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Chapter 19: Problem 70

Write the ion-product expressions for (a) lead(II) iodide; (b) strontium sulfate; (c) chromium(III) hydroxide.

Short Answer

Expert verified
(a) \[ K_{sp} = [Pb^{2+}] [I^{-}]^2 \] (b) \[ K_{sp} = [Sr^{2+}] [SO_4^{2-}] \] (c) \[ K_{sp} = [Cr^{3+}] [OH^{-}]^3 \]

Step by step solution

01

Lead(II) Iodide - Write the Dissociation Equation

Lead(II) iodide dissociates in water as follows: \[ PbI_2(s) \rightarrow Pb^{2+}(aq) + 2 I^{-}(aq) \]
02

Lead(II) Iodide - Write the Ion-Product Expression

The ion-product expression (Ksp) for lead(II) iodide is given by: \[ K_{sp} = [Pb^{2+}] [I^{-}]^2 \]
03

Strontium Sulfate - Write the Dissociation Equation

Strontium sulfate dissociates in water as follows: \[ SrSO_4(s) \rightarrow Sr^{2+}(aq) + SO_4^{2-}(aq) \]
04

Strontium Sulfate - Write the Ion-Product Expression

The ion-product expression (Ksp) for strontium sulfate is given by: \[ K_{sp} = [Sr^{2+}] [SO_4^{2-}] \]
05

Chromium(III) Hydroxide - Write the Dissociation Equation

Chromium(III) hydroxide dissociates in water as follows: \[ Cr(OH)_3(s) \rightarrow Cr^{3+}(aq) + 3 OH^{-}(aq) \]
06

Chromium(III) Hydroxide - Write the Ion-Product Expression

The ion-product expression (Ksp) for chromium(III) hydroxide is given by: \[ K_{sp} = [Cr^{3+}] [OH^{-}]^3 \]

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Key Concepts

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

Solubility Product Constant
Understanding the solubility product constant (Ksp) is key to grasping how salts dissolve in water. The Ksp value represents the level at which a solute dissolves in solution. Higher Ksp values mean greater solubility. To determine Ksp, you need the concentrations of the ions produced when a compound dissolves.
For example, for lead(II) iodide (PbI2), the dissociation produces lead ions (Pb虏鈦�) and iodide ions (I鈦�). The Ksp expression is: \[ K_{sp} = [Pb^{2+}] [I^{-}]^2 \] The Ksp equation helps predict whether a precipitate will form when solutions are mixed. If the ion product exceeds the Ksp, precipitation occurs. Otherwise, the solution remains unsaturated.
Dissociation Equations
Dissociation equations describe how ionic compounds split into their ions in aqueous solutions. These equations are essential for writing ion-product expressions.
For instance, when lead(II) iodide (PbI2) dissolves, the equation is: \[ PbI_2(s) \rightarrow Pb^{2+}(aq) + 2I^{-}(aq) \] Similarly, strontium sulfate (SrSO4) dissociates as: \[ SrSO_4(s) \rightarrow Sr^{2+}(aq) + SO_4^{2-}(aq) \] And chromium(III) hydroxide (Cr(OH)3) dissociates as: \[ Cr(OH)_3(s) \rightarrow Cr^{3+}(aq) + 3OH^{-}(aq) \] These equations show the ratio at which ions are produced, which is crucial for calculating the Ksp.
Chemical Equilibria
Chemical equilibria refer to the state where the concentrations of reactants and products remain constant over time. This is critical in understanding how dissolution and precipitation reactions balance.
When a salt dissolves, it reaches a dynamic equilibrium where the rate of dissolution equals the rate of precipitation. For example, in lead(II) iodide: \[ PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^{-}(aq) \] The Ksp expression derived from this equilibrium helps predict the concentrations of ions in solution. If an additional substance alters the ion concentrations, the system adjusts to maintain equilibrium.
Understanding equilibria ensures you can predict changes in solubility or when a precipitate will form, based on ion concentrations and other factors.

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Most popular questions from this chapter

Find the \(\mathrm{pH}\) during the titration of \(20.00 \mathrm{~mL}\) of \(0.1000 \mathrm{M}\) butanoic acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\left(K_{\mathrm{a}}=1.54 \times 10^{-5}\right),\) with \(0.1000 \mathrm{M}\) \(\mathrm{NaOH}\) solution after the following additions of titrant: (a) \(0 \mathrm{~mL}\) (b) \(10.00 \mathrm{~mL}\) (c) \(15.00 \mathrm{~mL}\) (d) \(19.00 \mathrm{~mL}\) (e) \(19.95 \mathrm{~mL}\) (f) \(20.00 \mathrm{~mL}\) (g) \(20.05 \mathrm{~mL}\) (h) \(25.00 \mathrm{~mL}\)

What are the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and the \(\mathrm{pH}\) of a buffer that consists of \(0.20 M \mathrm{HF}\) and \(0.25 M \mathrm{KF}\left(K_{\mathrm{a}}\right.\) of \(\left.\mathrm{HF}=6.8 \times 10^{-4}\right) ?\)

Gout is caused by an error in metabolism that leads to a buildup of uric acid in body fluids, which is deposited as slightly soluble sodium urate \(\left(\mathrm{C}_{5} \mathrm{H}_{3} \mathrm{~N}_{4} \mathrm{O}_{3} \mathrm{Na}\right)\) in the joints. If the extracellular \(\left[\mathrm{Na}^{+}\right]\) is \(0.15 \mathrm{M}\) and the solubility of sodium urate is \(0.085 \mathrm{~g} / 100 . \mathrm{mL}\) what is the minimum urate ion concentration (abbreviated [Ur-]) that will cause a deposit of sodium urate?

Tooth enamel consists of hydroxyapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\) \(\left(K_{\mathrm{sp}}=6.8 \times 10^{-37}\right) .\) Fluoride ion added to drinking water reacts with \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\) to form the more tooth decay-resistant fluorapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\left(K_{\mathrm{sp}}=1.0 \times 10^{-60}\right) .\) Fluoridated water has dramatically decreased cavities among children. Calculate the solubility of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\) and of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\) in water.

What are the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and the \(\mathrm{pH}\) of a benzoic acid-benzoate buffer that consists of \(0.33 \mathrm{MC}_{6} \mathrm{H}_{5} \mathrm{COOH}\) and \(0.28 \mathrm{MC}_{6} \mathrm{H}_{5} \mathrm{COONa}\) \(\left(K_{\mathrm{a}}\right.\) of benzoic acid \(\left.=6.3 \times 10^{-5}\right) ?\)
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