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Chapter 17: Problem 117

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) AgCl, (b) AgBr, (c) AgI.

Short Answer

Expert verified
The equilibrium concentrations of Ag⁺ in parts per billion (ppb) when in contact with each slightly soluble salt are: (a) AgCl: 13,400 ppb, (b) AgBr: 707 ppb, and (c) AgI: 3.87 ppb.

Step by step solution

01

Write the solubility equilibria for the salts

We first need to write the solubility equilibria for the slightly soluble salts AgCl, AgBr, and AgI. Each equation consists of the dissociation of the salt into its constituent ions: \(AgCl_{(s)} ⇌ Ag^+_{(aq)} + Cl^-_{(aq)}\) \(AgBr_{(s)} ⇌ Ag^+_{(aq)} + Br^-_{(aq)}\) \(AgI_{(s)} ⇌ Ag^+_{(aq)} + I^-_{(aq)}\)
02

Write the solubility product expressions

Next, we write the solubility product expressions based on the solubility equilibria. The solubility product (Ksp) is the equilibrium constant for a solid substance dissolving in an aqueous solution. For each salt, Ksp is equal to the product of ion concentrations raised to their stoichiometric coefficients: \(K_{sp}(AgCl) = [Ag^+][Cl^-]\) \(K_{sp}(AgBr) = [Ag^+][Br^-]\) \(K_{sp}(AgI) = [Ag^+][I^-]\)
03

Use Ksp values to find the concentration of Ag⁺

Using the Ksp values for AgCl, AgBr, and AgI (found in Appendix D), we can determine the concentration of Ag⁺ in equilibrium with these salts. Let x be equal to the concentration of Ag⁺ in the aqueous solution. Since each salt is a 1:1 compound, the concentration of the corresponding anion (Cl⁻, Br⁻, or I⁻) must also be equal to x. Therefore, we can write the Ksp expressions as: \(K_{sp}(AgCl) = x^2\) \(K_{sp}(AgBr) = x^2\) \(K_{sp}(AgI) = x^2\) We can now solve for x (concentration of Ag⁺) for each salt using their respective Ksp values: (a) \(AgCl: K_{sp} = 1.8 × 10^{-10} = x^2\) (b) \(AgBr: K_{sp} = 5.0 × 10^{-13} = x^2\) (c) \(AgI: K_{sp} = 1.5 × 10^{-16} = x^2\)
04

Calculate the concentration of Ag⁺ in ppb

After solving for x in each case, we can convert the Ag⁺ concentration from moles per liter (M) to parts per billion (ppb) by using the following conversion factor: \(1\:M = 10^9\:ppb\) (a) For AgCl: \(x = \sqrt{1.8 × 10^{-10}} = 1.34 × 10^{-5}\: M\) \(1.34 × 10^{-5}\: M × 10^9\: ppb = 13,400\: ppb\) (b) For AgBr: \(x = \sqrt{5.0 × 10^{-13}} = 7.07 × 10^{-7}\: M\) \(7.07 × 10^{-7}\: M × 10^9\: ppb = 707\: ppb\) (c) For AgI: \(x = \sqrt{1.5 × 10^{-16}} = 3.87 × 10^{-9}\: M\) \(3.87 × 10^{-9}\: M × 10^9\: ppb = 3.87\: ppb\)
05

Present the final results

Finally, we can present the equilibrium concentrations of Ag⁺ in parts per billion when in contact with each salt: (a) AgCl: 13,400 ppb (b) AgBr: 707 ppb (c) AgI: 3.87 ppb

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Most popular questions from this chapter

Suppose that a 10 -mL sample of a solution is to be tested for \(\mathrm{I}^{-}\) ion by addition of 1 drop \((0.2 \mathrm{~mL})\) of $0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ What is the minimum number of grams of \(\mathrm{I}^{-}\) that must be present for \(\mathrm{PbI}_{2}(s)\) to form?

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