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Consider the following equilibrium: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s) $$ Use Le Ch芒telier's principle to predict how the amount of solid silver chloride will change when the equilibrium is disturbed by: (a) Adding \(\mathrm{NaCl}\) (b) Adding \(\mathrm{AgNO}_{3}\) (c) Adding \(\mathrm{NO}_{3}\), which reacts with \(\mathrm{Ag}^{+}\) to form the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) (d) Removing \(\mathrm{Cl}^{-} ;\) also account for the change using the reaction quotient \(Q_{c}\)

Short Answer

Expert verified
(a) More AgCl forms; (b) More AgCl forms; (c) Less AgCl remains; (d) AgCl dissolves.

Step by step solution

01

Introduction to the Equilibrium

The equilibrium equation given is \( \mathrm{Ag}^{+}(aq)+\mathrm{Cl}^{-}(aq) \rightleftharpoons \mathrm{AgCl}(s) \). According to Le Ch芒telier's principle, if an external change is applied to a system at equilibrium, the system will adjust itself to partially counteract that change.
02

Step A: Adding NaCl

NaCl will dissociate in solution to form \( \mathrm{Na}^{+} \) and \( \mathrm{Cl}^{-} \). The increase in \( \mathrm{Cl}^{-} \) ions will shift the equilibrium to the left to reduce the effect of the change (i.e., to use up excess \( \mathrm{Cl}^{-} \)). Thus, more \( \mathrm{AgCl} \) will precipitate, increasing the amount of solid silver chloride.
03

Step B: Adding AgNO3

Adding \( \mathrm{AgNO}_3 \) provides more \( \mathrm{Ag}^{+} \) ions to the solution. This will shift the equilibrium to the left as well, causing more \( \mathrm{AgCl} \) to form and precipitate. Hence, the amount of solid silver chloride will increase.
04

Step C: Adding NH3

Addition of \( \mathrm{NH}_3 \) which reacts with \( \mathrm{Ag}^{+} \) to form \( \mathrm{Ag(NH}_3)_2^{+} \). This decreases the concentration of free \( \mathrm{Ag}^{+} \) ions. As a response, the equilibrium will shift to the right to produce more \( \mathrm{Ag}^{+} \) ions, decreasing the amount of solid silver chloride.
05

Step D: Removing Cl鈦

Removing \( \mathrm{Cl}^{-} \) from the solution will result in a leftward shift of the equilibrium to produce more \( \mathrm{Cl}^{-} \), which means more solid silver chloride will dissolve back in the solution. This reduces the amount of solid silver chloride.
06

Reaction Quotient Consideration for Step D

After \( \mathrm{Cl}^{-} \) is removed, calculate \( Q_c = \frac{[\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]}{1} \). If \( Q_c < K_{sp} \), the system will shift to the right to increase \( Q_c \) to equal \( K_{sp} \), thereby dissolving more \( \mathrm{AgCl} \).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is the state in a chemical reaction where the rates of the forward and reverse reactions are equal. This means that the concentration of reactants and products remains constant over time, although they are not necessarily equal. In a dynamic equilibrium, molecules are continuously reacting, but there's no net change in the concentrations of the reactants and products.

Equilibrium can be influenced by changes in concentration, temperature, and pressure. Le Ch芒telier's principle is a fundamental concept that helps predict how a reaction at equilibrium will respond to such changes.
  • Le Ch芒telier's principle states that if an external change is applied to a system at equilibrium, the system adjusts to partially counteract that change.
  • This principle allows for the prediction of the direction in which a reaction will shift when disturbed.
In the context of our given equilibrium with silver chloride, any addition or removal of ions influences the reaction's equilibrium position.
Reaction Quotient
The reaction quotient, denoted as \( Q_c \), is crucial for understanding chemical equilibria. It represents the ratio of the concentrations of the products to the reactants at any given point in time, not necessarily at equilibrium.

For the reaction \( \mathrm{Ag}^{+}(aq)+\mathrm{Cl}^{-}(aq) \rightleftharpoons \mathrm{AgCl}(s) \), the reaction quotient is calculated as:
\[ Q_c = \frac{[\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]}{1} \]
  • If \( Q_c \) is less than the equilibrium constant \( K_{sp} \), the reaction will shift to the right, forming more products to reach equilibrium.
  • If \( Q_c \) is greater than \( K_{sp} \), the reaction will shift to the left, forming more reactants.
This concept helps in predicting and explaining shifts in equilibrium, such as when elements are removed from or added to the system, as seen with silver ions and chloride ions.
Silver Chloride Solubility
Silver chloride (\( \mathrm{AgCl} \)) is an ionic compound with limited solubility in water. Its solubility is governed by its solubility product constant, \( K_{sp} \). When \( \mathrm{AgCl} \) is in equilibrium with its ions in solution, any changes to the system can affect its solubility.

