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Chapter 16: Problem 85

What is meant by the solubility product for a sparingly soluble salt? Choose a sparingly soluble salt and show how the salt ionizes when dissolved in water, and write the expression for its solubility product.

Short Answer

Expert verified
The solubility product (K_sp) is an equilibrium constant for the dissolution of a sparingly soluble salt in water, representing the product of the molar concentrations of the ions, each raised to the power of its stoichiometric coefficient. For example, consider silver chloride (AgCl). When it dissolves in water, it ionizes as follows: \[ AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq) \] The solubility product expression for AgCl is given by: \[ K_{sp} = [Ag^+][Cl^-] \]

Step by step solution

01

Definition of Solubility Product

Solubility product, denoted as K_sp, is an equilibrium constant for the dissolution of a sparingly soluble salt in water. It is the product of the molar concentrations of the ions formed when a saturated solution of the salt is in equilibrium with its undissolved solid, each concentration raised to the power of its stoichiometric coefficient in the dissolution equation.
02

Choose a Sparingly Soluble Salt

Let's consider the sparingly soluble salt, silver chloride (AgCl), as our example.
03

Ionization of the Salt in Water

When silver chloride dissolves in water, it dissociates into its constituent ions, silver (Ag+) and chloride (Cl-), in the following equilibrium reaction: \[ AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq) \]
04

Expression for Solubility Product

The solubility product (K_sp) of silver chloride is calculated as the product of the concentrations of the silver and chloride ions, each raised to the power of its stoichiometric coefficient in the dissolution equation. Thus, the solubility product expression for silver chloride can be written as: \[ K_{sp} = [Ag^+][Cl^-] \] Where [Ag+] and [Cl-] represent the molar concentrations of silver and chloride ions, respectively, in the saturated solution at equilibrium.

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Most popular questions from this chapter

Suppose a reaction has the equilibrium constant \(K=4.5 \times 10^{-6}\) at a particular temperature. If an experiment is set up with this reaction, will there be large relative concentrations of products present at equilibrium? Is this reaction useful as a means of producing the products? How might the reaction be made more useful?

For the reaction $$2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)$$ it is found at equilibrium at a certain temperature that the concentrations are \([\mathrm{CO}(g)]=2.7 \times 10^{-4} \mathrm{M}\) \(\left[\mathrm{O}_{2}(g)\right]=1.9 \times 10^{-3} \mathrm{M},\) and \(\left[\mathrm{CO}_{2}(g)\right]=1.1 \times 10^{-1} \mathrm{M}\) Calculate \(K\) for the reaction at this temperature.

Approximately \(1.5 \times 10^{-3} \mathrm{g}\) of iron(II) hydroxide, \(\mathrm{Fe}(\mathrm{OH})_{2}(s),\) dissolves per liter of water at \(18^{\circ} \mathrm{C} .\) Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Fe}(\mathrm{OH})_{2}(s)\) at this temperature.

At high temperatures, elemental bromine, \(\mathrm{Br}_{2}\), dissociates into individual bromine atoms. $$\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)$$ Suppose that in an experiment at \(2000^{\circ} \mathrm{C},\) it is found that \(\left[\mathrm{Br}_{2}\right]=0.97 \mathrm{M}\) and \([\mathrm{Br}]=0.034 \mathrm{M}\) at equilibrium. Calculate the value of \(K\).

Suppose a reaction has the equilibrium constant \(K=\) \(1.3 \times 10^{8} .\) What does the magnitude of this constant tell you about the relative concentrations of products and reactants that will be present once equilibrium is reached? Is this reaction likely to be a good source of the products?
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