Zadanie 1d, Paper 1B, maj 2025, TZ2, Chemia IB HL
Wzorowane na: Chemistry HL Paper 1B, May 2025, TZ2, pytanie 1(d). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
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Copper(II) ions, Cu2+, in the rinse water from a copper-plating workshop must be monitored before the water is discharged. Their concentration can be determined either by complexometric titration with a standard solution of EDTA or by atomic absorption spectroscopy (AAS).
Polecenie
For the AAS method, a series of standards of known concentration was prepared by diluting a stock solution of Cu2+ ions. The instrument was zeroed with deionized water and the absorbance of each standard was recorded.
Powiększ(i) Draw the line of best fit for these calibration points on the graph.
(ii) A sample of the rinse water gave an absorbance of 0.065. Determine, in ppm, the concentration of Cu2+ in the sample.
(iii) The absorbance is defined as , where I0 is the intensity of the radiation entering the sample and I the intensity leaving it. Determine what percentage of the incident radiation is absorbed by this sample.
(iv) To prepare the 6.00 ppm standard, 2.50 cm3 of the stock solution was made up to 50.00 cm3 with deionized water. Calculate, in ppm, the concentration of the stock solution.
(v) Calculate the molar concentration, in mol dm−3, of Cu2+ in the 6.00 ppm standard, given that 1 ppm corresponds to 1 mg dm−3.
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Pokaż rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) straight line of best fit passing through/close to the origin with the points evenly spread on either side of it ✔
(ii) «from the line of best fit, A ≈ 0.0125c, so » 5.2 «ppm» ✔
(iii) «» 0.861 «so 86.1 % of the radiation is transmitted» ✔
«percentage absorbed = 100 − 86.1 =» 13.9 «%» ✔
(iv) «» 120 «ppm» ✔
(v) «6.00 mg dm−3 = 6.00 × 10−3 g dm−3; » 9.44 × 10−5 «mol dm−3» ✔
Schemat punktowania
(i) [1] for a single straight line with a balanced spread of points; a dot-to-dot line scores 0. (ii) [1] for a value in the range 5.0 to 5.4 ppm read from the candidate's line. (iii) M1: I/I0 = 0.861 or I0/I = 1.16; M2: 13.9 % (accept 13.8 to 14.0 %); award [2] for the correct final answer; 6.5 % scores 0. (iv) [1]. (v) [1]; ECF for a wrong power of ten only if the working is shown.
Komentarz. The absorbance is a logarithm, so an absorbance of 0.065 does not mean that 6.5 % of the light is absorbed; the fraction transmitted is 10 to the power of minus A.