Problem 49 - Chemistry: The Central Science in SI Units, Global Edition

Question

Write the chemical equation and the \(K_{a}\) expression for the dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: \((\mathbf{a}) \mathrm{HNO}_{2},\) (b) \(\mathrm{ClH}_{2} \mathrm{CCOOH}\).

Step-by-Step Solution

Answer

(a) For HNO₂ dissociation with H⁺ as a product: \[\mathrm{HNO}_{2}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{NO}_{2}^{-}(aq)\] \(K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{NO}_{2}^{-}]}{[\mathrm{HNO}_{2}]}\) With hydronium ion as a product: \[\mathrm{HNO}_{2}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(aq) + \mathrm{NO}_{2}^{-}(aq)\] \(K_{a} = \frac{[\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{NO}_{2}^{-}]}{[\mathrm{HNO}_{2}]}\) (b) For ClH₂CCOOH dissociation with H⁺ as a product: \[\mathrm{ClH}_{2}\mathrm{CCOOH}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{ClH}_{2}\mathrm{CCOO}^{-}(aq)\] \(K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{ClH}_{2}\mathrm{CCOO}^{-}]}{[\mathrm{ClH}_{2}\mathrm{CCOOH}]}\) With hydronium ion as a product: \[\mathrm{ClH}_{2}\mathrm{CCOOH}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(aq) + \mathrm{ClH}_{2}\mathrm{CCOO}^{-}(aq)\] \(K_{a} = \frac{[\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{ClH}_{2}\mathrm{CCOO}^{-}]}{[\mathrm{ClH}_{2}\mathrm{CCOOH}]}\)

Step 1: Writing the chemical equation for HNO₂ dissociation with H⁺ as a product

To write the chemical dissociation reaction for HNO₂ in aqueous solution, we can represent it in the following form: \[\mathrm{HNO}_{2}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{NO}_{2}^{-}(aq)\]

Step 2: Writing the chemical equation for HNO₂ dissociation with hydronium ion as a product

For the dissociation reaction of HNO₂ with hydronium ion as a product, we can write the equation as: \[\mathrm{HNO}_{2}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(aq) + \mathrm{NO}_{2}^{-}(aq)\]

Step 3: Writing the \(K_{a}\) expression for the dissociation of HNO₂

The \(K_{a}\) expression is the equilibrium constant for the dissociation reaction of an acid. When the dissociation occurs with \(\mathrm{H}^{+}(aq)\) as a product, the \(K_{a}\) expression for HNO₂ is given by: \[K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{NO}_{2}^{-}]}{[\mathrm{HNO}_{2}]}\] And with the hydronium ion as a product, the expression is: \[K_{a} = \frac{[\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{NO}_{2}^{-}]}{[\mathrm{HNO}_{2}]}\]

Step 4: Writing the chemical equation for ClH₂CCOOH dissociation with H⁺ as a product

For the dissociation reaction of ClH₂CCOOH in aqueous solution, we can write the equation as follows: \[\mathrm{ClH}_{2}\mathrm{CCOOH}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{ClH}_{2}\mathrm{CCOO}^{-}(aq)\]

Step 5: Writing the chemical equation for ClH₂CCOOH dissociation with hydronium ion as a product

To write the dissociation reaction of ClH₂CCOOH with the hydronium ion as a product, we can represent it in the following form: \[\mathrm{ClH}_{2}\mathrm{CCOOH}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(aq) + \mathrm{ClH}_{2}\mathrm{CCOO}^{-}(aq)\]

Step 6: Writing the \(K_{a}\) expression for the dissociation of ClH₂CCOOH

The \(K_{a}\) expression is the equilibrium constant for the dissociation reaction of an acid. When the dissociation occurs with \(\mathrm{H}^{+}(aq)\) as a product, the \(K_{a}\) expression for ClH₂CCOOH is given by: \[K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{ClH}_{2}\mathrm{CCOO}^{-}]}{[\mathrm{ClH}_{2}\mathrm{CCOOH}]}\] And with the hydronium ion as a product, the expression is: \[K_{a} = \frac{[\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{ClH}_{2}\mathrm{CCOO}^{-}]}{[\mathrm{ClH}_{2}\mathrm{CCOOH}]}\]

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a central concept in chemistry that describes the state in which the concentrations of all reactants and products remain constant over time. This occurs when the rate of the forward reaction, where reactants convert into products, equals the rate of the reverse reaction, where products convert back into reactants.
This dynamic balance means that, even though the reaction appears to have stopped, reactions are still occurring albeit at equal rates in both directions.

• It's important to realize that equilibrium does not imply equal concentrations of reactants and products.
• Instead, it reflects a steady state where no observable changes in concentration occur.

Understanding chemical equilibrium is crucial because it influences various chemical processes, such as reactions in a biological system or industrial applications. This concept applies directly to the dissociation of acids in aqueous solutions, where certain concentrations are maintained due to this balance.

Equilibrium Constant (Ka)

The equilibrium constant, specifically the acid dissociation constant represented as \(K_a\), is a special value that quantifies the extent of dissociation of an acid in water. It reflects how strongly an acid releases its protons or forms \[ ext{H}^+ \] ions in a solution. Higher \(K_a\) values indicate stronger acids that dissociate more completely. This is crucial for predicting the behavior of acids in various chemical systems.
To construct the \(K_a\) expression, you divide the product of the concentrations of the dissociated ions by the concentration of the undissociated acid:

• For an acid \[ ext{HA}(aq) ightleftharpoons ext{H}^+(aq) + ext{A}^-(aq) \] the \(K_a\) expression is:

\[ K_a = \frac{[ ext{H}^+][ ext{A}^-]}{[ ext{HA}]} \]

• If the dissociation involves the formation of \[ ext{H}_3 ext{O}^+ \] ions, as in \[ ext{HA}(aq) + ext{H}_2 ext{O}(l) ightleftharpoons ext{H}_3 ext{O}^+(aq) + ext{A}^-(aq) \], the expression remains conceptually similar:

\[ K_a = \frac{[ ext{H}_3 ext{O}^+][ ext{A}^-]}{[ ext{HA}]} \] This expression helps chemists understand how an acid will behave in different solution environments, aiding in predictions about pH and reaction direction.

Aqueous Solution Chemistry

Aqueous solution chemistry involves the study of substances dissolved in water, creating a solution where water is the solvent. This branch of chemistry is fundamental because many chemical reactions, biological processes, and industrial operations occur in water or involve water solutions.
In the context of acid dissociation, understanding how acids behave in aqueous solutions is key. Water is a polar solvent, which means it can effectively stabilize ions produced during acid dissociation. This property makes water an excellent medium for studying ionic dissociation and allows different acids to release protons (\( ext{H}^+ \)) or form hydronium ions (\( ext{H}_3 ext{O}^+ \)).

• Aqueous chemistry explains phenomena like the conductivity of solutions, which increases as ions dissociate.
• It also contributes to the understanding of pH, a measure of acidity, which is particularly vital in processes like digestion and brewing.

Thus, aqueous solution chemistry provides a framework for analyzing numerous essential biochemical and industrial processes involving aqueous media.