This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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here are the solutions to Exercise 4 and Exercise 5. EXERCISE 4 a) (i) The piece of apparatus used for measuring the portion of Na_2CO_3 solution is a pipette. a) (ii) Three errors in the result presented are: The calculated volume for the rough titration (24.00 cm^3 - 0.00 cm^3 = 24.00 cm^3) does not match the recorded volume (24.10 cm^3). The second titration reading (22.5 cm^3) lacks precision; burette readings should be recorded to two decimal places (e.g., 22.50 cm^3). The average titre value calculation includes values that are not concordant (e.g., 22.75 cm^3 and 22.40 cm^3 differ by more than 0.10 cm^3). Only concordant values should be averaged. b) (i) The colour of phenolphthalein in the Na_2CO_3 solution (which is basic) is pink. b) (ii) Yes, phenolphthalein is a suitable indicator for the first equivalence point of this titration. Reason: Na_2CO_3 is a weak base, and HCl is a strong acid. Phenolphthalein changes colour in the pH range of 8.2-10.0, which corresponds to the pH at the first equivalence point where Na_2CO_3 is converted to NaHCO_3. c) (i) The acid would turn blue litmus paper red. c) (ii) After the burette had been washed and rinsed with water, the final cleaning step should be to rinse it with the acid (HCl) to be used before filling it. c) (iii) The type of reaction demonstrated by the experiment is an acid-base neutralization reaction. c) (iv) The apparatus used to measure the volume of: HCl solution is a burette. Na_2CO_3 solution is a pipette. EXERCISE 5 a) (i) Concentration of A in mol/dm^3 Step 1: Calculate the molar mass of NaHSO_4. Molar mass of NaHSO_4 = Na + H + S + (4 × O) = 23 g/mol + 1 g/mol + 32 g/mol + (4 × 16 g/mol) = 23 g/mol + 1 g/mol + 32 g/mol + 64 g/mol = 120 g/mol Step 2: Convert the mass concentration to molar concentration. Concentration of A (C_A) = Mass concentrationMolar mass C_A = 10.6 g/dm^3120 g/mol C_A = 0.08833 mol/dm^3 The concentration of A is 0.0883 mol/dm^3. a) (ii) Concentration of B in mol/dm^3 Step 1: Write down the balanced chemical equation and identify the mole ratio. NaHSO_4(aq) + NaOH(aq) Na_2SO_4(aq) + H_2O(l) From the equation, the mole ratio of NaHSO_4 to NaOH is 1:1. So, n_A = 1 and n_B = 1. Step 2: List the given values. Concentration of A (C_A) = 0.08833 mol/dm^3 (from part a(i)) Volume of A (V_A) = 22.50 cm^3 Volume of B (V_B) = 25.0 cm^3 Concentration of B (C_B) = ? Step 3: Use the titration formula to calculate C_B. C_AV_An_A = C_BV_Bn_B C_B = C_AV_An_BV_Bn_A C_B = (0.08833 mol/dm^3) × (22.50 cm^3) × 1(25.0 cm^3) × 1 C_B = 1.98742525.0 C_B = 0.079497 mol/dm^3 The concentration of B is 0.0795 mol/dm^3. Drop the next question!