To understand how different perturbations affect the solubility of \( \mathrm{AgCl} \), we can apply Le Ch芒telier's principle:
  • Adding compounds such as \( \mathrm{NaCl} \) or \( \mathrm{AgNO}_3 \), which dissociate to produce ions already present in the equilibrium, will shift the balance toward the formation of more solid \( \mathrm{AgCl} \).
  • Conversely, removing ions from the solution will cause more \( \mathrm{AgCl} \) to dissolve, as the equilibrium shifts to make up for the deficit.
In practice, examining ion concentrations and knowing \( K_{sp} \) are essential for predicting how \( \mathrm{AgCl} \) will behave in response to changes in the system.

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Most popular questions from this chapter

The following reaction, which has \(K_{c}=0.145\) at \(298 \mathrm{~K}\), takes place in carbon tetrachloride solution: $$ 2 \mathrm{BrCl}(\mathrm{soln}) \rightleftharpoons \mathrm{Br}_{2}(\mathrm{soln})+\mathrm{Cl}_{2}(\text { soln }) $$ A measurement of the concentrations shows \([\mathrm{BrCl}]=0.050 \mathrm{M}\), \(\left[\mathrm{Br}_{2}\right]=0.035 \mathrm{M}\), and \(\left[\mathrm{Cl}_{2}\right]=0.030 \mathrm{M}\) (a) Calculate \(\mathrm{Q}_{\circ}\) and determine the direction of reaction to attain equilibrium. (b) Determine the equilibrium concentrations of \(\mathrm{BrCl}, \mathrm{Br}_{\mathrm{t}}\), and \(\mathrm{Cl}_{2}\)

In a basic aqueous solution, chloromethane undergoes a substitution reaction in which CI is replaced by \(\mathrm{OH}\) : $$ \mathrm{CH}_{3} \mathrm{Cl}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cl}^{-}(a q) $$ The equilibrium constant \(K_{c}\) is \(1 \times 10^{16} .\) Calculate the equilibrium concentrations of \(\mathrm{CH}_{3} \mathrm{Cl}, \mathrm{CH}_{3} \mathrm{OH}, \mathrm{OH}^{-}\), and \(\mathrm{Cl}^{-}\) in a solu- tion prepared by mixing equal volumes of \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{Cl}\) and \(0.2\) M \(\mathrm{NaOH}\). (Hint: In defining \(x\), assume that the reaction goes \(100 \%\) to completion, and then take account of a small amount of the reverse reaction.)

The following reaction, catalyzed by iridium, is endothermic at \(700 \mathrm{~K}\) : $$ \mathrm{CaO}(s)+\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CaCO}_{3}(s)+4 \mathrm{H}_{2}(g) $$ For a reaction mixture at equilibrium at \(700 \mathrm{~K}\), how would the following changes affect the total quantity of \(\mathrm{CaCO}_{3}\) in the reaction mixture once equilibrium is re-established? (a) Increasing the temperature (b) Adding calcium oxide (c) Removing methane \(\left(\mathrm{CH}_{4}\right)\) (d) Increasing the total volume (e) Adding iridium

The reaction of iron(III) oxide with carbon monoxide is important in making steel. At \(1200 \mathrm{~K}, K_{\mathrm{p}}=19.9\) for the reaction $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \rightleftharpoons 2 \mathrm{Fe}(l)+3 \mathrm{CO}_{2}(g) $$ What are the equilibrium partial pressures of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) if \(\mathrm{CO}\) is the only gas present initially, at a partial pressure of \(0.978\) atm?

The following reaction is important in gold mining: $$ \begin{array}{r} 4 \mathrm{Au}(s)+8 \mathrm{CN}(a q)+\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \\ 4 \mathrm{Au}(\mathrm{CN})_{2}^{-}(a q)+4 \mathrm{OH}^{-}(a q) \end{array} $$ For a reaction mixture at equilibrium, in which direction would the reaction go to re-establish equilibrium after each of the following changes? (a) Adding gold (b) Increasing the hydroxide concentration (c) Increasing the partial pressure of oxygen (d) Adding \(\mathrm{Fe}^{3+}(a q)\), which reacts with cyanide to form \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q)\)

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