You're going to analyze a mineral sample that weighs 111.515 g for the silver (Ag, AW=107.87...


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?


By definition, the solubility of the compound is measured as its ability to dissolve in a solvent to form an aqueous solution. One of the main considerations in determining the solubility of a compound is the choice of solvent, temperature, and concentration of the solute. A solution that contains less amount of solute from its maximum is called unsaturated. A solution that contains just enough solute to reach the maximum ability of the solvent to dissolve the solute is called a saturated solution. Finally, a solution that contains solute beyond the maximum capability of the solvent to dissolve it is called supersaturated. In a supersaturated solution, a precipitate will form.

Answer and Explanation: 1

The problem above involved a sample that contains silver which was allowed to react with chloride ions. The balanced chemical reaction for this is shown below:

{eq}\rm Ag(s) + Cl^-(aq)\rightarrow AgCl(s) {/eq}

It was mentioned that it formed a solid precipitate (AgCl) with a mass of {eq}\rm 7.7\,\mu g {/eq}. Converting this to milligram:

{eq}\rm =7.7\,\mu g\left(\dfrac{1\,g}{1\times10^{6}\,\mu g}\right)\left(\dfrac{1\times10^{3}}{1\,g}\right)\\ =7.7\times10^{-3}\,mg {/eq}

A. We want to compute the mass of the silver in the sample. We do this by converting the mass obtained to moles and making use of the molar ratio from the balanced chemical equation.

  • {eq}\rm MM\,Ag=107.87\,mg/mmol\\ MM\,AgCl=143.32\,mg/mmol {/eq}

Computing for the moles of the solid silver chloride formed:

{eq}\rm moles\,AgCl=(7.7\times10^{-3}\,mg\left(\dfrac{1\,mmol}{143.32\,mg}\right)\\ moles\,AgCl=5.4\times10^{-5}\,mmol {/eq}

Now, we can use this in solving the number of moles of silver. From the balanced chemical equation, the ratio of silver to silver chloride is 1:1. Hence,

{eq}\rm moles\,AgCl=moles\,Ag=5.4\times10^{-5}\,mmol {/eq}

Converting this to mass:

{eq}\rm mass\,Ag=5.4\times10^{-5}\,mmol\left(\dfrac{107.87\,mg}{1\,mmol}\right)\\ mass\,Ag=5.8\times10^{-3}\,mg {/eq}

B. Now, we want to convert this to parts per billion. Parts-per-billion is defined as:

{eq}\rm ppb=\dfrac{mass\,of\,solute}{mass\,of\,sample}\times10^9 {/eq}

Converting the mass of silver to grams:

{eq}\rm =5.8\times10^{-3}\,mg\left(\dfrac{1\,g}{1\times10^3\,mg}\right)\\ =5.8\times10^{-6}\,g {/eq}

Solving for the ppb:

{eq}\rm ppb=\dfrac{5.8\times10^{-6}\,g}{111.515\,g}\times10^9\\ ppb=52 ppb {/eq}

C. Mercury could also co-precipitate with silver during the reaction. This is due to the fact that mercury can combine with the structure of silver. Moreover, mercury is also known to incorporate wide variety of metals in its structure including gold and silver. This might be the reason for the increase silver concentration in the solution.

Learn more about this topic:

What is Precipitation? - Definition, Types & Formation


Chapter 19 / Lesson 8

What is precipitation? Learn the definition, formation, types, and examples of precipitation. Also, learn the meaning of the term precipitate.

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