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Consider the reaction$$\mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{FeSCN}^{2+}(a q)$$.How will the equilibrium position shift if a. water is added, doubling the volume? b. \(\mathrm{AgNO}_{3}(a q)\) is added? (AgSCN is insoluble.) c. \(\mathrm{NaOH}(a q)\) is added? \(\left[\mathrm{Fe}(\mathrm{OH})_{3} \text { is insoluble. }\right]\) d. \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) is added? c. \(\mathrm{NaOH}(a q)\) is added?is insoluble.] d. \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) is added?

Short Answer

Expert verified
a. When water is added, doubling the volume, the concentration of all ions decreases, and the equilibrium shifts to the right, favoring the formation of \(\mathrm{FeSCN}^{2+}(a q)\). b. When \(\mathrm{AgNO}_{3}(a q)\) is added, the concentration of \(\mathrm{SCN}^-\) ions decreases due to the formation of insoluble \(\mathrm{AgSCN}\), and the equilibrium shifts to the left, favoring the formation of more \(\mathrm{SCN}^-\) ions. c. When \(\mathrm{NaOH}(a q)\) is added, the concentration of \(\mathrm{Fe}^{3+}\) ions decreases due to the formation of insoluble \(\mathrm{Fe}(\mathrm{OH})_{3}\), and the equilibrium shifts to the left, favoring the formation of more \(\mathrm{Fe}^{3+}\) ions. d. When \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) is added, the concentration of \(\mathrm{Fe}^{3+}\) ions increases, and the equilibrium shifts to the right, favoring the formation of more \(\mathrm{FeSCN}^{2+}(a q)\).

Step by step solution

01

Analyze the effect of volume change

Doubling the volume will dilute all ions in the reaction, effectively decreasing their concentration. According to Le Chatelier's Principle, the system will shift to counteract the change, which in this case means shifting towards the side of the reaction with more ions.
02

Determine the shift

Given the reaction, the products side has more ions than the reactants side. Therefore, the equilibrium will shift to the right, favoring the formation of \(\mathrm{FeSCN}^{2+}(a q)\). b. \(\mathrm{AgNO}_{3}(a q)\) is added
03

Analyze the effect of \(\mathrm{AgNO}_{3}(a q)\) addition

When \(\mathrm{AgNO}_{3}\) is added, \(\mathrm{Ag}^+\) ions will react with \(\mathrm{SCN}^-\) ions to form insoluble \(\mathrm{AgSCN}\). This reduces the concentration of \(\mathrm{SCN}^-\) ions.
04

Determine the shift

The decrease in \(\mathrm{SCN}^-\) ions concentration will cause the equilibrium to shift to the left, favoring the formation of more \(\mathrm{SCN}^-\) ions. c. \(\mathrm{NaOH}(a q)\) is added
05

Analyze the effect of \(\mathrm{NaOH}(a q)\) addition

When \(\mathrm{NaOH}\) is added, \(\mathrm{OH}^-\) ions will react with \(\mathrm{Fe}^{3+}\) ions to form insoluble \(\mathrm{Fe}(\mathrm{OH})_{3}\). This reduces the concentration of \(\mathrm{Fe}^{3+}\) ions.
06

Determine the shift

The decrease in \(\mathrm{Fe}^{3+}\) ions concentration will cause the equilibrium to shift to the left, favoring the formation of more \(\mathrm{Fe}^{3+}\) ions. d. \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) is added
07

Analyze the effect of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) addition

When \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) is added to the equilibrium, the concentration of \(\mathrm{Fe}^{3+}\) ions will increase.
08

Determine the shift

The increase in \(\mathrm{Fe}^{3+}\) ions concentration will cause the equilibrium to shift to the right, favoring the formation of more \(\mathrm{FeSCN}^{2+}(a q)\).

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry that describes how a system at equilibrium responds to changes. Imagine a chemical reaction that has reached equilibrium, meaning the rate of the forward reaction equals the rate of the reverse reaction. Now, if something disturbs this balance, how does the system respond?

