You’re going to analyze a mineral sample that weighs 111.515 g for the silver (A

Question

You’re going to analyze a mineral sample that weighs 111.515 g for the silver (Ag, AW=107.87 g/mol) content. You dissolve the mineral sample in acid solution. To that solution, you add excess chloride ion to form the solid precipitate, silver chloride (AgCl, FW=143.32 g/mol). After drying, the solid precipitate obtained weighs 7.7 ug. A) What mass of silver was present in the mineral sample? B) What was the concentration of silver in the mineral sample in parts-per-billion? C) A subsequent analysis on a similar mineral sample showed that a significant amount of mercury was present. Mercury also precipitates with chloride from solution. If mercury caused the method to detect increased “silver” concentration, what term would describe that phenomenon and how might you solve it?

Explanation / Answer

Ans.

A). Calculation of Mass of silver present in the mineral sample

Weight of sample = 111.515 g

AW of Ag= 107.87 g/mol

Weight of silver chloride obtained = 7.7 ug or 7.7 x 10-6gm or 7.7 ppm

Mass of silver present in the mineral sample = 143.32/ 107.87x 7.7 x 10-6gm

= 1.02305 x 10-6gm or 1.02305 ppm

B) The concentration of silver in the mineral sample in parts-per-billion is 1.02305 x 1000= 1023.05 ppb.

C) Silver get amalgamated with Mercury and Mercury act as a good collector for Gold and silver. Subsequently the concentration of Silver may increase.

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