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Chapter 16: Problem 17

List some ways one can increase the solubility of a salt in water.

Short Answer

Expert verified
To increase the solubility of a salt in water, one can: (1) increase the temperature of the solution, (2) stir the solution, (3) grind the salt into smaller particles, (4) add a solubility-enhancing agent, and (5) adjust the pH of the solution. However, some methods may not be effective for all salts, and caution should be taken when employing these methods to avoid adverse effects on the environment or other properties of the solution.

Step by step solution

01

1. Increase the temperature of the solution

One way to increase the solubility of a salt in water is by increasing the temperature of the solution. At higher temperatures, the kinetic energy of the water molecules increases, allowing more salt to dissolve. However, note that this method may not be effective for all salts, as some salts have solubility that decreases with temperature.
02

2. Stir the solution

Stirring a solution can speed up the dissolution process and allow more salt to dissolve. This is because stirring increases the contact between the solute (salt) particles and the solvent (water) molecules, facilitating the interactions between them and promoting dissolution.
03

3. Grind the salt into smaller particles

Smaller particles of salt have a larger surface area, which means more surface for the water molecules to interact with the salt, promoting faster dissolution. Grinding the salt into smaller particles can therefore increase the solubility of a salt in water.
04

4. Add a solubility-enhancing agent

Some salts have decreased solubility in the presence of certain ions or molecules. Adding a solubility-enhancing agent, such as a chelating agent or a complexing agent, can help bind these interfering ions or molecules and increase the solubility of the salt in water. However, this method should be employed with caution, as it might have adverse effects on the environment or downstream processes.
05

5. Adjust the pH

The solubility of some salts can be influenced by pH. In such cases, adjusting the pH of the solution may increase the salt's solubility. For example, for salts containing basic anions, decreasing the pH (adding acid) can help increase the solubility as it shifts the equilibrium in favor of dissolution. Conversely, salts containing acidic cations can have increased solubility by increasing the pH (adding a base). This method should also be employed with caution, as changing the pH may cause unintended changes in other properties of the solution.

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Most popular questions from this chapter

Tooth enamel is composed of the mineral hydroxyapatite. The \(K_{\mathrm{sp}}\) of hydroxyapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\) is \(6.8 \times 10^{-37}\) . Calculate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\) , forms. The \(K_{\mathrm{sp}}\) of this substance is \(1 \times 10^{-60}\) . Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?

The concentration of Mg \(^{2+}\) in seawater is 0.052\(M .\) At what pH will 99\(\%\) of the \(\mathrm{Mg}^{2+}\) be precipitated as the hydroxide salt? \(\left[K_{\mathrm{sp}} \text { for } \mathrm{Mg}(\mathrm{OH})_{2}=8.9 \times 10^{-12} .\right]\)

The active ingredient of Pepto-Bismol is the compound bismuth subsalicylate, which undergoes the following dissociation when added to water: $$\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}^{2-}(a q) +\mathrm{Bi}^{3+}(a q)+\mathrm{OH}^{-}(a q) \qquad K=?$$ If the maximum amount of bismuth subsalicylate that reacts by this reaction is \(3.2 \times 10^{-19} \mathrm{mol} / \mathrm{L}\) , calculate the equilibrium constant for the preceding reaction.

Sodium chloride is listed in the solubility rules as a soluble compound. Therefore, the \(K_{\mathrm{sp}}\) value for \(\mathrm{NaCl}\) is infinite. Is this statement true or false? Explain.

The copper(l) ion forms a complex ion with \(\mathrm{CN}^{-}\) according to the following equation: $$\mathrm{Cu}^{+}(a q)+3 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q) \quad K=1.0 \times 10^{11}$$ a. Calculate the solubility of \(\mathrm{CuBr}(s)\left(K_{\mathrm{sp}}=1.0 \times 10^{-5}\right)\) in \(1.0 \mathrm{L}\) of \(1.0 \mathrm{M} \mathrm{NaCN}\) . b. Calculate the concentration of \(\mathrm{Br}^{-}\) at equilibrium. c. Calculate the concentration of \(\mathrm{CN}^{-}\) at equilibrium.
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