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Write the expression for the solubility-product constant for each of the following ionic compounds: \(\mathrm{BaCrO}_{4}, \mathrm{CuS}, \mathrm{PbCl}_{2}\) and \(\mathrm{LaF}_{3}\).

Short Answer

Expert verified
For \( \mathrm{BaCrO}_{4} \), \( K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{CrO}_{4}^{2-}] \). For \( \mathrm{CuS} \), \( K_{sp} = [\mathrm{Cu}^{2+}][\mathrm{S}^{2-}] \). For \( \mathrm{PbCl}_{2} \), \( K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^{2} \). For \( \mathrm{LaF}_{3} \), \( K_{sp} = [\mathrm{La}^{3+}][\mathrm{F}^{-}]^{3} \).

Step by step solution

01

Understanding the Solubility Product Constant

The solubility product constant, denoted as \( K_{sp} \), is a constant for a given substance at a specific temperature. It represents the level at which a solute dissolves in solution. The expression for \( K_{sp} \) is derived from the equilibrium concentrations of the ions produced when the compound dissolves in water.
02

Write the Dissolution Equation for BaCrOâ‚„

The dissolution of \( \mathrm{BaCrO}_{4} \) in water can be represented as:\[ \mathrm{BaCrO}_{4} (s) \rightleftharpoons \mathrm{Ba}^{2+} (aq) + \mathrm{CrO}_{4}^{2-} (aq) \]The solubility-product expression for this is:\[ K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{CrO}_{4}^{2-}] \]
03

Write the Dissolution Equation for CuS

The dissolution of \( \mathrm{CuS} \) in water is given by:\[ \mathrm{CuS} (s) \rightleftharpoons \mathrm{Cu}^{2+} (aq) + \mathrm{S}^{2-} (aq) \]Its solubility-product expression is:\[ K_{sp} = [\mathrm{Cu}^{2+}][\mathrm{S}^{2-}] \]
04

Write the Dissolution Equation for PbClâ‚‚

The dissolution of \( \mathrm{PbCl}_{2} \) can be represented as:\[ \mathrm{PbCl}_{2} (s) \rightleftharpoons \mathrm{Pb}^{2+} (aq) + 2\mathrm{Cl}^{-} (aq) \]The solubility-product expression for this equilibrium is:\[ K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^{2} \]
05

Write the Dissolution Equation for LaF₃

The dissolution of \( \mathrm{LaF}_{3} \) is described by:\[ \mathrm{LaF}_{3} (s) \rightleftharpoons \mathrm{La}^{3+} (aq) + 3\mathrm{F}^{-} (aq) \]Therefore, the solubility-product expression is:\[ K_{sp} = [\mathrm{La}^{3+}][\mathrm{F}^{-}]^{3} \]

