This question is about sulfuric acid, \( H_2SO_4 \).
(a) Dilute sulfuric acid and aqueous sodium hydroxide can be used to prepare sodium sulfate crystals using a method that involves titration.
The apparatus for titration is shown in Fig. 4.1.
Thymolphthalein is used as an indicator for this titration.
(i) State the colour change of thymolphthalein at the end-point of this titration.
(ii) Suggest why universal indicator is not used for this titration.
(b) 25.0 cm³ of aqueous sodium hydroxide, NaOH, of concentration 0.100 mol/dm³ is neutralised by 20.0 cm³ of dilute sulfuric acid, \( H_2SO_4 \).
The equation for the reaction is shown.
\[2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O\]
Calculate the concentration of \( H_2SO_4 \) using the following steps.
– Calculate the number of moles of NaOH used.
– Determine the number of moles of \( H_2SO_4 \) that react with the NaOH.
– Calculate the concentration of \( H_2SO_4 \).
(c) A student is provided with an aqueous solution of sodium sulfate.
Describe how to prepare a pure sample of sodium sulfate crystals from this solution.
(d) Potassium hydrogen sulfate, \( KHSO_4 \), can be prepared by a reaction between aqueous potassium hydroxide and dilute sulfuric acid. Water is the only other product.
Write a symbol equation for this reaction.
(e) Potassium hydrogen sulfate, \( KHSO_4 \), dissolves in water to form solution X.
Solution X contains \( K^+ \), \( H^+ \) and \( SO_4^{2-} \) ions.
(i) Name the type of solution that contains \( H^+ \) ions.
(ii) State the observations when the following tests are done.
A flame test is carried out on X.
- Solid copper(II) carbonate is added to X.
- Aqueous barium nitrate acidified with dilute nitric acid is added to X.
(f) 0.325 g of Zn is added to dilute sulfuric acid which contains 0.0100 moles of \( H_2SO_4 \).
The equation for this reaction is shown.
\[Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2\]
(i) Determine whether Zn or \( H_2SO_4 \) is the limiting reactant.
Explain your answer.
(ii) In another experiment, 48.0 cm³ of hydrogen gas, \( H_2 \), is produced. The experiment is carried out at room temperature and pressure, r.t.p.
Calculate the number of molecules in 48.0 cm³ of \( H_2 \) gas measured at r.t.p.
The value of the Avogadro constant is \( 6.02 \times 10^{23} \).
▶️ Answer/Explanation
(a)(i) from blue to colourless
Thymolphthalein is blue in basic solutions (pH > 9.3) and colorless in acidic solutions (pH < 8.0). The end-point occurs when the solution changes from basic to acidic.
(a)(ii) universal indicator has too many colour changes
Universal indicator shows a range of colors for different pH values, making it difficult to identify the exact end-point of the titration.
(b)
Step 1: Moles of NaOH = concentration × volume = 0.100 mol/dm³ × 0.025 dm³ = 0.0025 mol
Step 2: From the equation, 2 moles NaOH react with 1 mole \( H_2SO_4 \), so moles \( H_2SO_4 \) = 0.0025 ÷ 2 = 0.00125 mol
Step 3: Concentration \( H_2SO_4 \) = moles ÷ volume = 0.00125 mol ÷ 0.020 dm³ = 0.0625 mol/dm³
(c)
1. Heat the solution to evaporate water until saturation point is reached (when crystals start to form)
2. Allow the solution to cool slowly, allowing crystals to form
3. Filter the crystals and dry them between filter papers or in a warm oven
(d) \( KOH + H_2SO_4 \rightarrow KHSO_4 + H_2O \)
(e)(i) acid / acidic solution
(e)(ii)
● Flame test: lilac flame (characteristic of potassium ions)
● Copper(II) carbonate added: solid dissolves, bubbles/fizzing observed (from \( CO_2 \) gas), and solution turns blue (from \( Cu^{2+} \) ions)
● Barium nitrate added: white precipitate forms (barium sulfate)
(f)(i)
Moles of Zn = mass ÷ molar mass = 0.325 g ÷ 65 g/mol = 0.005 mol
From the equation, 1 mole Zn reacts with 1 mole \( H_2SO_4 \). Since there are 0.0100 moles \( H_2SO_4 \) and only 0.005 moles Zn, Zn is the limiting reactant.
