Question

A concentration of \(10-100\) parts per billion (by mass) of \(\mathrm{Ag}^{+}\) is an effective disinfectant in swimming pools. However, if the concentration exceeds this range, the \(\mathrm{Ag}^{+}\) can cause adverse health effects. One way to maintain an appropriate concentration of \(\mathrm{Ag}^{+}\) is to add a slightly soluble salt to the pool. Using \(K_{s p}\) values from Appendix \(D\), calculate the equilibrium concentration of \(\mathrm{Ag}^{+}\) in parts per billion that would exist in equilibrium with (a) \(\mathrm{AgCl},(\mathbf{b})\) \(\mathrm{AgBr},(\mathrm{c}) \mathrm{AgI}\)

Short answer

The equilibrium concentration of \(\mathrm{Ag}^{+}\) in parts per billion that exists when in equilibrium are: (a) for \(\mathrm{AgCl}\) is too high (approx. \(1.33 \times 10^{-5}\, \frac{mol}{L}\)), (b) for \(\mathrm{AgBr}\) falls within the desired range (approx. \(7.07 \times 10^{-7}\, \frac{mol}{L}\)) and (c) for \(\mathrm{AgI}\) is too low (approx. \(2.92 \times 10^{-9}\, \frac{mol}{L}\)). So, adding \(\mathrm{AgBr}\) to the swimming pool would maintain an appropriate concentration of \(\mathrm{Ag}^{+}\).

Step-by-step solution

Convert the concentration range to molar units

First, we need to convert the \(\mathrm{Ag}^{+}\) concentration range from parts per billion (ppb) to molar units (mol/L). To do this, we can use the following conversion factors: - 1 ppb = 1 µg/L (micrograms per liter) - To obtain the molar concentration, we'll divide by the molar mass of \(\mathrm{Ag}^{+}\), which is 107.87 g/mol. Thus, we have: \(10 \, \text{ppb} = 10 \times 10^{-6} \, \frac{g}{L} = 10 \times 10^{-6} \times \frac{1}{107.87} \, \frac{mol}{L} \approx 9.27 \times 10^{-8} \, \frac{mol}{L} \) \(100 \, \text{ppb} = 100 \times 10^{-6} \, \frac{g}{L} = 100 \times 10^{-6} \times \frac{1}{107.87} \, \frac{mol}{L} \approx 9.27 \times 10^{-7} \, \frac{mol}{L} \) Therefore, the concentration range of \(\mathrm{Ag}^+\) for disinfection is approximately from \(9.27 \times10^{-8} \, \frac{mol}{L}\) to \(9.27 \times 10^{-7} \, \frac{mol}{L}\).

Write the balanced equations and \(K_{sp}\) expressions for the salts

For each salt, we'll write the balanced dissolution equation and the corresponding \(K_{sp}\) expression: a) \(\mathrm{AgCl:}\) \(\mathrm{AgCl (s) \rightleftharpoons Ag^{+} (aq) + Cl^{-} (aq)}\) \(K_{sp}(\mathrm{AgCl}) = [\mathrm{Ag^{+}}][\mathrm{Cl^{-}}]\) b) \(\mathrm{AgBr:}\) \(\mathrm{AgBr (s) \rightleftharpoons Ag^{+} (aq) + Br^{-} (aq)}\) \(K_{sp}(\mathrm{AgBr}) = [\mathrm{Ag^{+}}][\mathrm{Br^{-}}]\) c) \(\mathrm{AgI:}\) \(\mathrm{AgI (s) \rightleftharpoons Ag^{+} (aq) + I^{-} (aq)}\) \(K_{sp}(\mathrm{AgI}) = [\mathrm{Ag^{+}}][\mathrm{I^{-}}]\)

Use the given \(K_{sp}\) values to find the equilibrium concentrations of \(\mathrm{Ag}^{+}\)

