Question 1

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  1. Higher Paper Practice, Question 1 Empirical formula is Na2S2O3. (Note Mr not provided, can’t find molecular formula the usual way) Sodium thiosulfate is an ionic compound, hence molecular formula is likely the same as empirical formula, hence formula is Na2S2O3. If compound is covalent, then must find n from Mr and not assume empirical formula = molecular formula.

  2. Q1 (b) • X: sulfur or S(Q says X is a yellow solid) • Y: sulfur dioxide or SO2 (Recall SO2 turns acidified K2Cr2O7 from orange to green) • H2SO4 (aq) + Na2S2O3 (aq) → Na2SO4 (aq) + H2O (l) + S (s) + SO2 (g) • Acidic • SO2 + 2NaOH → Na2SO3 + H2O • Or SO2 + H2O → H2SO3 • Bond angle is 120 • as the shape is Trigonal planar • Since X is S, Z is SO3 3 bond pairs, 0 lone pair Shape: trigonal planar

  3. Question 2 • i) covalent bond (Q says between atoms, so likely asking for chem bond; If Q says between molecules, then it is asking for intermolecular forces such as id-id) • ii) simple molecular structure as SO2Cl2 has a low melting and boiling point (Use the info provided to answer!!)

  4. Q2 (a) iii) SO2Cl2 + 2H2O → 2HCl + H2SO4 iv) (1) No. of mol of SO2Cl2 = 1 mol of SO2Cl2 produces 2 mol of HCl and 1 mol of H2SO4 HCl is monobasic while H2SO4 is dibasic 1 mol of HCl gives 1 mol of H+ while 1 mol of H2SO4 gives 2 mol of H+ Hence 1 mol of SO2Cl2 produces a total of 2+2 = 4 mol of H+ 0.500 mol of SO2Cl2 produce 0.500x4=2 mol of H+ Concentration of H+ = 2 ÷ 5 = 0.400 mol/dm3 pH = -lg [H+] = -lg 0.400 =0.398

  5. Q 2 (a) (iv) (2) You must understand that solution X is a mixture of acids, hence the ion that is reacting with aq. NaOH must be H+ H+ + OH-→ H2O No. of mol of H+ in 5 dm3 = 2 1 mol of H+ reacts with 1 mol of OH- 2 mol of H+ react with 2 mol of OH- No. of mol = conc in mol/dm3 x vol in dm3 2 = 1.00 x volume Volume = 2.00 dm3

  6. Q 2 (b) Around B (central atom) 3 bond pairs, 0 lone pair Shape: trigonal planar Since the ion has a -1 charge, Iodine gains 1 e- and has a total of 8 e- It uses 4 e- to form 4 bonds with 4 Cl atoms, left 4 e- Hence it has 4 bond pairs and 2 lone pair Shape: square planar

  7. Unreacted acid Std soln with known conc Find no of mol of unreacted HCl Find no. of mol of HCl reacted with CaCO3 by subtracting no. of mol of unreacted acid from initial no. of mol of acid used Find mass of CaCO3 Find no of mol of CaCO3 Q3 (a) CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l) Excess HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) Back titration

  8. Question 3 (a) HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) No. of mol of NaOH = No. of mol of NaOH = 1 mol of NaOH reacts with 1 mol of HCl 0.001217 mol of NaOH react with 0.001217 mol of HCl No. of mol of unreacted HCl in 25.0 cm3 = 0.001217 Hence in 250 cm3, no. of mol of unreacted HCl = 0.001217x10 = 0.01217 Initial no. of mol of HCl added to CaCO3 = No. of mol of HCl reacted with CaCO3 = 0.0250-0.01217 = 0.01283

  9. Question 3 (a) Cont’d CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l) 2 mol of HCl react with 1 mol of CaCO3 0.01283 mol of HCl react with 0.006415 mol of CaCO3 Mass of CaCO3 = 0.006415 x [40+12+3(16)] = 0.6415 g % of CaCO3 =

  10. H H H H Question 3 (b) Thought process Atomic radius of halogen increases down a group F Cl Bond length H-X increases Br Bond strengthdecreases I Ease of dissociation of H+ increases down the group

  11. Question 3 (b) Rank (weakest to strongest bond strength): H-I, H-Br, H-Cl, H-F Down a group, the atomic radius of Group VII element increases but the radius of hydrogen atom remains constant, hence the covalent bond length of hydrogen halides down the group increases. The longer the covalent bond length, the weaker the H-X bond, hence HI has the weakest H-X bond while HF has the strongest H-X bond. Thus it is easiest for the H-I bond to be broken to release the H+ and HI is the strongest acid.

  12. Question 4 (a) Mix samples of the contents of the bottles in pairs. The following observations will be made. Identity of the bottles can be made from the observations above. For example, if effervescence is observed when 2 bottles are mixed, the bottles must contain hydrochloric acid and sodium carbonate. 2HCl + Na2CO3  2NaCl + H2O + CO2 The remaining bottle must be zinc sulfate. (Continue next slide)

  13. Then add content of remaining bottle to one of the bottles. If a white precipitate is observed, the bottle must contain sodium carbonate. If no visible change is observed, then the bottle contains hydrochloric acid. ZnSO4 (aq) + Na2CO3 (aq)  ZnCO3 (s) + Na2SO4 (aq)

  14. Question 4 (b) • Effervescence is observed because ascorbic acid reacts with sodium bicarbonate to produce carbon dioxide gas. • H+ + HCO3- CO2 + H2O • (Note: sodium bicarbonate = sodium hydrogencarbonate

  15. Question 4 (b) (ii) Mass of ascorbic acid = 1000 mg = 1 g No. of mol of ascorbic acid (C6H8O6) = Mass of NaHCO3 = 700 mg = 0.700 g No. of mol NaHCO3 = HA + NaHCO3  NaA + H2O + CO2 1 mol HA reacts with 1 mol NaHCO3 Hence 5.681 x 10-3 mol of HA react with 5.681 x 10-3 mol of NaHCO3 Hence HA is limiting and NaHCO3 is in excess. 1 mol HA produces 1 mol of CO2 5.681 x 10-3 mol HA produce 5.681 x 10-3 mol of CO2 Volume of CO2 = 5.681 x 10-3 x 24 = 0.136 dm3

  16. Question 4 (b) (iii) CaO + 2HA  CaA2 + H2O No. of mol of CaO = No. of mol HA = 1 mol of CaO reacts with 2 mol of HA. 0.1428 mol of CaO react with 0.2856 mol of HA Hence CaO is limiting; HA is in excess 1 mol of CaO produces 1 mol CaA2 0.1428 mol CaO produce 0.1428 mol of CaA2 Given that formula of ascorbic acid = C6H8O6 Formula of ascorbate (formed when ascorbic acid loses 1 H): C6H7O6- Hence calcium ascorbate has the formula: Ca(C6H7O6)2 Mass of calcium ascorbate = 0.1428 x [40+2(12x6 + 7 + 6x16)] = 55.7 g

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