(f)(ii)
At r.t.p., 1 mole of gas occupies 24 dm³ (24,000 cm³)
Moles of \( H_2 \) = 48.0 cm³ ÷ 24,000 cm³/mol = 0.002 mol
Number of molecules = moles × Avogadro’s constant = 0.002 × 6.02 × 10²³ = 1.20 × 10²¹ molecules
Silver bromide, AgBr, is made when aqueous silver ethanoate, CH3COOAg, is added to aqueous sodium bromide, NaBr.
The equation for the reaction is shown in equation 1.
CH3COOAg + NaBr → CH3COONa + AgBr
The method includes the following steps.
step 1 – Add 200.0 cm3 of 0.0500 mol/dm3 CH3COOAg to a beaker. This volume contains 0.0100 mol of Ag+ ions.
step 2 – Add 50.0 cm3 of aqueous NaBr. This volume contains 0.0100 mol of Br– ions. A precipitate forms.
step 3 – Filter the mixture.
step 4 – Dry the solid residue until all the water is removed.
step 5 – Record the mass of the dry residue.
(a) Complete the ionic equation for the reaction by adding the missing state symbols.
\[ \text{Ag}^+(……) + \text{Br}^-(……) \rightarrow \text{AgBr}(……) \]
(b) Name a different aqueous silver salt which could be used in step 1.
(c) Use the information in step 2 to calculate the concentration of aqueous NaBr.
(d) State the colour of the precipitate which forms in step 2.
(e) Use the information in step 1, step 2 and equation 1 to determine the number of moles of AgBr formed. Use this value to calculate the mass of AgBr formed.
(f) Name the salt dissolved in the filtrate in step 3.
(g) The recorded mass of the dry residue in step 5 is greater than the mass calculated in (e) because a step is missing from the procedure.
(i) Suggest the missing step.
(ii) Name the substance responsible for the greater mass of the dry residue.
(h) Barium sulfate can be made by the same method but with different aqueous solutions.
(i) Suggest two aqueous solutions which can be added together to make barium sulfate.
(ii) Write the balanced symbol equation for this reaction.
▶️ Answer/Explanation
(a) \[ \text{Ag}^+(aq) + \text{Br}^-(aq) \rightarrow \text{AgBr}(s) \]
The silver ions and bromide ions are in aqueous solution before reacting, and silver bromide forms a solid precipitate.
(b) silver nitrate
Silver nitrate is another common soluble silver salt that could provide Ag+ ions for the reaction.
(c) 0.200 mol/dm3
Calculation: concentration = moles/volume = 0.0100 mol / (50.0 cm3/1000) = 0.200 mol/dm3
(d) cream
Silver bromide forms a cream-colored precipitate, distinct from the white of silver chloride or yellow of silver iodide.
(e) number of moles of AgBr = 0.0100
mass of AgBr = 1.88 g
Calculation: The limiting reactant is either Ag+ or Br–, both at 0.0100 mol. Molar mass of AgBr = 107.9 (Ag) + 79.9 (Br) = 187.8 g/mol. Mass = 0.0100 mol × 187.8 g/mol = 1.878 g ≈ 1.88 g.
(f) sodium ethanoate
CH3COONa is the soluble byproduct of the reaction that remains in the filtrate.
(g)(i) rinsing of residue
The missing step is washing the precipitate to remove any soluble impurities.
(g)(ii) (crystals of) sodium ethanoate
The extra mass comes from sodium ethanoate that wasn’t washed away, remaining with the precipitate.
(h)(i) barium chloride and sodium sulfate
Any soluble barium salt (e.g., barium chloride, barium nitrate) can react with any soluble sulfate (e.g., sodium sulfate, sulfuric acid) to form barium sulfate precipitate.
(h)(ii) BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
This is one possible balanced equation for the precipitation reaction forming barium sulfate.