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Key Concepts

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

Ionic Compounds
Ionic compounds are substances formed from the combination of positively and negatively charged ions. These ions are held together by strong electrostatic forces known as ionic bonds. This formation typically occurs between metals and non-metals. For instance, in the compound \( \text{BaCrO}_4 \), barium \( (\text{Ba}^{2+}) \) is a metal, and chromate \( (\text{CrO}_4^{2-}) \) is a polyatomic ion consisting of non-metals.
These compounds dissolve in water to release their constituent ions. In solution, ionic compounds dissociate completely or partially into their ions, depending on their solubility. The behavior and solubility of these compounds in water are crucial for determining the concentration of the ions, which directly influences the solubility product constant \( (K_{sp}) \). Understanding this concept is important when considering reactions in aqueous solutions in various chemical processes.
Dissolution Equation
A dissolution equation represents the process where an ionic compound dissolves in water to form its constituent ions. This equation is essential in visualizing the equilibrium that exists between the undissolved solid and the ions in solution.
For instance, for the compound \( \text{BaCrO}_4 \), the dissolution equation is: \[ \text{BaCrO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{CrO}_4^{2-} (aq) \]
This equation signifies that the solid barium chromate \( \text{(BaCrO}_4 \text{)} \) dissolves to form barium ions \( \text{(Ba}^{2+} \text{)} \) and chromate ions \( \text{(CrO}_4^{2-} \text{)} \). Each dissolution equation varies depending on the compound in question, showcasing the unique stoichiometry and resulting ions.
Equilibrium Concentrations
Equilibrium concentrations refer to the concentrations of ions in a solution when a dissolution reaction reaches equilibrium. At equilibrium, the rate at which the solid dissolves to form ions equals the rate at which ions recombine to form the solid.
Let's take \( \text{PbCl}_2 \) as an example: \[ \text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Cl}^{-} (aq) \]
Here, the equilibrium concentrations of the ions \([\text{Pb}^{2+}]\) and \([\text{Cl}^{-}]\) are used to write the solubility product constant expression, which helps to determine the extent of dissolution. Specifically, equilibrium concentrations are crucial in calculating the \( K_{sp} \), the solubility product constant, which quantifies this balance.
Chemical Equilibrium
Chemical equilibrium is the state in a chemical reaction where the rates of the forward and reverse reactions are equal. For dissolution reactions like those of ionic compounds, equilibrium signifies that the process of dissolving and precipitating ions occurs at the same rate. This balance is key to understanding reactions in closed systems.
Take the dissolution of \( \text{LaF}_3 \) for instance: \[ \text{LaF}_3 (s) \rightleftharpoons \text{La}^{3+} (aq) + 3\text{F}^{-} (aq) \]
Once this reaction reaches chemical equilibrium, no net change in concentration occurs for \( \text{La}^{3+} \) and \( \text{F}^{-} \) ions. They continue to form and dissolve at equal rates, maintaining their concentrations constant over time. Understanding equilibrium is crucial for predicting reaction behavior and calculating the solubility product constant \( K_{sp} \). This concept applies broadly across numerous chemical reactions and solutions.

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Most popular questions from this chapter

The solubility-product constant for barium permanganate, \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\), is \(2.5 \times 10^{-10}\), Assume that solid \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\) is in equilibrium with a solution of \(\mathrm{KMnO}_{4}\). What concentration of \(\mathrm{KMnO}_{4}\) is required to establish a concentration of \(2.0 \times 10^{-8} \mathrm{M}\) for the \(\mathrm{Ba}^{2+}\) ion in solution?

A buffer contains a weak acid, HA, and its conjugate base. The weak acid has a \(\mathrm{pK}_{a}\) of 4.5 , and the buffer has a pH of 4.7 . Without doing a calculation, state which of these possibilities are correct at pH 4.7 . (a) \([\mathrm{HA}]>\left[\mathrm{A}^{-}\right]\) (b) \([\mathrm{HA}]=\left[\mathrm{A}^{-}\right],\) or \((\mathbf{c})[\mathrm{HA}]<\left[\mathrm{A}^{-}\right]\).

A buffer, consisting of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\), helps control the pH of physiological fluids, Many carbonated soft drinks also use this buffer system. What is the \(\mathrm{pH}\) of a soft drink in which the major buffer ingredients are \(10.0 \mathrm{~g}\) of \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) and \(10.0 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{HPO}_{4}\) per \(0.500 \mathrm{~L}\) of solution?

A solution containing several metal ions is treated with dilute HCl; no precipitate forms. The pH is adjusted to about \(1,\) and \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled through. Again, no precipitate forms. The pH of the solution is then adjusted to about 8 . Again, \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled through. This time a precipitate forms. The filtrate from this solution is treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\). No precipitate forms. Which of these metal cations are either possibly present or definitely absent: \(\mathrm{Al}^{3+}, \mathrm{Na}^{+}, \mathrm{Ag}^{+}, \mathrm{Mg}^{2+} ?\)

Aspirin has the structural formula CC(=O)Oc1ccccc1C(=O)O At body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{a}\) for aspirin equals \(3 \times 10^{-5}\). If two aspirin tablets, each having a mass of \(325 \mathrm{mg}\), are dissolved in a full stomach whose volume is \(1 \mathrm{~L}\) and whose \(\mathrm{pH}\) is \(2,\) what percent of the aspirin is in the form of neutral molecules?

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