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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

Expert verified
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

01

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: \[ K_{sp}(\mathrm{CrF_3}) = [\mathrm{Cr^{3+}}][\mathrm{F^-}]^3 \]
02

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: \[ K_{sp}[\mathrm{Au2(C_2O_4)_3}] = [\mathrm{Au^{3+}}]^2[\mathrm{(C_2O_4)^{-}}]^3 \]
03

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: \[ K_{sp}[\mathrm{Cd_3(PO_4)_2}] = [\mathrm{Cd^{2+}}]^3[(PO_4)^{3-}]^2 \]
04

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: \[ K_{sp}(\mathrm{SrF_2}) = [\mathrm{Sr^{2+}}][\mathrm{F^-}]^2 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 CrF鈧, it dissociates into one Cr鲁鈦 ion and three F鈦 ions.
  • For Au鈧(C鈧侽鈧)鈧, two Au鲁鈦 ions and three C鈧侽鈧劼测伝 ions form upon dissolution.
  • Cd鈧(PO鈧)鈧 dissociates into three Cd虏鈦 ions and two PO鈧劼斥伝 ions.
  • And for SrF鈧, we observe dissociation into one Sr虏鈦 ion and two F鈦 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 CrF鈧 and SrF鈧 involve metal cations like Cr鲁鈦 and Sr虏鈦, and non-metal anions like F鈦.
  • Complex compounds like Au鈧(C鈧侽鈧)鈧 include more complicated polyatomic ions like C鈧侽鈧劼测伝.
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 SrF鈧 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.
  • CrF鈧's expression, \[K_{sp} = [\text{Cr}^{3+}][\text{F}^-]^3\], uses Cr鲁鈦 and F鈦 concentrations.
  • Au鈧(C鈧侽鈧)鈧 has \[ K_{sp} = [\text{Au}^{3+}]^2[\text{(C鈧侽鈧)}^{-}]^3 \], 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.

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Most popular questions from this chapter

Assume that, to be visible to the unaided eye, a precipitate must weigh more than \(1 \mathrm{mg}\). If you add \(1.0 \mathrm{mL}\) of \(1.0 \mathrm{M} \mathrm{NaCl}(\mathrm{aq})\) to \(100.0 \mathrm{mL}\) of a clear saturated aqueous AgCl solution, will you be able to see \(\mathrm{AgCl}(\mathrm{s})\) precipitated as a result of the common-ion effect? Explain.

The solubility of \(\mathrm{CdCO}_{3}(\mathrm{s})\) in \(1.00 \mathrm{M} \mathrm{KI}(\mathrm{aq})\) is \(1.2 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Given that \(K_{\mathrm{sp}}\) of \(\mathrm{CdCO}_{3}\) is \(5.2 \times 10^{-12},\) what is \(K_{\mathrm{f}}\) for \(\left[\mathrm{CdI}_{4}\right]^{2-} ?\)

The solubility of \(\mathrm{AgCN}(\mathrm{s})\) in \(0.200 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq})\) is \(8.8 \times 10^{-6} \mathrm{mol} / \mathrm{L} .\) Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{AgCN}\).

Write net ionic equations for the following qualitative cation analysis procedures. (a) precipitation of \(\mathrm{PbCl}_{2}(\mathrm{s})\) from a solution containing \(\mathrm{Pb}^{2+}\) (b) dissolution of \(\mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{s})\) in a solution of \(\mathrm{NaOH}(\mathrm{aq})\) (c) dissolution of \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) in \(\mathrm{HCl}(\mathrm{aq})\) (d) precipitation of \(\mathrm{CuS}(\mathrm{s})\) from an acidic solution of \(\mathrm{Cu}^{2+}\) and \(\mathrm{H}_{2} \mathrm{S}\)

Calculate the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in \(1.00 \mathrm{M}\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})\).

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