Question

Write the equilibrium equation and the \(K_{s p}\) expression for each of the following. (a) \(\mathrm{Co}_{2} \mathrm{~S}_{3}\) (b) \(\mathrm{PbCl}_{2}\) (c) \(\mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\) (d) \(\mathrm{Sc}(\mathrm{OH})_{3}\)

Short answer

Question: Write the equilibrium equations and solubility product constant expressions (\(K_{sp}\)) for each of the following compounds: (a) \(\mathrm{Co}_{2} \mathrm{S}_{3}\), (b) \(\mathrm{PbCl}_{2}\), (c) \(\mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\), and (d) \(\mathrm{Sc}(\mathrm{OH})_{3}\). Answer: (a) \(\mathrm{Co}_{2} \mathrm{S}_{3} \leftrightharpoons 2 \mathrm{Co}^{3+} + 3 \mathrm{S}^{2-}\) \(K_{sp} = [\mathrm{Co}^{3+}]^2[\mathrm{S}^{2-}]^3\) (b) \(\mathrm{PbCl}_{2} \leftrightharpoons \mathrm{Pb}^{2+} + 2 \mathrm{Cl}^{-}\) \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^-]^2\) (c) \(\mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7} \leftrightharpoons 2 \mathrm{Zn}^{2+} + 2 \mathrm{P} \mathrm{O}_{4}^{3-}\) \(K_{sp} = [\mathrm{Zn}^{2+}]^2[\mathrm{P} \mathrm{O}_{4}^{3-}]^2\) (d) \(\mathrm{Sc}(\mathrm{OH})_{3} \leftrightharpoons \mathrm{Sc}^{3+} + 3 \mathrm{OH}^-\) \(K_{sp} = [\mathrm{Sc}^{3+}][\mathrm{OH}^-]^3\)

Step-by-step solution

(a) Writing the equilibrium equation for Co2S3

For the compound \(\mathrm{Co}_{2} \mathrm{S}_{3}\), it dissociates in water as follows: \(\mathrm{Co}_{2} \mathrm{S}_{3} \leftrightharpoons 2 \mathrm{Co}^{3+} + 3 \mathrm{S}^{2-}\)

(a) Writing the \(K_{sp}\) expression for Co2S3

The solubility product constant expression for \(\mathrm{Co}_{2} \mathrm{S}_{3}\) is: \(K_{sp} = [\mathrm{Co}^{3+}]^2[\mathrm{S}^{2-}]^3\)

(b) Writing the equilibrium equation for PbCl2

For the compound \(\mathrm{PbCl}_{2}\), it dissociates in water as follows: \(\mathrm{PbCl}_{2} \leftrightharpoons \mathrm{Pb}^{2+} + 2 \mathrm{Cl}^{-}\)

(b) Writing the \(K_{sp}\) expression for PbCl2

The solubility product constant expression for \(\mathrm{PbCl}_{2}\) is: \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^-]^2\)

(c) Writing the equilibrium equation for Zn2P2O7

For the compound \(\mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\), it dissociates in water as follows: \(\mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7} \leftrightharpoons 2 \mathrm{Zn}^{2+} + 2 \mathrm{P} \mathrm{O}_{4}^{3-}\)

(c) Writing the \(K_{sp}\) expression for Zn2P2O7

The solubility product constant expression for \(\mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\) is: \(K_{sp} = [\mathrm{Zn}^{2+}]^2[\mathrm{P} \mathrm{O}_{4}^{3-}]^2\)

(d) Writing the equilibrium equation for Sc(OH)3

For the compound \(\mathrm{Sc}(\mathrm{OH})_{3}\), it dissociates in water as follows: \(\mathrm{Sc}(\mathrm{OH})_{3} \leftrightharpoons \mathrm{Sc}^{3+} + 3 \mathrm{OH}^-\)

(d) Writing the \(K_{sp}\) expression for Sc(OH)3

The solubility product constant expression for \(\mathrm{Sc}(\mathrm{OH})_{3}\) is: \(K_{sp} = [\mathrm{Sc}^{3+}][\mathrm{OH}^-]^3\)

Key concepts

Equilibrium Equations

In chemistry, equilibrium equations are essential to understanding how a compound dissociates into ions in a solution. This process, known as ionic dissociation, reveals the balance of compounds breaking apart and recombining. For example, when a compound like cobalt(III) sulfide \(\mathrm{Co}_2\mathrm{S}_3\) is added to water, it dissociates into its constituent ions:\(\mathrm{Co}_2\mathrm{S}_3 \leftrightharpoons 2\mathrm{Co}^{3+} + 3\mathrm{S}^{2-}\).Similarly, lead(II) chloride \(\mathrm{PbCl}_2\) will dissociate in water:\(\mathrm{PbCl}_2 \leftrightharpoons \mathrm{Pb}^{2+} + 2\mathrm{Cl}^-\).This reaction shows how complex compounds can split into simpler ions, establishing an equilibrium where the forward and backward reactions occur at the same rate. In this state, the concentrations of reactants and products remain constant over time, illustrating the dynamic yet balanced nature of chemical reactions.

Ionic Dissociation

Ionic dissociation is a vital concept in predicting how ionic compounds behave in a solution. When an ionic compound like zinc pyrophosphate \(\mathrm{Zn}_2\mathrm{P}_2\mathrm{O}_7\) enters water, it breaks into individual ions:\(\mathrm{Zn}_2\mathrm{P}_2\mathrm{O}_7 \leftrightharpoons 2\mathrm{Zn}^{2+} + 2\mathrm{PO}_4^{3-}\).This process is crucial because it determines the availability of ions for chemical reactions. Ionic compounds are typically crystalline, and their dissociation in water leads to the separation of these ions from their solid form into the liquid phase.Understanding ionic dissociation helps us comprehend how solutes dissolve, affecting solution properties like conductivity, reactivity, and stability. Life processes, like nerve function and muscle contraction, rely heavily on ionic dissociation to ensure that ions are available for biological reactions.

Chemical Solubility

Chemical solubility describes how well a substance can dissolve in a solvent, impacting how compounds interact in a solution. The solubility of a compound is influenced by factors such as temperature, pressure, and the nature of both the solute and solvent. For example, scandium hydroxide \(\mathrm{Sc}(\mathrm{OH})_3\) can dissolve in water to some extent:\(\mathrm{Sc}(\mathrm{OH})_3 \leftrightharpoons \mathrm{Sc}^{3+} + 3\mathrm{OH}^-\).Substances with high solubility dissolve completely, leaving no solid residue, while those with low solubility may result in a saturated solution with an equilibrium between the dissolved ions and the undissolved solid. Understanding solubility is crucial for predicting reaction outcomes, especially in industrial processes and laboratory settings where precise concentrations are necessary for desired reactions.

Ksp Expressions

The solubility product constant \(K_{sp}\) is a vital part of chemical solubility calculations, providing a quantitative measure of a compound's solubility. This constant is derived from the equilibrium concentrations of ions in a saturated solution. For example, the \(K_{sp}\) expression for cobalt(III) sulfide is:\(K_{sp} = [\mathrm{Co}^{3+}]^2[\mathrm{S}^{2-}]^3\).This expression represents the maximum product of ion concentrations that can be present in solution without forming a precipitate. For lead(II) chloride, the calculation follows:\(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^-]^2\).These expressions are critical in predicting whether a precipitate will form when solutions are mixed, making them a cornerstone of analytical chemistry, especially in qualitative analysis and the separation of ions in complex solutions.