One way to determine the predominant species at equilibrium for an acid-base reaction is to say that the reaction arrow points to the acid with the higher value of \(\mathrm{p} K_{a}\). For example, $$ \begin{array}{cr} \mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \longleftarrow \mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+} \\ \mathrm{p} K_{\mathrm{a}} 9.24 & \mathrm{p} K_{\mathrm{a}}-1.74 \\ \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{p} K_{\mathrm{a}} 9.24 & \mathrm{p} K_{\mathrm{a}} 15.7 \end{array} $$ Explain why this rule works.

Short Answer

Expert verified
Answer: The rule helps to determine the predominant species at equilibrium because it is a shortcut for comparing the strengths of the acids involved in the reaction. By pointing towards the acid with the higher pKa value, we are indicating the side containing the weaker acid, which is favored in the equilibrium. Consequently, the predominant species will be those on the side with the weaker acid as determined by the higher pKa value.

Step by step solution

01

Understand acid-base equilibria

In an acid-base equilibrium, the reaction involves an acid, which donates a hydrogen ion (H+) to the base, which then accepts the hydrogen ion. The strength of an acid or a base is determined by its ability to donate or accept hydrogen ions respectively. A stronger acid will more readily donate a hydrogen ion than a weaker acid, which means that the equilibrium will favor the side of the weaker acid. The pKa value is useful in comparing the strength of acids: the lower the pKa value, the stronger the acid.
02

Relate pKa values to reaction arrows

Now that we have established that lower pKa values indicate stronger acids and higher pKa values indicate weaker acids, we can see how this relationship is connected to the placement of reaction arrows. The rule states that the reaction arrow will point to the acid with the higher pKa value, which means that the reaction arrow points to the side of the weaker acid, meaning the side that is favored in the equilibrium reaction.
03

Apply the rule to the example equations

In the given example, we have two equations: 1. \(NH_4^+ + H_2O \longleftarrow NH_3 + H_3O^+\) \(pK_a~9.24 \qquad pK_a~-1.74\) 2. \(NH_4^+ + OH^- \longrightarrow NH_3 + H_2O\) \(pK_a~9.24 \qquad pK_a~15.7\) In the first equation, the reaction arrow points to the left, towards the \(NH_4^+\) with a higher pKa value of 9.24, indicating that this reaction favors the production of the weaker acid, \(NH_4^+\). In the second equation, the reaction arrow points to the right, towards \(H_2O\) with a higher pKa value of 15.7, indicating that this reaction favors the side of the weaker acid, \(H_2O\).
04

Conclude with an explanation for why the rule works

This rule works because it is essentially a shortcut for determining the favored side of an acid-base equilibrium reaction. By assessing whether the reaction arrow points to the acid with the higher or lower pKa value, we are comparing the strengths of the acids involved and determining which side of the reaction is favored. In this way, we can more easily determine the predominant species at equilibrium - the side containing the weaker acid, as indicated by the higher pKa value, will be favored.

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Key Concepts

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

Understanding pKa Values
A crucial concept in acid-base chemistry is the pKa value, which is inherently linked to the strengths of acids. The pKa is the negative logarithm of the acid dissociation constant (Ka) and it measures how readily an acid donates its hydrogen ion in a solution. To put it simply, the lower the pKa value of an acid, the stronger that acid is, and vice versa.

In practical terms, pKa values serve as a numerical scale to compare the relative strengths of different acids. They are essential for predicting the direction of acid-base equilibrium reactions. If you have two acids in a reaction, you can predict that at equilibrium, the solution will predominantly contain the weaker acid, which has the higher pKa value. Therefore, when evaluating a reaction, knowing the pKa values allows you to determine which side is more stable and consequently favored in an acid-base equilibrium.
Assessing Acid Strength
The acid strength can influence a variety of chemical and biological processes, making its understanding essential for predicting the outcome of acid-base reactions. Generally, the strength of an acid is gauged by its ability to lose a proton (H+). The dissociation of an acid in water is a prime example; a strong acid such as hydrochloric acid (HCl) will almost completely dissociate into H+ and Cl-, displaying a low pKa value.

