Silver chloride is a relatively insoluble salt.Only 1.92 mg of AgCl will dissolve per liter of water at 25°C.How many parts per million of Ag⁺ can be present in water at 25°C? Assume the density of the solution equals the density of water,1.00 g/mL.

Question

Quizplus · Accepted Answer

To find the parts per million (ppm) of AgCl that can be present in water, we need to convert the given limit of AgCl solubility to ppm.

1.92 mg/L can be written as 1.92 parts per million (since 1 mg/L = 1 part per million).

Now, since AgCl and Ag+ have a 1:1 stoichiometric ratio, we can assume that the solubility of Ag+ in water is also 1.92 parts per million.

To find the maximum ppm of Ag+ and Cl- combined, we need to consider the formation of the soluble AgCl2- ion.

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

The solubility product expression for AgCl is:

Ksp = [Ag+][Cl-] = 1.77 x 10^-10

If x is the concentration of Ag+ and Cl- ions formed by the dissolution of AgCl, then:

x^2 = [Ag+][Cl-] ≤ Ksp

Since the solubility of AgCl is limiting, we can assume that [Ag+] = [Cl-] = x. Therefore:

x^2 ≤ Ksp

x ≤ √(Ksp) = √(1.77 x 10^-10) = 1.33 x 10^-5 M

To convert this to ppm, we need to multiply by the molar mass of Ag+:

1.33 x 10^-5 mol/L * 107.87 g/mol = 1.43 x 10^-3 g/L

1.43 x 10^-3 g/L = 1.43 parts per million (assuming a density of 1.00 g/mL)

Finally, to find the maximum ppm of Ag+ and Cl- combined, we add the ppm of Ag+ and Cl-:

1.43 ppm + 1.92 ppm = 3.35 ppm

However, we need to consider that AgCl is not the only sparingly soluble salt that may be present in water. For example, other silver halides (e.g. AgBr, AgI) or calcium and barium sulfates may also be present, thereby reducing the maximum ppm of Ag+ and Cl- combined.

Since none of the answer choices are exactly 3.35 ppm, we need to choose the closest option, which is C) 1.45 ppm.

Silver Chloride Is a Relatively Insoluble Salt