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The following two equilibrium reactions can be written for aqueous carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) $$ \begin{array}{ll} \mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{HCO}_{3}^{-}(\mathrm{aq}) & K_{1} \\ \mathrm{HCO}_{3}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}(\mathrm{aq}) & K_{2} \end{array} $$ For each reaction write the equilibrium constant expression. By using Le Châtelier's principle we may naively predict that by adding \(\mathrm{H}_{2} \mathrm{CO}_{3}\) to the system, the concentration of \(\mathrm{CO}_{3}^{2-}\) would increase. What we observe is that after adding \(\mathrm{H}_{2} \mathrm{CO}_{3}\) to the equilibrium mixture, an increase in the concentration of \(\mathrm{CO}_{3}^{2-}\) occurs when \(\left[\mathrm{CO}_{3}^{2-}\right] \ll \mathrm{K}_{2}\) however, the concentration of \(\mathrm{CO}_{3}^{2-}\) will decrease when \(\left[\mathrm{CO}_{3}^{2-}\right] \gg K_{2} .\) Show that this is true by considering the ratio of \(\left[\mathrm{H}^{+}\right] /\left[\mathrm{HCO}_{3}^{-}\right]\) before and after adding a small amount of \(\mathrm{H}_{2} \mathrm{CO}_{3}\) to the solution, and by using that ratio to calculate the \(\left[\mathrm{CO}_{3}^{2-}\right]\)

Short Answer

Expert verified
The equilibrium constant expressions are \(K_1 = \frac{{[H^{+}][HCO_{3}^{-}]}}{{[H_{2}CO_{3}]}}\) and \(K_2 = \frac{{[H^{+}][CO_{3}^{2-}]}}{{[HCO_{3}^{-}]}}\). The ratio of \(H^{+}\) to \(HCO_{3}^{-}\) before and after the addition of \(H_2CO_{3}\) will determine if the concentration of \(CO_{3}^{2-}\) increases or decreases. If the concentration of \(CO_{3}^{2-}\) is less than \(K_2\), the concentration increases. If it is larger, the concentration decreases.

Step by step solution

01

Establish Equilibrium Constant Expressions

The equilibrium constant expressions for the given reactions can be written as \(K_1 = \frac{{[H^{+}][HCO_3^{-}]}}{{[H_{2}CO_{3}]}}\) and \(K_2 = \frac{{[H^{+}][CO_{3}^{2-}]}}{{[HCO_{3}^{-}]}}\).
02

Derive the Concentration Ratio

The concentration ratio is expressed by the concentration of \(H^{+}\) over the concentration of \(HCO_3^{-}\). Substituting \(K_1\) and \(K_2\) in this equilibrium we get: \(\frac{{[H^{+}]}}{{[HCO_3^{-}]} = \frac{{[H_{2}CO_{3}]}}{{[HCO_{3}^{-}]}}\cdot K_1 = K_2 \cdot [CO_{3}^{2-}]\)
03

Predict the Concentration Changes

Applying Le Châtelier's Principle: As we add \(H_2CO_3\), the system will react to consume it, shifting the reaction to the right. However, this shift can also affect the second reaction. By substituting our established concentration ratio into the equation, we get \(\frac{{[H^{+}]}}{{[HCO_{3}^{-}]} = K_2 \cdot [CO_{3}^{2-}]\). This means the concentration of \(CO_3^{2-}\) is proportional to this ratio. If \(CO_{3}^{2-} << K_{2}\), increasing this ratio will increase \(CO_{3}^{-2}\) concentrations. On the other hand, if \(CO_{3}^{2-} >> K_{2}\), same increase will make the \(CO_{3}^{2-}\) concentrations decrease.

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Key Concepts

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

Equilibrium Constant Expression
Understanding the equilibrium constant expression is crucial when studying chemical equilibria. Take the example of carbonic acid in water, which can dissociate in two steps, each characterized by its own equilibrium constant, namely K1 and K2. The equilibrium constant for a reaction represents the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients.

For the first dissociation of carbonic acid, the equilibrium constant expression is written as:
\(K_1 = \frac{{[H^+][HCO_3^-]}}{{[H_2CO_3]}}\).
Similarly, for the second dissociation, we write:
\(K_2 = \frac{{[H^+][CO_{3}^{2-}]}}{{[HCO_{3}^-]}}\).
These expressions are pivotal because they tell us the concentration of ions we can expect at equilibrium in a given solution. If you encounter a situation where you need to calculate the concentrations at equilibrium, just remember these expressions and apply them accordingly!
Le Châtelier's Principle
Le Châtelier's principle is like a rule of thumb that predicts how a system at equilibrium will respond to disturbances, such as changes in concentration, pressure, or temperature. The principle states that if an external change is imposed on a system at equilibrium, the system will adjust in a way that counteracts the change and a new equilibrium will be established.