Given the solubility product constant, \(K_{sp}\) values, we can find the equilibrium concentration of \(\mathrm{Ag}^{+}\) in aqueous solutions for each salt. In each case, we'll have equal concentrations of \(\mathrm{Ag^{+}}\) and corresponding anions, so we can say that \(K_{sp} = x^2\), where \(x\) is the equilibrium concentration of \(\mathrm{Ag}^{+}\). a) For \(\mathrm{AgCl}\) \(K_{sp}(\mathrm{AgCl}) = 1.77 \times 10^{-10}\) \(1.77 \times 10^{-10} = x^2\) \(x = \sqrt{1.77 \times 10^{-10}}\) \(x \approx 1.33 \times 10^{-5}\, \frac{mol}{L}\) b) For \(\mathrm{AgBr}\) \(K_{sp}(\mathrm{AgBr}) = 5.0 \times 10^{-13}\) \(5.0 \times 10^{-13} = x^2\) \(x = \sqrt{5.0 \times 10^{-13}}\) \(x \approx 7.07 \times 10^{-7}\, \frac{mol}{L}\) c) For \(\mathrm{AgI}\) \(K_{sp}(\mathrm{AgI}) = 8.51 \times 10^{-17}\) \(8.51 \times 10^{-17} = x^2\) \(x = \sqrt{8.51 \times 10^{-17}}\) \(x \approx 2.92 \times 10^{-9}\, \frac{mol}{L}\)

Compare the resulting equilibrium concentrations with the desired range

Now, we must compare the equilibrium concentrations of \(\mathrm{Ag}^{+}\) with the range needed for effective disinfection in swimming pools. a) \(1.33 \times 10^{-5}\, \frac{mol}{L}\) (from \(\mathrm{AgCl}\)) is too high. b) \(7.07 \times 10^{-7}\, \frac{mol}{L}\) (from \(\mathrm{AgBr}\)) falls within the range (from \(9.27 \times 10^{-8}\, \frac{mol}{L}\) to \(9.27 \times 10^{-7}\, \frac{mol}{L}\)). c) \(2.92 \times 10^{-9}\, \frac{mol}{L}\) (from \(\mathrm{AgI}\)) is too low. Hence, the equilibrium concentration of \(\mathrm{Ag}^{+}\) in parts per billion that exists when in equilibrium with (a) \(\mathrm{AgCl}\) is too high, (b) \(\mathrm{AgBr}\) is within the desired range, and (c) \(\mathrm{AgI}\) is too low. Adding \(\mathrm{AgBr}\) to the swimming pool would maintain an appropriate concentration of \(\mathrm{Ag}^{+}\).

Key concepts

Equilibrium Concentration

In the fascinating world of chemistry, the concept of equilibrium concentration relates to the status a reversible chemical reaction achieves when the rate of the forward reaction equals that of the reverse reaction. In this state of dynamic equilibrium, the concentrations of reactants and products remain constant over time, even though the reactions continue to proceed.

For instance, when a slightly soluble salt dissolves in water, it establishes a solubility equilibrium. At this point, the amount of salt dissolving is equal to the amount of salt re-forming from its ions in solution. Equilibrium concentration is represented using square brackets around the chemical symbol or formula, i.e., [Ag⁺], to indicate the molarity of a solute in a solution at equilibrium.

Solubility Equilibrium

Diving deeper into the pool of knowledge, we encounter the process of solubility equilibrium. This is a specific type of chemical equilibrium that occurs when a solute's dissolution and precipitation occur at the same rate. The solubility product constant (K_sp) is a key player here, representing the extent to which a compound will dissolve in water.

The K_sp value is unique for every solid and is temperature-dependent. In our swimming pool scenario with silver salts, the K_sp tells us how much silver ion, Ag⁺, will be in solution when the salt is at equilibrium with its undissolved form.

Parts Per Billion (ppb)

When we jump into the metric system, we meet a unit of measure known as parts per billion (ppb). This unit is akin to a drop of water in a swimming pool and is used to quantify exceedingly small concentrations of a substance. One part per billion signifies one part of a solute per billion parts of the solution, typically by mass or volume.

It's akin to finding a precious locket at the bottom of a vast lake. Relating this to our silver ion example, if a swimming pool sanitizer level is measured in ppb, it reflects the miniscule amount of silver ions dispersed in the water. Maintaining the delicate balance of this low concentration ensures effective disinfection without any adverse effects.

Molar Mass

Now, let's talk about molar mass, the atomic weightliner of chemistry. This value represents the mass of one mole of a given substance and is typically measured in grams per mole (g/mol). It's like the unique ID for each chemical compound, acting as a bridge between the microscopic world of atoms and the macroscopic world where we make observations.

Molar mass is pivotal when converting between mass and moles, as seen with the Ag⁺ ion in our swimming pool problem. The molar mass of silver ions helps us calculate the equilibrium concentrations of silver in the water, which is an essential step when determining which silver salt to use to achieve the desired disinfectant effect.