The concentration of aqueous chloride ions can be found by titration with aqueous silver nitrate, AgNO3(aq). Ag+(aq) + Cl-(aq) → AgCl(s) The indicator used is aqueous potassium chromate(VI), K2CrO4(aq). As AgNO3(aq) is added to aqueous chloride ions, a white precipitate of AgCl(s) is formed. When all the chloride ions have reacted, further addition of AgNO3(aq) leads to the formation of a red precipitate of silver chromate(VI), Ag2CrO4(s). The first appearance of the red precipitate shows the end-point of the titration. A student carries out an experiment to determine the number of molecules of water of crystallisation, x, in hydrated barium chloride, BaCl2•xH2O(s).
- (a)(i)Calculate the mass of solid silver nitrate, AgNO3(s), needed to make 250.0 cm³ of 0.0500 mol dm⁻³ AgNO3(aq). Give your answer to two decimal places.[1]
- (a)(ii)Describe how the student should make 250.0 cm³ of 0.0500 mol dm⁻³ AgNO3(aq) starting from the mass of AgNO3(s) calculated in (a)(i) in a 50 cm³ beaker. Give the name and size of any key apparatus used. Write your answer using a series of numbered steps.[3]
- (b)The student uses the following method. step 1 Dissolve 1.58g of BaCl2•xH2O(s) to form 250 cm³ of aqueous solution. Label this solution A. step 2 Transfer 20.0 cm³ of solution A into a conical flask. step 3 Add aqueous sodium sulfate, Na2SO4(aq), to the flask and swirl the mixture to remove barium ions from the solution. step 4 Add 2-3 drops of K2CrO4(aq) indicator to the flask. step 5 Titrate the contents of the flask against 0.0500 mol dm⁻³ AgNO3(aq). step 6 Repeat steps 2 to 5 to collect sufficient data for analysis.
- (b)(i)Suggest a suitable piece of apparatus for transferring 20.0 cm³ of solution A in step 2.[1]
- (b)(ii)Suggest why barium ions are removed in step 3 before performing the titration.[1]
- (b)(iii)Suggest why chemically resistant gloves should be worn to carry out step 4.[1]
- (c)The student's results are shown in Table 1.1. Table 1.1 | | rough titration | titration 1 | titration 2 | titration 3 | | :-------------------- | :------------- | :----------- | :----------- | :----------- | | burette reading (final)/cm³ | 20.10 | 40.55 | 20.75 | 20.90 | | burette reading (initial)/cm³ | 0.00 | 20.25 | 0.05 | 0.30 | | titre/cm³ | 20.10 | 20.30 | 20.70 | 20.60 | The student uses the titres from titrations 2 and 3 shown in Table 1.1 to calculate a mean titre value of 20.65 cm³.
- (c)(i)Explain why only these two values are used.[1]
- (c)(ii)Calculate the percentage error in the titre volume for titration 3. Show your working.[1]
- (d)The equation for the reaction of silver nitrate with barium chloride is shown. 2AgNO3(aq) + BaCl2(aq) → Ba(NO3)2(aq) + 2AgCl(s)
- (d)(i)Calculate the amount, in mol, of AgNO3(aq) in the mean titre of 20.65 cm³.[1]
- (d)(ii)Calculate the amount, in mol, of BaCl2(aq) in 250 cm³ of solution A.[1]
- (d)(iii)Calculate the value of x in the formula BaCl2•xH2O.[2]
- (e)Another student uses a different experimental method to check the value of x obtained by the method described in (b). Give a brief description of another method, not involving titration, that could be used to determine the value of x in the formula BaCl2•xH2O(s). Write your answer using a series of numbered steps. Your plan should include details of the following: • the apparatus and method you would use • the measurements you would make. You are provided with standard laboratory apparatus.[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.
- [1] 2.12 g
- [3] M1 add a (small) volume of (distilled) water (to the small beaker) AND dissolve the solid / AgNO3 (S)
- M2 Transfer the solution and washings (using a funnel) AND into a 250 cm³ volumetric flask.
- M3 Make up to the (calibration) mark / line with distilled water. AND then invert the (stoppered) flask (several times)
- [1] (20(.0) cm³) pipette.
- [1] barium (ions) could react with chromate (ions).
- [1] Potassium chromate(VI) / K2CrO4(aq) is an irritant / corrosive
- [1] They are (the only ones that are) concordant.
- [1] 2×0.05 / 20.60 ×100 = 0.48(54%) Correct working must be shown.
- [1] 20.65 × 0.05 / 1000 = 0.00103(25)
- [1] = ((d)(i) / 2) *(250 / 20)
- = (d)(i) x 6.25
- [2] M1 Mr = 1.58 / (d)(ii)
- M2 x= (M1 – 208.3) / 18 AND x must be an integer. Some working must be shown.
- [3] Heat to drive off water of crystallisation.
- M1 Apparatus 1 Bunsen burner. 2 Crucible (supported by pipeclay triangle)
- M2 Measurements 1 Measure mass of crucible (+ lid) (empty). 2 Measure mass crucible (+ lid) + solid before heating 3 Measure mass of crucible (+ lid) + residue (after heating.)
- M3 Key step in method • Description of heat to constant mass.
- OR
- Produce a suitable precipitate using a viable reaction.
- M1 Apparatus and method • Chooses suitable reagent. • Filter the precipitate. • Wash and (allow to) dry
- M2 Measurements • Measure mass of hydrated barium chloride used. • Measure mass of precipitate.
- M3 Key step in method • Description of dry to constant mass.
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