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You are browsing through the Handbook of Hypothetical Chemistry when you come across a solid that is reported to have a \(K_{\text {sp }}\) value of zero in water at \(25^{\circ} \mathrm{C}\). What does this mean?

Short Answer

Expert verified
A \(K_{\text{sp}}\) value of zero for a solid in water at \(25^{\circ} \mathrm{C}\) means that the compound is completely insoluble and does not dissolve in water at all, as its ion concentrations in the solution would be zero.

Step by step solution

01

Understanding Solubility Product Constant

The solubility product constant, \(K_{\text{sp}}\), is an equilibrium constant that describes the degree to which a solid compound dissolves in water. It is specifically used for sparingly soluble salts and represents the product of the concentrations of the ions of the salt in its saturated solution, each raised to the power of their stoichiometric coefficients. For a general equation of a sparingly soluble salt, \(\text{A}_{\text{x}}\text{B}_{\text{y}} \rightleftharpoons \text{x A}^{\text{+}} + \text{y B}^{\text{-}}\), the \(K_{\text{sp}}\) can be represented as the following expression: \[K_{\text{sp}} = [\text{A}^+]^{\text{x}} [\text{B}^-]^{\text{y}}\] A higher value of \(K_{\text{sp}}\) indicates that more of the solid will dissolve in water, while a lower value indicates that less of the solid will dissolve.
02

Interpreting \(K_{\text{sp}}\) Value of Zero

Given that the solid has a \(K_{\text{sp}}\) value of zero in water at \(25^{\circ} \mathrm{C}\), this means that the solid does not dissolve in water at all. The concentrations of the ions in the solution would be zero, indicating that the solid is completely insoluble in water.
03

Conclusion

In conclusion, a \(K_{\text{sp}}\) value of zero for a solid means that the compound does not dissolve in water at all and is considered insoluble at \(25^{\circ} \mathrm{C}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
The equilibrium constant is a fundamental concept in chemistry that represents the balance between the reactants and products in a closed system at equilibrium. It reflects how far the reaction proceeds before reaching this state. For a general chemical reaction:\[\text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD}\]The equilibrium constant, denoted as \(K\), is calculated using the concentrations of the products and reactants raised to the power of their stoichiometric coefficients:\[K = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}\]This value is specific to each reaction and only depends on temperature. The equilibrium constant tells us about the ratio of the concentrations of products to reactants at equilibrium. If \(K\) is large, the position of equilibrium favors the formation of products. Conversely, if \(K\) is small, the equilibrium favors the reactants. This concept is crucial for understanding the behavior of chemicals in solution, including those involving sparingly soluble salts.
Sparingly Soluble Salts
Sparingly soluble salts are compounds that do not readily dissolve in water. These salts produce only a small amount of dissolved ions in solution. The solubility product constant, \(K_{\text{sp}}\), is used to describe the extent to which a sparingly soluble salt can dissolve in water.When such a salt dissolves, it exists in equilibrium between its solid form and the ions in solution. Consider a general salt \(\text{A}_x\text{B}_y\), which partially dissociates in water:\[\text{A}_x\text{B}_y \rightleftharpoons x \text{A}^+ + y \text{B}^-\]The \(K_{\text{sp}}\) is the product of the concentrations of its ions:\[K_{\text{sp}} = [\text{A}^+]^x[\text{B}^-]^y\]Only salts with very low \(K_{\text{sp}}\) are considered sparingly soluble. These salts have minimal impact on the concentration of ions in water. Understanding the \(K_{\text{sp}}\) helps predict the conditions under which a salt may precipitate from a solution or remain dissolved.
Insolubility in Water
Insolubility in water describes the inability of a substance to dissolve and form a homogeneous solution. In the context of the solubility product constant \(K_{\text{sp}}\), a value of zero indicates absolute insolubility. This means that the compound does not dissolve, and no ions are present in the water from the salt.When a compound is completely insoluble:
  • The concentration of ions in the solution is effectively zero.
  • The solid remains intact without dissociation.
In a hypothetical situation where a compound reports a \(K_{\text{sp}}\) of zero at \(25^{\circ} \mathrm{C}\), it signifies that the compound does not interact with water to form ions. This solid characteristic is useful for determining conditions under which certain reactions and mixtures are stable without forming aqueous ions. It is important in selecting materials that will not dissolve or react adversely in water-based environments.

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Most popular questions from this chapter

The copper(I) ion forms a chloride salt that has \(K_{\text {sp }}=1.2 \times 10^{-6}\). Copper(I) also forms a complex ion with \(\mathrm{Cl}^{-}\) : $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \quad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in \(0.10 \mathrm{M} \mathrm{NaCl}\).

A solution is prepared by mixing \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) with \(50.0 \mathrm{~mL}\) of \(1.0 M \mathrm{KCl}\). Calculate the concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) at equilibrium. \(\left[K_{\text {sp }}\right.\) for \(\mathrm{PbCl}_{2}(s)\) is \(\left.1.6 \times 10^{-5} .\right]\)

As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mixture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write equations for all the changes mentioned above. What conclusions can you draw concerning the sizes of the \(K_{\mathrm{sp}}\) values for \(\mathrm{AgCl}, \mathrm{AgBr}\), and \(\mathrm{AgI}\) ?

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(0.10 \mathrm{M} \mathrm{KIO}_{3}\) solution is \(2.6 \times\) \(10^{-11} \mathrm{~mol} / \mathrm{L}\). Calculate \(K_{\text {sp }}\) for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s) .\)

The overall formation constant for \(\mathrm{HgI}_{4}{ }^{2-}\) is \(1.0 \times 10^{30} .\) That is, $$1.0 \times 10^{30}=\frac{\left[\mathrm{HgI}_{4}^{2-}\right]}{\left[\mathrm{Hg}^{2+}\right]\left[1^{-}\right]^{4}}$$ What is the concentration of \(\mathrm{Hg}^{2+}\) in \(500.0 \mathrm{~mL}\) of a solution that was originally \(0.010 \mathrm{M} \mathrm{Hg}^{2+}\) and \(0.78 \mathrm{M} \mathrm{I}^{-}\) ? The reaction is $$\mathrm{Hg}^{2+}(a q)+4 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}{ }^{2-}(a q)$$

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