According to Le Chatelier, if an external change is applied to a system at equilibrium, the system will adjust itself to counteract or "oppose" that change. There are several ways equilibrium can be disturbed:
  • Concentration changes
  • Pressure changes (for gases)
  • Temperature changes


For example, adding more reactants will shift the equilibrium to produce more products as the system tries to "use up" the added reactants. Conversely, removing a product will also shift the balance to the right to make more product. This intuitive behavior helps predict how chemical reactions will respond to various changes.
Equilibrium Shift
The concept of equilibrium shift describes the direction a reaction moves when disturbed. It's a way to understand how a reaction attempts to restore balance after a change. Consider the reaction: \[\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq) \]When the concentration of one part of the equilibrium changes, like adding more reactants or products, the entire system shifts to restore equilibrium.

  • Adding more \( \text{SCN}^- \) ions shifts the equilibrium to the right, producing more \( \text{FeSCN}^{2+} \).
  • Removing \( \text{Fe}^{3+} \) ions by reacting them with a base like \( \text{NaOH} \) would shift the balance to favor more \( \text{Fe}^{3+} \) formation.


In essence, an "equilibrium shift" involves the reaction moving left or right along the reaction equation to accommodate changes in concentration.
Precipitation Reactions
Precipitation reactions involve the formation of an insoluble compound from dissolved ions in a solution. They are crucial in many areas of chemistry and often occur when two aqueous solutions mix. The key to understanding precipitation reactions is recognizing that when certain ions combine, they can create an insoluble solid, or precipitate, that will settle out of solution.

Consider the following scenario:
  • When \( \text{AgNO}_3 \) is added to a solution containing \( \text{SCN}^- \) ions, \( \text{Ag}^+ \) ions react with \( \text{SCN}^- \) to form \( \text{AgSCN} \), which is a white, insoluble solid.
  • Similarly, adding \( \text{NaOH} \) to a solution with \( \text{Fe}^{3+} \) ions causes an insoluble compound, \( \text{Fe(OH)}_3 \), to form.


Precipitation reactions are not only fascinating to watch but they also serve as a practical way to remove impurities or recover specific elements from liquid mixtures. They play a vital role in qualitative chemical analysis as well as in industrial applications.

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Most popular questions from this chapter

Given \(K=3.50\) at \(45^{\circ} \mathrm{C}\) for the reaction $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)$$ and \(K=7.10\) at \(45^{\circ} \mathrm{C}\) for the reaction $$2 \mathrm{A}(g)+\mathrm{D}(g) \rightleftharpoons \mathrm{C}(g)$$what is the value of \(K\) at the same temperature for the reaction $$ \mathrm{C}(g)+\mathrm{D}(g) \rightleftharpoons 2 \mathrm{B}(g)$$.What is the value of \(K_{\mathrm{p}}\) at \(45^{\circ} \mathrm{C}\) for the reaction? Starting with 1.50 atm partial pressures of both \(\mathrm{C}\) and \(\mathrm{D},\) what is the mole fraction of B once equilibrium is reached?

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\).Calculate values of \(K\) for the following reactions at this temperature. a. \(\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\). b. \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) c. \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\). d. \(2 \mathrm{N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\).

"Old-fashioned "smelling salts" consist of ammonium carbon=ate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} .\) The reaction for the decomposition of ammomium carbonate $$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$,is endothermic. Would the smell of ammonia increase or decrease as the temperature is increased?

At \(125^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.25\) for the reaction $$2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$.A 1.00 -L flask containing \(10.0 \mathrm{g}\) NaHCO \(_{3}\) is evacuated and heated to \(125^{\circ} \mathrm{C}\) a. Calculate the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) after equilibrium is established. b. Calculate the masses of \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) present at equilibrium. c. Calculate the minimum container volume necessary for all of the \(\mathrm{NaHCO}_{3}\) to decompose.

At \(2200^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.050\) for the reaction $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$.What is the partial pressure of NO in equilibrium with \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) that were placed in a flask at initial pressures of 0.80 and 0.20 atm, respectively?

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