Solubility is an equilibrium position, whereas \(K_{\mathrm{sp}}\) is an equilibrium constant. Explain the difference.

Short Answer

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Solubility represents the equilibrium position where the rate of dissolving equals the rate of precipitation and is usually expressed in units such as g/L or M. It can change depending on temperature, pressure, and other factors. On the other hand, \(K_{\mathrm{sp}}\) is a dimensionless constant used to quantify the solubility of a substance in water and helps predict precipitation. It accounts for the common ion effect, unlike the solubility value. In summary, solubility indicates the amount of solute that can dissolve, while \(K_{\mathrm{sp}}\) predicts when precipitation will occur.

Step by step solution

01

Define Solubility

Solubility is the maximum amount of a solute that can dissolve in a solvent at a given temperature and produce a saturated solution. It's an equilibrium position, which means that when the system reaches solubility, the forward reaction (dissolving) is equal to the reverse reaction (precipitation). In other words, the rate of solute dissolving equals the rate of solute precipitating.
02

Define \(K_{\mathrm{sp}}\) (Solubility product constant)

The solubility product constant, denoted as \(K_{\mathrm{sp}}\), is an equilibrium constant associated with the dissolution of a sparingly soluble compound. It is used to quantify the solubility of a substance in water. The \(K_{\mathrm{sp}}\) can be calculated using the concentrations of the ions that constitute the saturated solution at equilibrium.
03

Equilibrium expression of \(K_{\mathrm{sp}}\)

For a general reaction involving a sparingly soluble compound AB with the dissociation as follows: \(AB_{(s)} \leftrightarrows A^{+}_{(aq)} + B^{-}_{(aq)}\) The equilibrium expression for \(K_{\mathrm{sp}}\) is: \(K_{\mathrm{sp}} = [A^{+}][B^{-}]\)
04

Difference between Solubility and \(K_{\mathrm{sp}}\)

The main differences between the solubility and the solubility product constant are: 1. Solubility represents the equilibrium position where the rate of dissolving equals the rate of precipitation, while \(K_{\mathrm{sp}}\) is the equilibrium constant that is used to quantify the solubility of a substance in water. 2. Solubility is usually expressed in different units such as grams per liter (g/L) or moles per liter (molarity, M), while \(K_{\mathrm{sp}}\) is a dimensionless constant (it has no units). 3. The value of solubility can change depending on temperature, pressure, and the presence of other ions in the solution. The value of \(K_{\mathrm{sp}}\) is determined by the stoichiometry of the reaction and the concentrations of the ions at equilibrium. 4. Solubility is a measure of how much solute can dissolve in a solvent to create a saturated solution, while \(K_{\mathrm{sp}}\) is useful for predicting whether a precipitate will form or not when solutions are mixed. 5. The solubility does not necessarily account for the common ion effect, whereas the \(K_{\mathrm{sp}}\) does. The common ion effect occurs when the solubility of a salt is reduced when it is dissolved in a solution that already contains one of the ions in the salt. In summary, solubility and \(K_{\mathrm{sp}}\) are related concepts, but they represent different aspects of the dissolution of a sparingly soluble compound. Solubility determines the amount of a solute that can dissolve in a solvent, while \(K_{\mathrm{sp}}\) is a constant that can help predict when precipitation will occur.

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Most popular questions from this chapter

Calculate the solubility of \(\mathrm{AgCN}(s)\left(K_{\mathrm{sp}}=2.2 \times 10^{-12}\right)\) in a solution containing \(1.0 M \mathrm{H}^{+} .\left(K_{\mathrm{a}} \text { for } \mathrm{HCN} \text { is } 6.2 \times 10^{-10} .\right)\)

A solution contains 0.018 molel each of \(\mathrm{I}^{-}, \mathrm{Br}^{-},\) and \(\mathrm{Cl}^{-}\). When the solution is mixed with \(200 . \mathrm{mL}\) of \(0.24\) \(M\) \(\mathrm{AgNO}_{3}\), what mass of \(\mathrm{AgCl}(s)\) precipitates out, and what is \(\left[\mathrm{Ag}^{+}\right] ?\) Assume no volume change. $$\begin{aligned} \operatorname{AgI}: K_{\mathrm{sp}} &=1.5 \times 10^{-16} \\ \operatorname{AgBr}: K_{\mathrm{sp}} &=5.0 \times 10^{-13} \\ \mathrm{AgCl}: K_{\mathrm{sp}} &=1.6 \times 10^{-10} \end{aligned}$$

For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. \(\mathrm{FeC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.1 \times 10^{-7},\) or \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}, K_{\mathrm{sp}}=1.4 \times 10^{-7}\) b. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}, K_{\mathrm{sp}}=8.1 \times 10^{-12},\) or \(\mathrm{Mn}(\mathrm{OH})_{2}\) \(K_{\mathrm{sp}}=2 \times 10^{-13}\)

A solution contains \(1.0 \times 10^{-5} M \mathrm{Ag}^{+}\) and \(2.0 \times 10^{-6} M \mathrm{CN}^{-}\) Will AgCN( \(s\) ) precipitate? \(\left(K_{\mathrm{sp}} \text { for } \mathrm{AgCN}(s) \text { is } 2.2 \times 10^{-12} .\right)\)

Which of the following will affect the total amount of solute that can dissolve in a given amount of solvent? a. The solution is stirred. b. The solute is ground to fine particles before dissolving. c. The temperature changes.

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