Zadanie 1d, Paper 1B, maj 2025, TZ2: Chemia IB HL

Zadania w arkuszu

Informacja do zadań 1a-1d.

Copper(II) ions, $\ce{Cu^2+}$

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, 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).

For the AAS method, a series of standards of known concentration was prepared by diluting a stock solution of $\ce{Cu^2+}$ ions. The instrument was zeroed with deionized water and the absorbance of each standard was recorded.

scatter graph of absorbance against concentration of Cu2+ in ppm, four calibration points, axes from 0 to 10 ppm and from 0 to 0.12

(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 $\ce{Cu^2+}$ in the sample.

(iii) The absorbance is defined as $A = \log_{10}\dfrac{I_0}{I}$ , where $I_0$

(iv) To prepare the 6.00 ppm standard, 2.50 $\mathrm{cm^3}$ of the stock solution was made up to 50.00 $\mathrm{cm^3}$ with deionized water. Calculate, in ppm, the concentration of the stock solution.

(v) Calculate the molar concentration, in $\mathrm{mol\,dm^{-3}}$ , of $\ce{Cu^2+}$ in the 6.00 ppm standard, given that 1 ppm corresponds to 1 $\mathrm{mg\,dm^{-3}}$

Sprawdzane umiejętności

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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, $A \approx 0.0125c$ , so $c = \dfrac{0.065}{0.0125} =$ » 5.2 «ppm» ✔

(iii) « $\dfrac{I}{I_0} = 10^{-0.065} =$ » 0.861 «so 86.1 % of the radiation is transmitted» ✔

«percentage absorbed = 100 − 86.1 =» 13.9 «%» ✔

(iv) « $c_1 = \dfrac{c_2 V_2}{V_1} = \dfrac{6.00 \times 50.00}{2.50} =$

(v) «6.00 $\mathrm{mg\,dm^{-3}}$ = $6.00 \times 10^{-3}\ \mathrm{g\,dm^{-3}}$

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.

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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.

.

» 120 «ppm» ✔

;

\dfrac{6.00 \times 10^{-3}}{63.55} =

»

9.44 \times 10^{-5}

«

\mathrm{mol\,dm^{-3}}

» ✔