Problem 2
Question
Complete the changes in concentrations for each of the following reactions: (a) \(\operatorname{BaSO}_{4}(s) \rightarrow \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\) (b) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s) \longrightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\) (c) \(\mathrm{Al}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q)\) (d) \(\operatorname{Pb}(\mathrm{OH}) \mathrm{Cl}(s) \longrightarrow \mathrm{Pb}^{2+}(a q)+\mathrm{OH}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) (e) \(\operatorname{Ca}_{3}\left(\mathrm{AsO}_{4}\right)_{2}(s) \longrightarrow 3 \mathrm{Ca}^{2+}(a q)+2 \mathrm{AsO}_{4}^{3-}(a q)\)
Step-by-Step Solution
Verified Answer
(a) For each mole of \text{BaSO}_4 that dissolves, [Ba^{2+}] increases by 1 mole and [SO_{4}^{2-}] increases by 1 mole. (b) For each mole of \text{Ag}_2\text{SO}_4 that dissolves, [Ag^+] increases by 2 moles and [SO_{4}^{2-}] increases by 1 mole. (c) For each mole of \text{Al(OH)}_3 that dissolves, [Al^{3+}] increases by 1 mole and [OH^{-}] increases by 3 moles. (d) For each mole of \text{Pb(OH)Cl} that dissolves, [Pb^{2+}] increases by 1 mole, [OH^{-}] increases by 1 mole, and [Cl^{-}] increases by 1 mole. (e) For each mole of \text{Ca}_3(\text{AsO}_4)_2 that dissolves, [Ca^{2+}] increases by 3 moles and [AsO_{4}^{3-}] increases by 2 moles.
1Step 1: Identify the Solids and the Ions Formed
Determine the ions produced when each solid dissolves in water, keeping in mind the conservation of charge and the stoichiometry of the reaction.
2Step 2: Write the Balanced Equations for the Dissolution
Write equations for each substance dissolving, including coefficients to balance the charges on both sides.
3Step 3: List the concentration changes for each ion
For each equation, identify how the concentration of each ion changes relative to the solid. For each mole of solid that dissolves, the changes in concentration of the ions are based on the stoichiometry of the reaction.
Key Concepts
Chemical StoichiometrySolubilityConservation of ChargeIonic Compounds
Chemical Stoichiometry
Chemical stoichiometry is the study of the quantitative relationships between the reactants and products in a chemical reaction. Understanding stoichiometry is crucial to solving problems like dissolution reactions because it involves the balancing of elements and charge to satisfy the laws of conservation of mass and charge.
In dissolution reactions, stoichiometry tells us how many ions of each kind will be produced from the dissociation of a certain amount of solid compound. For instance, in the case of \(\mathrm{Ag}_2\mathrm{SO}_4(s) \longrightarrow 2\mathrm{Ag}^+(aq) + \mathrm{SO}_4^{2-}(aq)\), the stoichiometry indicates that one mole of silver sulfate yields two moles of silver ions and one mole of sulfate ions upon dissolution.
In dissolution reactions, stoichiometry tells us how many ions of each kind will be produced from the dissociation of a certain amount of solid compound. For instance, in the case of \(\mathrm{Ag}_2\mathrm{SO}_4(s) \longrightarrow 2\mathrm{Ag}^+(aq) + \mathrm{SO}_4^{2-}(aq)\), the stoichiometry indicates that one mole of silver sulfate yields two moles of silver ions and one mole of sulfate ions upon dissolution.
Solubility
Solubility is a measure of how well a substance can dissolve in a solvent to form a solution. In the context of dissolution reactions involving ionic compounds like \(\operatorname{BaSO}_4(s)\), solubility is a key concept because it affects the concentrations of ions in the solution.
Each substance has its inherent solubility limit, beyond which no more solute can dissolve in the solvent. The solubility product constant \(K_{sp}\) is a useful expression to quantify the solubility, which is particularly relevant for sparingly soluble salts such as \(\mathrm{BaSO}_4\). The lower the \(K_{sp}\), the less soluble the compound is. When a saturated solution is in equilibrium, the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficient, equals the \(K_{sp}\).
Each substance has its inherent solubility limit, beyond which no more solute can dissolve in the solvent. The solubility product constant \(K_{sp}\) is a useful expression to quantify the solubility, which is particularly relevant for sparingly soluble salts such as \(\mathrm{BaSO}_4\). The lower the \(K_{sp}\), the less soluble the compound is. When a saturated solution is in equilibrium, the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficient, equals the \(K_{sp}\).
Conservation of Charge
The conservation of charge is a fundamental principle stating that the total electric charge in an isolated system remains constant over time. This principle is essential in writing chemical equations for dissolution reactions because the sum of charges on the reactant and product side must be equal.
When an ionic compound like \(\mathrm{Al}(\mathrm{OH})_3(s)\) dissolves, it splits into its respective ions: \(\mathrm{Al}^{3+}(aq)\) and \(3\mathrm{OH}^{-}(aq)\). There must be a balance in the number of positively charged ions and negatively charged ions to ensure that the overall charge is conserved. In this example, for every aluminum ion with a +3 charge, there must be three hydroxide ions each with a −1 charge to maintain a neutral overall charge.
When an ionic compound like \(\mathrm{Al}(\mathrm{OH})_3(s)\) dissolves, it splits into its respective ions: \(\mathrm{Al}^{3+}(aq)\) and \(3\mathrm{OH}^{-}(aq)\). There must be a balance in the number of positively charged ions and negatively charged ions to ensure that the overall charge is conserved. In this example, for every aluminum ion with a +3 charge, there must be three hydroxide ions each with a −1 charge to maintain a neutral overall charge.
Ionic Compounds
Ionic compounds are chemical compounds composed of cations (positively charged ions) and anions (negatively charged ions) that are held together by ionic bonds. These bonds result from the electrostatic attractions between ions with opposite charges.
When ionic compounds dissolve in water, they dissociate into their constituent ions as shown in the reaction \(\operatorname{Pb}(\mathrm{OH})\mathrm{Cl}(s) \longrightarrow \mathrm{Pb}^{2+}(aq) + \mathrm{OH}^{-}(aq) + \mathrm{Cl}^{-}(aq)\). The solid lead hydroxychloride separates into lead, hydroxide, and chloride ions. The solubility and the dissolution process of ionic compounds can be affected by factors such as the ionic strength of the solution, the presence of other ions, and temperature.
When ionic compounds dissolve in water, they dissociate into their constituent ions as shown in the reaction \(\operatorname{Pb}(\mathrm{OH})\mathrm{Cl}(s) \longrightarrow \mathrm{Pb}^{2+}(aq) + \mathrm{OH}^{-}(aq) + \mathrm{Cl}^{-}(aq)\). The solid lead hydroxychloride separates into lead, hydroxide, and chloride ions. The solubility and the dissolution process of ionic compounds can be affected by factors such as the ionic strength of the solution, the presence of other ions, and temperature.
Other exercises in this chapter
Problem 1
Complete the changes in concentrations for each of the following reactions: (a) \(\operatorname{AgI}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{I}^{-}(a q)
View solution Problem 9
Write the ionic equation for the dissolution and the \(K_{\mathrm{sp}}\) expression for each of the following slightly soluble ionic compounds: (a) \(\mathrm{La
View solution Problem 10
The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities of the following compounds in grams per \(100 \mathrm{mL}\) o
View solution Problem 11
The Handbook of Chemistry and Physics (http://openstaxcollege.org///16Handbook) gives solubilities of the following compounds in grams per \(100 \mathrm{mL}\) o
View solution