You are in the silver recovery business. You receive 55‐gal drums of waste silver solution from photography laboratories, metal plating operations, and other industrial companies. Currently, you have a drum of waste from a silver‐plating factory with the silver concentration at 1 oz per gal. (Precious metals are measured in troy ounces, which are 31.103 g.) You precipitate the silver as the chloride, which you sell for $2.00 per gram. The sodium chloride, which is the source of the chloride ion, costs $0.46 per kilogram.

Question

You are in the silver recovery business. You receive 55‐gal drums of waste silver solution from photography laboratories, metal plating operations, and other industrial companies. Currently, you have a drum of waste from a silver‐plating factory with the silver concentration at 1 oz per gal. (Precious metals are measured in troy ounces, which are 31.103 g.) You precipitate the silver as the chloride, which you sell for $2.00 per gram. The sodium chloride, which is the source of the chloride ion, costs $0.46 per kilogram.

a) Calculate the mass of sodium chloride needed to react completely with the silver present in one drum of waste.
b) If more sodium chloride is added, more silver chloride precipitates, in accordance with the common ion effect. Eventually, the value of the silver precipitated is lower than the cost of the sodium chloride added. Determine the mass of silver left in the drum after the sodium chloride (computed in part a) was added by performing a solubility product calculation. Determine the monetary value of this quantity of silver.
c) Silver is one of eight metals covered by the Resource Conservation and recovery Act (RCRA). Express the concentration of silver present after the precipitation reaction in terms of g mL–1. The legal limit for silver‐containing discharges is 5 μg mL–1. Is the residual silver concentration above or below this limit? If the silver concentration is above the limit, calculate how much additional sodium chloride must be added to one drum to use the common ion effect to reduce it to 5.0 μg mL–1.

Answer 1

A long question. How much do you know how to do? I'm not inclined to work such a long multipart question for you but here is a link that will allow you to calculate the first part.

http://www.jiskha.com/science/chemistry/stoichiometry.html

The other parts are based on Ksp for AgCl.

Answer 2

a) To calculate the mass of sodium chloride needed to react completely with the silver present in one drum of waste, you'll first need to determine the amount of silver in the drum.

Given that the silver concentration is 1 oz per gallon and there are 55 gallons in a drum, you can calculate the total amount of silver in the drum as follows:

Silver in one drum = silver concentration (oz/gal) x volume of the drum (gal)

Since we're working with troy ounces and grams, we'll need to convert the silver concentration to grams:

1 troy ounce = 31.103 grams

Therefore, the silver concentration in grams per gallon is:

1 oz/gal x 31.103 g/oz = 31.103 g/gal

Now, you can substitute the values into the equation to calculate the total amount of silver in one drum:

Silver in one drum = 31.103 g/gal x 55 gal

b) Now, let's determine the mass of silver left in the drum after the sodium chloride was added, by performing a solubility product calculation.

The solubility product constant (Ksp) for silver chloride (AgCl) can be used to determine its solubility in water. The Ksp value for AgCl is 1.8 x 10^-10.

Using the chemical equation for the precipitation reaction:

Ag+ + Cl- → AgCl

We can see that 1 mole of silver chloride is formed for every mole of silver precipitated.

The molar mass of AgCl is approximately 143.32 g/mol, and the molar mass of silver (Ag) is 107.87 g/mol.

Knowing this, we can set up an equation to determine the concentration of silver ions (Ag+) in the solution:

Ksp = [Ag+][Cl-]

Since the concentration of silver ions is equal to the concentration of silver chloride (AgCl) formed, we can simplify the equation:

Ksp = [AgCl]^2

Now, we can solve for the concentration of silver chloride (AgCl):

[AgCl] = √(Ksp)

Substituting the Ksp value into the equation:

[AgCl] = √(1.8 x 10^-10)

Next, we need to convert the concentration of silver chloride to grams per milliliter (g/mL).

The density of the waste silver solution is not given, so we'll assume its density is close to that of water, which is 1.00 g/mL.

Multiplying the concentration of silver chloride by the density of the solution will give us the concentration of silver chloride in grams per milliliter:

[AgCl] g/mL = [AgCl] x density of solution (g/mL)

Now we can calculate the mass of silver chloride precipitated:

Mass of AgCl = [AgCl] g/mL x volume of solution (mL)

c) To express the concentration of silver present after the precipitation reaction in terms of g/mL, you'll need to consider the total volume of the solution.

The concentration of silver after the precipitation reaction can be calculated by dividing the mass of silver remaining in the drum by the total volume of the solution (drum).

Finally, you can compare the residual silver concentration to the legal limit for silver-containing discharges (5 μg mL–1). If the concentration is above the limit, you'll need to calculate how much additional sodium chloride must be added to reduce it to 5.0 μg mL–1. To do this, you'll need to consider the common ion effect and determine the additional amount of chloride ions required to react with the remaining silver ions.