Chemistry Alternative To Practical 2024 Cover

Answer all the questions

1. A is a solution of a monobasic acid HX, suspected to be HCl or HNO₃, containing 3.15g of the acid in 500 cm³ of solution.
   
   B is 0.100 mol·dm⁻³ NaOH.

   25.0 cm³ portions of B were titrated with A using methyl orange as indicator.
   
   The following result was presented by one of the students:

   Complete the table below by listing four excretory products of plants and stating one use of each to humans.
   
   

   Average vol. of A used:

   $$\text{Average}=\frac{25.1+24.8}{2}$$
   
   $$=\frac{49.9}{2}$$

                        = 24.95 cm³

   (a) List three errors in the results recorded by the student.

   (b) Using the average titre value of the student:

   1. Calculate the concentration of A in mol·dm⁻³
   
   
   2. Calculate the molar mass of HX
   
   
   3. Identify HX
   
   

   The reaction equation is:
   
   HX_(aq) + NaOH_(aq) → NaX_(aq) + H₂O_(l)

   [H = 1, N = 14, O = 16, Cl = 35.5, Na = 23]

   [20 marks]

   (a) Three errors in the results recorded by the student.

   1. The first titre (25.1 cm³) from an initial reading of 0.0 cm³ indicates a rough titration and should not be used in the average.
   
   
   2. The initial burette readings are inconsistent, possibly due to incorrect setup or air bubbles in the burette tip.
   
   
   3. The average titre includes a rough result instead of just the two concordant values (24.8 and 24.9 cm³).
   
   

   (b) Using the average titre value of the student:

   1. Concentration of A in mol·dm⁻³:

      $$\text{Moles of NaOH}=\frac{0.100\times 25.0}{1000}$$
      $$=0.00250 \mathrm{mol}$$
      Since HX + NaOH → NaX + H₂O,
      
      mole ratio is 1:1,
      
      Therefore, moles of HX = 0.00250 mol
      
      Volume of HX used = 24.85 cm³
      
      = 0.02485 dm³
      
      $$\text{Concentration}=\frac{0.00250}{0.02485}$$
      
      = 0.1006 mol·dm⁻³

   
   

   2. Molar mass of HX:

      Moles in 500 cm³ = 0.1006 × 0.500
      
      = 0.0503 mol
      
      Mass = 3.15 g
      
      $$\text{Molar mass}=\frac{3.15}{0.0503}$$
      
      = 62.63 g/mol

   
   

   3. Identity of HX:

      Molar mass of HCl = 36.5 g/mol
      
      Molar mass of HNO₃ = 1 + 14 + (16 × 3)
      
      = 63 g/mol
      
      
      Since 62.63 ≈ 63
      
      HX is HNO₃ (Nitric Acid.)

   Balanced chemical equation:
   
   HX + NaOH → NaX + H₂O





2. P, Q, R and S are aqueous solutions of ZnCl₂, K₂CO₃, Pb(NO₃)₂ and H₂SO₄ but not necessarily in that order.
   
   The results obtained when these solutions were reacted with each other are as shown in the table below:

   	P	Q	R	S
   P	X	ppt	ppt	ppt
   Q	ppt	X	No ppt	effervescence
   R	ppt	No ppt	X	ppt
   S	ppt	effervescence	ppt	X

   From the results and your knowledge of the solubility of salts in water and the effect of acids on trioxocarbonates, identify, with reasons, each of the solutions.

   [16 marks]

   
   
   

   Identification and Reasoning:

   • S gives effervescence with Q, indicating that Q is a carbonate (K₂CO₃) and S is an acid (H₂SO₄).
   
   
   • P forms precipitates with Q, R, and S – matching the behavior of Pb(NO₃)₂ which reacts with CO₃²⁻ and SO₄²⁻.
   
   
   • R does not react with Q, matching ZnCl₂ and the moderate solubility of ZnCO₃.
   
   
   • Q is K₂CO₃ (as it reacts with acid to produce effervescence but does not form a precipitate with ZnCl₂).
   
   

   Final Identifications:

   Label	Identified Solution	Chemical Formula
   P	Lead(II) nitrate	Pb(NO3)2
   Q	Potassium carbonate	K2CO3
   R	Zinc chloride	ZnCl2
   S	Sulfuric acid	H2SO4





3. (a) Describe one chemical test to distinguish between aqueous solutions of:

   1. calcium trioxonitrate(V) and zinc trioxonitrate(V);
   
   
   2. potassium chloride and potassium trioxocarbonate(IV).

   (b) Explain why the molar heat of neutralization of a strong base by a strong acid is constant.

   (c) Name two gases that are soluble in water.

   [14 marks]

   
   
   

   Chemical tests to distinguish between given aqueous solutions:

   (i) Calcium trioxonitrate(V) vs. Zinc trioxonitrate(V):

   Test: Add aqueous sodium hydroxide (NaOH) dropwise, then in excess.

   • Calcium trioxonitrate(V): A white precipitate of calcium hydroxide forms and does not dissolve in excess NaOH.
     
     Equation:
     
     Ca²⁺ + 2OH^- → Ca(OH)₂ (s)
     
     
   • Zinc trioxonitrate(V): A white precipitate forms which dissolves in excess NaOH to form a colorless solution.
     
     Equations:
     
     Zn²⁺ + 2OH^- → Zn(OH)₂ (s)
     
     Zn(OH)₂ + 2OH^- → [Zn(OH)₄]^2-

   (ii) Potassium chloride vs. Potassium trioxocarbonate(IV):

   Test: Add dilute hydrochloric acid (HCl).

   • Potassium chloride: No observable reaction (neutral salt).
   
   
   • Potassium trioxocarbonate(IV): Effervescence due to carbon dioxide gas release.
     
     Equation:
     
     CO₃^2- + 2H⁺ → CO₂ (g) + H₂O (l)

   Molar heat of neutralization of a strong base by a strong acid:

   It is constant (≈ –57 kJ/mol) because strong acids and bases completely ionize in water.
   
   The net reaction is:
   
   H⁺ (aq) + OH^- (aq) → H₂O (l).
   
   This reaction is the same in all strong acid–strong base reactions, so the heat change is always the same.

   Two gases that are soluble in water:

   • Ammonia (NH₃)
   
   
   • Carbon dioxide (CO₂)

END OF PAPER