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

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.