Question

The following \(K_{\mathrm{sp}}\) values are found in a handbook. Write the solubility product expression to which each one applies. For example, \(K_{\mathrm{sp}}(\mathrm{AgCl})=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=\) \(1.8 \times 10^{-10}\). (a) \(K_{\mathrm{sp}}\left(\mathrm{Cr} \mathrm{F}_{3}\right)=6.6 \times 10^{-11}\) (b) \(K_{\mathrm{sp}}\left[\mathrm{Au}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]=1 \times 10^{-10}\) (c) \(K_{\mathrm{sp}}\left[\mathrm{Cd}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=2.1 \times 10^{-33}\) (d) \(K_{\mathrm{sp}}\left(\mathrm{Sr} \mathrm{F}_{2}\right)=2.5 \times 10^{-9}\)

Short answer

The solubility product expressions to which each of the stated \(K_{sp}\) values applies are: \[ K_{sp}(\mathrm{CrF_3}) = [\mathrm{Cr^{3+}}][\mathrm{F^-}]^3 \], \[ K_{sp}[\mathrm{Au_2(C_2O_4)_3}] = [\mathrm{Au^{3+}}]^2[\mathrm{(C_2O_4)^{-}}]^3 \], \[ K_{sp}[\mathrm{Cd_3(PO_4)_2}] = [\mathrm{Cd^{2+}}]^3[(PO_4)^{3-}]^2 \], \[ K_{sp}(\mathrm{SrF_2}) = [\mathrm{Sr^{2+}}][\mathrm{F^-}]^2 \]

Step-by-step solution

Determine the Dissociation Reaction and Solubility Product for CrF3

The dissociation of CrF3 in water is represented as: \[ \mathrm{CrF_3} \rightarrow \mathrm{Cr^{3+}} + 3\mathrm{F^-} \] Therefore, the solubility product expression is: \[ K_{sp}(\mathrm{CrF_3}) = [\mathrm{Cr^{3+}}][\mathrm{F^-}]^3 \]

Determine the Dissociation Reaction and Solubility Product for Au2(C2O4)3

The dissociation of Au2(C2O4)3 in water is represented as: \[ \mathrm{Au_2(C_2O_4)_3} \rightarrow 2\mathrm{Au^{3+}} + 3\mathrm{(C_2O_4)^{-}} \] Therefore, the solubility product expression is: \[ K_{sp}[\mathrm{Au2(C_2O_4)_3}] = [\mathrm{Au^{3+}}]^2[\mathrm{(C_2O_4)^{-}}]^3 \]

Determine the Dissociation Reaction and Solubility Product for Cd3(PO4)2

The dissociation of Cd3(PO4)2 in water is represented as: \[ \mathrm{Cd_3(PO_4)_2} \rightarrow 3\mathrm{Cd^{2+}} + 2\mathrm{(PO_4)^{3-}} \] Therefore, the solubility product expression is: \[ K_{sp}[\mathrm{Cd_3(PO_4)_2}] = [\mathrm{Cd^{2+}}]^3[(PO_4)^{3-}]^2 \]

Determine the Dissociation Reaction and Solubility Product for SrF2

The dissociation of SrF2 in water is represented as: \[ \mathrm{SrF_2} \rightarrow \mathrm{Sr^{2+}} + 2\mathrm{F^-} \] Therefore, the solubility product expression is: \[ K_{sp}(\mathrm{SrF_2}) = [\mathrm{Sr^{2+}}][\mathrm{F^-}]^2 \]

Key concepts

Dissolution Reactions

Dissolution reactions involve the process where an ionic compound breaks down into its individual ions when it is placed in water. This process is key to understanding why some compounds dissolve better than others. When you drop a substance like table salt into some water, it dissolves because it separates into sodium and chloride ions.
This step is crucial because these ions only form because the water molecules pull them apart from the solid. The extent to which a compound can dissolve is specific to each substance, and it governs how we calculate its solubility product, a concept we will discuss later.

• For CrF₃, it dissociates into one Cr³⁺ ion and three F⁻ ions.
• For Au₂(C₂O₄)₃, two Au³⁺ ions and three C₂O₄²⁻ ions form upon dissolution.
• Cd₃(PO₄)₂ dissociates into three Cd²⁺ ions and two PO₄³⁻ ions.
• And for SrF₂, we observe dissociation into one Sr²⁺ ion and two F⁻ ions.

Each ionic aspect of these reactions plays a fundamental role in the broader concept of chemical equilibrium.

Chemical Equilibrium

Chemical equilibrium is a vital concept in chemistry where the rate of the forward reaction equals the rate of the backward reaction. Imagine a system where the amount of substance that gets dissolved is exactly the same as the amount that returns to its solid form. This balance is where the solution reaches equilibrium. It's like a busy highway where cars are entering and leaving the road at the same rates, keeping the number of cars constant.
This equilibrium is not static but dynamic; ions continually dissolve and re-form into solids at equal rates. For the dissolution of ionic compounds, equilibrium is achieved when the concentrations of the ions in solution reach a point where their rates of dissolution and precipitation match.
Understanding this balance is essential in calculating the solubility product constant, which relies heavily on these equilibrium concentrations of dissolved ions.

Ionic Compounds

Ionic compounds are made up of positively charged cations and negatively charged anions. They form when metals transfer electrons to non-metals, and the resulting opposite charges attract each other.
In the solid state, ionic compounds organize themselves in a repeating array called a lattice. However, when placed in water, these compounds dissociate into individual ions, making them free to move and conduct electricity. This behavior is what distinguishes ionic compounds from other types of chemical compounds.
Each ionic compound has its unique dissolution reaction, and this reaction is crucial in understanding how they dissolve in water.

• Compounds like CrF₃ and SrF₂ involve metal cations like Cr³⁺ and Sr²⁺, and non-metal anions like F⁻.
• Complex compounds like Au₂(C₂O₄)₃ include more complicated polyatomic ions like C₂O₄²⁻.

These compounds' behavior in solution forms the basis for setting chemical equilibrium in a system.

Ksp Calculations

Ksp, or the solubility product constant, plays a fundamental role in determining how much of the compound can dissolve in water. It's a mathematical representation of the equilibrium reached in dissolution reactions. Ksp expressions are formed using the concentrations of the ions, each raised to the power of their respective stoichiometric coefficients.
For instance, the Ksp expression of a compound like SrF₂ is \[ K_{sp} = [\text{Sr}^{2+}][\text{F}^-]^2 \]This equation helps chemists understand the solubility characteristics and predict whether a precipitate will form in a given solution.

• CrF₃'s expression, \[K_{sp} = [\text{Cr}^{3+}][\text{F}^-]^3\], uses Cr³⁺ and F⁻ concentrations.
• Au₂(C₂O₄)₃ has \[ K_{sp} = [\text{Au}^{3+}]^2[\text{(C₂O₄)}^{-}]^3 \], reflecting its complex cation-to-anion ratio.

These calculations are pivotal in predicting and analyzing solubility in various chemical contexts, aiding in both lab experiments and industrial applications.