Question

To determine the concentration of chloride ion (Cl-) in a 100 mL sample of ground water, a chemist adds a large enough volume of AgNO3 solution to precipitate all Cl- as AgCl. The mass of the resulting precipitate is 93.9 mg. What is the chloride ion concentration in milligrams of chloride per liter of ground water

Answer 1

Answer:

$[Cl^-]=232.3\frac{mgCl^-}{L}$

Explanation:

Hello,

In this case, we can represent the chemical reaction as:

$Cl^-(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NO_3^-(aq)$

In such a way, since the mass of the obtained silver chloride is 93.9 mg, we can compute the chloride ions in the ground water by using the following stoichiometric procedure whereas the molar mass of chloride ions and silver chloride are 35.45 g/mol and 143.32 g/mol respectively:

$m_{Cl^-}=93.3mgAgCl*\frac{1mmolAgCl}{143.32mgAgCl}*\frac{1mmolCl^-}{1mmolAgCl} *\frac{35.45mgCl^-}{1mmolCl^-} =23.23mgCl^-$

Finally, for the given volume of water in liters (0.100L), we compute the required concentration:

$[Cl^-]=\frac{23.2mgCl^-}{0.100L}$

$[Cl^-]=232.3\frac{mgCl^-}{L}$

Best regards.