An investigation is done to find the percentage of copper in a brass nail. The brass nail is dissolved in concentrated nitric acid and the resulting solution is diluted with distilled water. Cu²⁺ ions are formed in this reaction.
- (a)Use the electrode potentials shown to write a balanced ionic equation for the reaction between the copper in the brass nail and the concentrated nitric acid. equation electrode potential, E*/V 2H⁺(aq) + 2e⁻ ⇌ H₂(g) 0.00 Cu²⁺(aq) + e⁻ ⇌ Cu⁺(aq) +0.15 Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) +0.34 NO₃⁻(aq) + 2H⁺(aq) + e⁻ ⇌ NO₂(g) + H₂O(l) +0.81[2]
- (b)A balance accurate to two decimal places was used to weigh a small beaker and the mass of the beaker recorded. The brass nail was placed in the beaker and the mass increased by 3.76 g. Calculate the percentage error in measuring the mass of this nail. percentage error = %[1]
- (c)50 cm³ of concentrated nitric acid was added to the nail in the beaker. When the reaction finished and the nail had dissolved, the solution was completely transferred to a 250.0 cm³ volumetric flask along with the washings. The solution was made up to the mark with distilled water. Concentrated nitric acid is corrosive. State one precaution related to this hazard that must be taken when using concentrated nitric acid.[1]
- (d)(i)A colorimeter can be used to determine concentrations of metal ions in solution. When light passes through solutions of metal ions some of the light may be absorbed. The quantity of light absorbed is called the absorbance and is measured by a colorimeter. The absorbance of solutions of known concentration of copper(II) ions can be used to determine an unknown concentration of copper(II) ions. A solution of copper(II) nitrate of concentration 0.800 mol dm⁻³ is required for this investigation. Calculate the mass of copper(II) nitrate, Cu(NO3)2, needed to prepare 100.0 cm³ of 0.800 mol dm⁻³ solution. Give your answer to three significant figures. [A₁: Cu, 63.5; N, 14.0; Ο, 16.0] mass of copper(II) nitrate, Cu(NO3)2 = g[2]
- (d)(ii)Volumes of 0.800 mol dm⁻³ copper(II) nitrate are diluted with distilled water to prepare a series of ten solutions of different concentrations of copper(II) ions. The total volume needed of each solution is 20.00 cm³. The absorbance of each solution is recorded in a colorimeter at a wavelength of light of 630 nm and recorded in the table. Complete the table to show the volumes of 0.800 mol dm⁻³ copper(II) nitrate and distilled water needed to prepare each solution. Give all volumes to two decimal places. solution volume of 0.800 mol dm-3 volume of concentration absorbance copper(II) nitrate/cm³ distilled water/cm³ of Cu²+(aq)/mol dm-3 1 0.00 20.00 0.00 0.00 2 0.04 0.12 3 0.12 0.36 4 0.16 0.48 5 0.22 0.66 6 0.30 0.89 7 0.36 1.06 8 0.42 1.25 9 0.48 1.41 10 0.60 1.76[1]
- (d)(iii)Name a suitable piece of apparatus for accurately measuring the volumes you have calculated in (d)(ii).[1]
- (e)(i)The plot produced by comparing the absorbance of each solution with its concentration is referred to as a calibration graph. Plot a calibration graph of absorbance (y-axis) against concentration of Cu²+(aq) (x-axis). Use a cross (x) to plot each data point. Draw a line of best fit. [Figure X.X][2]
- (e)(ii)State the relationship between concentration of Cu2+(aq) and absorbance.[1]
- (f)(i)The absorbance of the solution prepared in (c) from the brass nail was found to be 0.56. Use your calibration graph to find the concentration of Cu²+(aq) in this solution. concentration of Cu²+(aq) = moldm-³[1]
- (f)(ii)Use your answer to (f)(i) and the information in (b) and (c) to calculate the percentage of copper, by mass, in the brass nail. Give your answer to three significant figures. (If you were unable to obtain an answer to (f)(i) you may use the value 0.22 moldm⁻³. This is not the correct value.) [A₁: Cu, 63.5] percentage of copper in the brass nail = %[2]
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Mark scheme
Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.
- [1] M1 for correct products
- [1] M2 for balanced correct equation
- [1] (2 × 0.005)/ 3.76 × 100 = 0.27%
- [1] Wear gloves/goggles
- [1] M1 Mr = 187.5 and n = 100 × 0.8 /1000 = 0.08
- [1] M2 mass = 0.08 × 187.5 = 15.0 g
- [1] All values recorded to 2 DP
- 0.00, 20.00, 0.00, 0.00
- 1.00, 19.00, 0.04, 0.12
- 3.00, 17.00, 0.12, 0.36
- 4.00, 16.00, 0.16, 0.48
- 5.50, 14.50, 0.22, 0.66
- 7.50, 12.50, 0.30, 0.89
- 9.00, 11.00, 0.36, 1.06
- 10.50, 9.50, 0.42, 1.25
- 12.00, 8.00, 0.48, 1.41
- 15.00, 5.00, 0.60, 1.76
- [1] Burette
- [1] M1 Plots
- [1] M2 Line
- [1] Directly proportional
- [1] 0.19 mol dm-3
- [1] M1 0.19 × 250 / 1000 × 63.5 = 3.01625
- [1] M2 3.01625/3.76 × 100 = 80.2%
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