Question

9) In a study of the poisoning of fish, 0.500 g of samples of fish tissue were analysed for barium. Each sample was digested with a permanganate-sulfuric acid mixture, and the volume adjusted to 10 mL and to construct calibration curve, 0.500-g samples of uncontaminated fish tissue were injected with various amounts of Ba+2, and corresponding absorbance as follows:

Standard 0: 0.00μg Ba+2; 0.015
Standard 1: 50 μg Ba+2; 0.220
Standard 2: 100 μg Ba+2; 0.415
Standard 3: 150 μg Ba+2; 0.602


If all samples and standards were treated in identical manner, what was the concentration of barium (in μg Ba per g of fish) in a fish sample whose absorbance value was 0.421 and 0.378, respectively?

Answer 1

#65320 Chemistry, Other

9) In a study of the poisoning of fish, $0.500\mathrm{g}$ of samples of fish tissue were analysed for barium. Each sample was digested with a permanganate-sulfuric acid mixture, and the volume adjusted to $10~\mathrm{mL}$ and to construct calibration curve, $0.500\mathrm{-g}$ samples of uncontaminated fish tissue were injected with various amounts of $\mathrm{Ba}^{+2}$ , and corresponding absorbance as follows:

Standard 0: $0.00\mu \mathrm{g}\mathrm{Ba}^{+2}$ ; 0.015

Standard 1: $50\mu \mathrm{g}\mathrm{Ba}^{+2}$ ; 0.220

Standard 2: $100\mu \mathrm{g}\mathrm{Ba}^{+2}$ ; 0.415

Standard 3: $150\mu \mathrm{g}\mathrm{Ba}^{+2}$ ; 0.602

If all samples and standards were treated in identical manner, what was the concentration of barium (in $\mu$ g Ba per g of fish) in a fish sample whose absorbance value was 0.421 and 0.378, respectively?

Answer:

From the data provided calibration curve must be built.

Figure 1 - Calibration curve for $\mathrm{Ba}^{+2}$ test

According to this curve, absorbance value 0.421 corresponds to $\mathrm{Ba}^{+2}$ concentration $100\mu \mathrm{g}$ . That is why, $\mathrm{Ba}^{+2}$ concentration in the fish sample is: $100 / 0.500 = 200\mu \mathrm{g} / \mathrm{g}$ of fish tissue.

Absorbance value 0.378 corresponds to $\mathrm{Ba}^{+2}$ concentration $86\mu \mathrm{g}$ .

That is why, $\mathrm{Ba}^{+2}$ concentration in the fish sample is: $86 / 0.500 = 172\mu \mathrm{g} / \mathrm{g}$ of fish tissue.

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