When carbonic acid is added to an equilibrium mixture of its dissociated forms, Le Châtelier's principle suggests the system will shift to reduce this addition by moving the reaction towards the formation of more products. Essentially, the equilibrium shifts to the right, meaning more HCO3- and CO32- ions are formed. The precise effect on the concentration of carbonate ions depends on their initial concentration relative to the equilibrium constant, demonstrating the nuanced predictions made possible by this principle.
Aqueous Carbonic Acid Reactions
Aqueous reactions of carbonic acid, such as the ones shown in our exercise, play a significant role in natural processes like the carbon cycle and the buffering of blood in our bodies. Carbonic acid (H2CO3) is weak and tends to dissociate into bicarbonate (HCO3-) and hydrogen ions (H+) in its first dissociation step. The bicarbonate can further dissociate into carbonate ions (CO32-) and additional hydrogen ions.

The reversibility of these reactions enables them to reach a state of dynamic equilibrium, where the rates of the forward and reverse reactions are equal, maintaining constant relative concentrations of all species involved. The subsequent behavior of these species upon perturbation is governed by Le Châtelier's principle, which ensures that the system works to restore equilibrium.
Concentration Ratio Calculation
Calculating concentration ratios is a methodological approach to quantifying the relationship between species in equilibrium. It provides a way to understand how one concentration affects another. In the context of the dissociation of carbonic acid, we particularly look at the ratio of H+ to HCO3- before and after adding carbonic acid.

The ratio is defined by the established equilibrium constants:
\(\frac{{[H^+]}}{{[HCO_3^-]}} = \frac{{[H_{2}CO_{3}]}}{{[HCO_{3}^-]}} \times K_1 = K_2 \times [CO_{3}^{2-}]\).
This calculation ultimately dictates the concentration of carbonate ions present in the solution. Depending on whether the carbonate ions are much less or much greater than the equilibrium constant, the direction in which their concentration will shift varies. This insightful calculation helps bridge the gap between the theoretical predictions of Le Châtelier's principle and the actual outcomes observed in the laboratory.

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Most popular questions from this chapter

In the gas phase, iodine reacts with cyclopentene \(\left(\mathrm{C}_{5} \mathrm{H}_{8}\right)\) by a free radical mechanism to produce cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) and hydrogen iodide. Explain how each of the following affects the amount of \(\mathrm{HI}(\mathrm{g})\) present in the equilibrium mixture in the reaction \begin{array}{r} \mathrm{I}_{2}(\mathrm{g})+\mathrm{C}_{5} \mathrm{H}_{8}(\mathrm{g}) \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{6}(\mathrm{g})+2 \mathrm{HI}(\mathrm{g}) \\ \Delta H^{\circ}=92.5 \mathrm{kJ} \end{array} (a) Raising the temperature of the mixture; (b) introducing more \(\mathrm{C}_{5} \mathrm{H}_{6}(\mathrm{g}) ;\) (c) doubling the volume of the container holding the mixture; (d) adding an appropriate catalyst; (e) adding an inert gas such as He to a constant-volume reaction mixture.

A classic experiment in equilibrium studies dating from 1862 involved the reaction in solution of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) to produce ethyl acetate and water. $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O}$$ The reaction can be followed by analyzing the equilibrium mixture for its acetic acid content. $$\begin{array}{r} 2 \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightleftharpoons \\ \mathrm{Ba}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In one experiment, a mixture of 1.000 mol acetic acid and 0.5000 mol ethanol is brought to equilibrium. A sample containing exactly one-hundredth of the equilibrium mixture requires \(28.85 \mathrm{mL} 0.1000 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. Calculate the equilibrium constant, \(K_{c}\), for the ethanol-acetic acid reaction based on this experiment.

An equilibrium mixture of \(\mathrm{SO}_{2}, \mathrm{SO}_{3},\) and \(\mathrm{O}_{2}\) gases is maintained in a \(2.05 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=35.5\) for the reaction $$2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})$$ (a) If the numbers of moles of \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\) in the flask are equal, how many moles of \(\mathrm{O}_{2}\) are present? (b) If the number of moles of \(\mathrm{SO}_{3}\) in the flask is twice the number of moles of \(\mathrm{SO}_{2}\), how many moles of \(\mathrm{O}_{2}\) are present?

The formation of nitrosyl chloride is given by the following equation: \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g})\) \(K_{\mathrm{c}}=4.6 \times 10^{4}\) at \(298 \mathrm{K} .\) In a \(1.50 \mathrm{L}\) flask, there are \(4.125 \mathrm{mol}\) of \(\mathrm{NOCl}\) and \(0.1125 \mathrm{mol}\) of \(\mathrm{Cl}_{2}\) present at equilibrium (298 K). (a) Determine the partial pressure of \(\mathrm{NO}\) at equilibrium. (b) What is the total pressure of the system at equilibrium?

Given the equilibrium constant values $$\begin{aligned} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(\mathrm{g}) \quad K_{\mathrm{c}}=2.7 \times 10^{-18} \\ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) & \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) K_{\mathrm{c}}=4.6 \times 10^{-3} \\ \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.1 \times 10^{-9} \end{aligned}$$ Determine a value of \(K_{\mathrm{c}}\) for the reaction $$ 2 \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) $$

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