Question

Write the balanced chemical equation and the \(K_{\text {sp }}\) expression for the slight solubility of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{~s})\)

Short answer

\( \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{s}) \rightarrow 2 \mathrm{Fe}^{3+}(\mathrm{aq}) + 3 \mathrm{SO}_{4}^{2-}(\mathrm{aq}) \); \( K_{\text{sp}} = [\mathrm{Fe}^{3+}]^2 [\mathrm{SO}_{4}^{2-}]^3 \).

Step-by-step solution

Understand the Dissolution

The dissolution of a salt involves breaking it into its constituent ions in water. For \( \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{~s}) \), it dissolves into iron ions (\( \mathrm{Fe}^{3+} \)) and sulfate ions (\( \mathrm{SO}_{4}^{2-} \)).

Write the Dissolution Equation

The dissolution equation for \( \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3} \) is: \[ \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{s}) \rightarrow 2 \mathrm{Fe}^{3+}(\mathrm{aq}) + 3 \mathrm{SO}_{4}^{2-}(\mathrm{aq}) \]. This equation shows the solid breaking into two iron ions and three sulfate ions.

Determine the Stoichiometry in \( K_{\text{sp}} \)

The stoichiometry directly influences the \( K_{\text{sp}} \) expression. Here, for every formula unit of \( \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3} \), two iron ions and three sulfate ions appear in the solution.

Write the \( K_{\text{sp}} \) Expression

The \( K_{\text{sp}} \) expression is based on the concentration of ions at equilibrium. Therefore, \[ K_{\text{sp}} = [\mathrm{Fe}^{3+}]^2 [\mathrm{SO}_{4}^{2-}]^3 \]. It shows how the product of the concentrations of dissolved ions raised to their stoichiometric coefficients equals the solubility product.

Key concepts

Solubility Product Constant

In the world of chemistry, the Solubility Product Constant, commonly known as \( K_{\text{sp}} \), plays a pivotal role in understanding solubility equilibrium. It is a mathematical expression that describes how soluble a compound is in a solution. The \( K_{\text{sp}} \) specifically applies to sparingly soluble salts, which dissociate into ions in solution. For a given nondissolved salt, such as \( \text{Fe}_2(\text{SO}_4)_3 \), the \( K_{\text{sp}} \) tells us about the extent to which the salt can dissolve to maintain equilibrium. At equilibrium, the rate at which the salt dissolves equals the rate at which it precipitates, meaning no net change occurs. The \( K_{\text{sp}} \) helps in calculating the maximum concentrations of ions in a saturated solution, signifying the point where no more solid can dissolve.The formula for writing a \( K_{\text{sp}} \) expression is:

• Identify the ions formed from the salt in the solution.
• Substitute the concentration of ions into the expression.
• Raise each ion's concentration to the power of its stoichiometric coefficient in the balanced equation.

These steps are essential to understanding the role of \( K_{\text{sp}} \) in predicting whether a precipitate will form in a chemical solution.

Dissolution Equations

Dissolution equations are essentially balanced chemical equations that show how a solid ionic compound dissociates into its ions in solution. Taking the example of \( \text{Fe}_2(\text{SO}_4)_3 \), the dissolution equation is crucial for illustrating how this compound breaks apart in water.The dissolution equation for \( \text{Fe}_2(\text{SO}_4)_3 \) is:\[ \text{Fe}_2(\text{SO}_4)_3(\text{s}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 3\text{SO}_4^{2-}(\text{aq}) \]This equation signifies that two iron ions and three sulfate ions are produced from one formula unit of iron(III) sulfate dissolving. The subscripts in the formula indicate the number of each ion produced, which is vital for predicting the outcome of reactions in solution.Understanding dissolution equations is key because they help predict the concentration of ions that result when solids dissolve. They are also necessary for setting up the \( K_{\text{sp}} \) expressions that determine solubility.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantification of reactants and products in chemical reactions. It is an essential aspect when discussing solubility and dissolution equations because it helps calculate how much of one reactant is required or how much product is formed.In the case of \( \text{Fe}_2(\text{SO}_4)_3 \), stoichiometry provides the ratios in which the ions appear in solution:- From one formula unit, two \( \text{Fe}^{3+} \) ions are produced.- Three \( \text{SO}_4^{2-} \) ions are released.These stoichiometric coefficients are crucial for writing dissolution equations and \( K_{\text{sp}} \) expressions accurately. For a \( K_{\text{sp}} \) calculation, each coefficient is used as an exponent to indicate the concentration of that ion in the solution. Therefore, a clear understanding of stoichiometry helps in predicting and calculating concentrations of ions in a solution, guiding both experimental and theoretical calculations.

Ion Concentration in Solutions

Concentration of ions in solution refers to the amount of ions present in a specific volume of solution. It is usually expressed in terms of molarity, which is moles of solute per liter of solution.When \( \text{Fe}_2(\text{SO}_4)_3 \) dissolves, the concentration of \( \text{Fe}^{3+} \) and \( \text{SO}_4^{2-} \) ions can be calculated using the dissolution equation and the solubility product expression:

• The molarity of \( \text{Fe}^{3+} \) ions will be twice the solubility of the compound, as per its stoichiometry.
• The molarity of \( \text{SO}_4^{2-} \) ions will be three times due to its stoichiometric ratio.

This relationship is crucial because understanding ion concentration determines how reactions proceed in solutions, helps with calculations involving equilibrium, and aids in predicting whether precipitation will occur under given conditions. Calculating these concentrations accurately is fundamental to working with chemical solutions and understanding their behaviors.