On the other hand, a weak acid like acetic acid doesn't dissociate completely and keeps most of its hydrogen atoms, indicating a higher pKa value. This concept is vital when assessing chemical reactions because the stronger acid tends to drive the reaction toward the production of the weaker acid, achieving a state of equilibrium. This understanding provides a window into the acid-base properties of substances and helps students grasp why certain reactions proceed in the direction they do.
Equilibrium Reaction Dynamics
Equilibrium reactions are a core concept in understanding not just acid-base chemistry, but all reversible chemical reactions. An equilibrium reaction is characterized by its reversible nature, where the forward and reverse reactions occur at the same rate, leading to a constant ratio of products to reactants over time.

This dynamic in acid-base chemistry is especially interesting as it involves the transfer of protons between species. In any given acid-base equilibrium reaction, the position of equilibrium is established by the relative strengths of the acids and bases involved. The equilibrium will favor the formation of the weaker acid and base, as they are less inclined to donate or accept protons, respectively. The principle of this equilibrium can be intuitively understood through Le Chatelier's principle, which states that a system at equilibrium will adjust to counteract any changes imposed on it, again favoring formation of the weaker acid and base.

This concept essentially governs the balance of the reaction, ensuring that a stable state is achieved where neither the forward nor the backward reaction is favored over the other in the long term.

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Most popular questions from this chapter

For each conjugate acid-base pair, identify the first species as an acid or base and the second species as its conjugate acid or base. In addition, draw Lewis structures for each species, showing all valence electrons and any formal charge. (a) \(\mathrm{HCOOH} \mathrm{HCOO}^{-}\) (b) \(\mathrm{NH}_{4}^{+} \mathrm{NH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-} \quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{HCO}_{3}{ }^{-} \quad \mathrm{CO}_{3}^{2-}\) (e) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} \mathrm{HPO}_{4}{ }^{2-}\) (f) \(\mathrm{CH}_{3} \mathrm{CH}_{3} \mathrm{CH}_{3} \mathrm{CH}_{2}{ }^{-}\) (g) \(\mathrm{CH}_{3} \mathrm{~S}^{-} \quad \mathrm{CH}_{3} \mathrm{SH}\)

4-Methylphenol, \(\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{OH}\left(\mathrm{p} K_{\mathrm{a}} 10.26\right)\), is only slightly soluble in water, but its sodium salt, \(\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}^{-} \mathrm{Na}^{+}\), is quite soluble in water. In which solution(s) will 4-methylphenol dissolve? (a) Aqueous \(\mathrm{NaOH}\) (b) Aqueous \(\mathrm{NaHCO}_{3}\) (c) Aqueous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\)

For each conjugate acid-base pair, identify the first species as an acid or base and the second species as its conjugate acid or conjugate base. In addition, draw Lewis structures for each species, showing all valence electrons and any formal charges. (a) \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{HSO}_{4}{ }^{-}\) (b) \(\mathrm{NH}_{3}, \mathrm{NH}_{2}{ }^{-}\) (c) \(\mathrm{CH}_{3} \mathrm{OH}, \mathrm{CH}_{3} \mathrm{O}^{-}\)

In acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), the \(\mathrm{OH}\) hydrogen is more acidic than the \(\mathrm{CH}_{3}\) hydrogens. Explain.

Acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), is a weak organic acid, \(\mathrm{p} K_{\mathrm{a}} 4.76\). Write an equation for the equilibrium reaction of acetic acid with each base. Which equilibria lie considerably toward the left? Which lie considerably toward the right? (a) \(\mathrm{NaHCO}_{3}\) (b) \(\mathrm{NH}_{3}\) (c) \(\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{NaOH}\)

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