The element Se, dispersed in 2 ml sample of detergent for dandruff control, was determined by suspending it in the warm ammoniacal solution that contained 45 ml of 0.02 M-AgNO₃. 6Ag⁺ + 3Se(s) + 6NH₃ +

Question

The element Se, dispersed in 2 ml sample of detergent for dandruff control, was determined by suspending it in the warm ammoniacal solution that contained 45 ml of 0.02 M-AgNO₃. 6Ag⁺ + 3Se(s) + 6NH₃ + 3H₂O $\to$ 2Ag₂Se(s) + Ag₂SeO₃(s) + 6NH₄⁺. The mixture was now treated with excess nitric acid which dissolved the Ag₂SeO₃ but not Ag₂Se. The Ag⁺ from the Ag₂SeO₃ and the excess of AgNO₃ consumed 10 ml of 0.01 N-KSCN in Volhard titration. How many milligrams of Se was contained per ml of sample? (Se = 80)

Answer:

24.00

Solution:

AgNO₃ taken = 45×0.02 = 0.9 mmol. For n mmol Se, Ag locked in Ag₂Se = 4n/3 mmol (per 3Se $\to$ 2Ag₂Se). The remaining Ag (excess + 2n/3 from Ag₂SeO₃) = 10×0.01 = 0.1 mmol. So 0.9 = 4n/3 + 0.1, giving n = 0.6 mmol Se = 0.6×80 = 48 mg in 2 ml. Per ml = 24 mg. Ans 0024.

TestHub · Accepted Answer

AgNO₃ taken = 45×0.02 = 0.9 mmol. For n mmol Se, Ag locked in Ag₂Se = 4n/3 mmol (per 3Se → 2Ag₂Se). The remaining Ag (excess + 2n/3 from Ag₂SeO₃) = 10×0.01 = 0.1 mmol. So 0.9 = 4n/3 + 0.1, giving n = 0.6 mmol Se = 0.6×80 = 48 mg in 2 ml. Per ml = 24 mg. Ans 0024.

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