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Calculate the percent ionization of hydrazoic acid \((\mathrm{HN}_{3})\) in solutions of each of the following concentrations \((K_{a}\) is given in Appendix \(\mathrm{D} ) :(\mathbf{a}) 0.400 M (\mathbf{b}) 0.100 M,(\mathbf{c}) 0.0400 M\)

Short Answer

Expert verified
The percent ionization of hydrazoic acid \((\mathrm{HN}_{3})\) can be found using the equilibrium expression and the given \(K_{a}\) value of 1.9 × 10⁻⁵. For each concentration, percent ionization is calculated as: Percent ionization = \(\frac{[\mathrm{H^+}]_{\text{equilibrium}}}{[\mathrm{HN}_{3}]_{\text{initial}}} × 100\%\) After solving for the equilibrium concentration of \(\mathrm{H^+}\) ions (y) using the \(K_{a}\) expression, we find the percent ionization for each given concentration: - (a) 0.400 M: Percent ionization = \(1.94\%\) - (b) 0.100 M: Percent ionization = \(4.37\%\) - (c) 0.0400 M: Percent ionization = \(6.77\%\)

Step by step solution

01

Write the chemical equilibrium expression

First, we need to write the chemical equation for the ionization of hydrazoic acid and the corresponding equilibrium expression. For hydrazoic acid, the ionization reaction is: \[\mathrm{HN}_{3} \rightleftharpoons \mathrm{H^+} + \mathrm{N}_3^{-}\] The equilibrium expression for this reaction is given by: \[K_a = \frac{[\mathrm{H^+}][\mathrm{N}_{3}^{-}]}{[\mathrm{HN}_3]}\]
02

Set up an ICE table to solve for the concentrations of ions

Next, we will set up an ICE (Initial, Change, Equilibrium) table to find the concentrations of \(\mathrm{H^+}\) and \(\mathrm{N}_{3}^{-}\) ions at equilibrium. Based on the initial concentrations given in the exercise, the table should look as follows: | | \(\mathrm{HN}_{3}\) | \(\mathrm{H^+}\) | \(\mathrm{N}_{3}^-\) | |---|---------------|-----------|----------------| | I | x | 0 | 0 | | C | -y | +y | +y | | E | x-y | y | y | Here, x is the initial concentration of hydrazoic acid, and y is the change in concentration during the ionization process.
03

Plug the equilibrium concentrations into the \(K_{a}\) expression

Now, we will plug the equilibrium concentrations from our ICE table into the \(K_{a}\) expression: \[K_a = \frac{y^2}{x-y}\] Since \(K_{a}\) for hydrazoic acid is small, we can assume that the change in concentration (y) is also small compared to the initial concentration (x). Therefore, we can simplify our \(K_{a}\) expression into: \[K_a = \frac{y^2}{x}\]
04

Obtain \(K_{a}\) value for hydrazoic acid and solve for y

From Appendix D, the value of \(K_{a}\) for hydrazoic acid is 1.9 × 10⁻⁵. Now we would need to find y for each given concentration (x): - (a) x = 0.400 M - (b) x = 0.100 M - (c) x = 0.0400 M By plugging in the \(K_{a}\) value and the initial concentrations (x) into the simplified \(K_{a}\) expression above, we can solve for y in each case.
05

Calculate the percent ionization

Once we have the concentration of \(\mathrm{H^+}\) ions (y) at equilibrium, we can calculate the percent ionization of hydrazoic acid using the following formula: Percent ionization = \(\frac{[\mathrm{H^+}]_{\text{equilibrium}}}{[\mathrm{HN}_{3}]_{\text{initial}}} × 100\%\) Plugging in the calculated values for y and the initial concentrations of hydrazoic acid for each case, we can calculate the percent ionization for each concentration.

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Key Concepts

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

Hydrazoic Acid
Hydrazoic acid, represented chemically as \( ext{HN}_3\), is a weak acid. This means it only partially ionizes in water. A weak acid, like hydrazoic acid, does not completely release its hydrogen ions into solution, which is critical for calculating its percent ionization.
Understanding the ionization behavior of hydrazoic acid is essential for solving chemical equilibrium problems involving this compound. In its aqueous solution, it reaches an equilibrium state where only a fraction of the \( ext{HN}_3\) molecules dissociate into hydrogen ions \( ext{H}^+\) and azide ions \( ext{N}_3^-\).
This partial dissociation is influenced by the acid's ionization constant \(K_a\), which measures the acid's strength. For hydrazoic acid, \(K_a\) is a small value, indicating its relatively low ability to release \( ext{H}^+\) ions in comparison to strong acids.
Equilibrium Expression
To understand the behavior of weak acids, it's crucial to grasp the concept of the equilibrium expression. For hydrazoic acid, when it ionizes, the chemical equation is: \[\text{HN}_3 \rightleftharpoons \text{H}^+ + \text{N}_3^-\]This reversible chemical equation reaches a point where the forward and reverse reactions occur at equal rates, establishing a dynamic balance or equilibrium.
The equilibrium expression is formulated based on this chemical equilibrium: \[K_a = \frac{[\text{H}^+][\text{N}_3^-]}{[\text{HN}_3]}\]Here, \(K_a\) is the equilibrium constant for the ionization of hydrazoic acid, and the brackets \([...]\) denote the concentration of each species in moles per liter (M).
This expression helps predict the extent of ionization and is pivotal in determining concentrations of ions at equilibrium, which is necessary for calculating the percent ionization.
ICE Table
An ICE table is an organized way to calculate the changes in concentration during a chemical reaction at equilibrium. ICE stands for Initial, Change, and Equilibrium. It's an invaluable tool when dealing with weak acids like hydrazoic acid.
For hydrazoic acid's ionization:
  • Initial (I): Start with the initial concentration of \(\text{HN}_3\) and assume \([\text{H}^+]\) and \([\text{N}_3^-]\) are initially zero.
  • Change (C): As the reaction progresses towards equilibrium, note the amount of \(\text{HN}_3\) that ionizes, represented by \(-y\), and the amount of \(\text{H}^+\) and \(\text{N}_3^-\) that forms, represented by \(+y\).
  • Equilibrium (E): Record the concentrations at equilibrium: \([\text{HN}_3] = x - y\), \([\text{H}^+] = y\), and \([\text{N}_3^-] = y\).
Using the ICE table helps set up the equilibrium expression systematically, aiding in the calculation of equilibrium concentrations.
Chemical Equilibrium
Chemical equilibrium is a state where the concentrations of reactants and products remain constant over time in a closed system. For weak acids like hydrazoic acid, achieving equilibrium involves establishing a balance between ionized and unionized forms.
At equilibrium, the rate of forward reaction (ionization) is equal to the rate of the reverse reaction (recombination). This balance allows for the calculation of important parameters such as \([\text{H}^+]\) and \([\text{N}_3^-]\) for percent ionization.
The position of equilibrium is described quantitatively by the equilibrium constant \(K_a\). Understanding chemical equilibrium is crucial, as it reflects how the initial concentration of hydrazoic acid affects the proportion of the acid that ionizes. This understanding is pivotal for calculating and predicting the behavior of acidic solutions.

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Most popular questions from this chapter

Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) Acid strength in a series of \(\mathrm{H}-\) A molecules increases with increasing size of \(\mathrm{A} .\) (b) For acids of the same general structure but differing electronegativities of the central atoms, acid strength decreases with increasing electronegativity of the central atom.(c) The strongest acid known is HF because fluorine is the most electronegative element.

Identify the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each equation, and also identify the conjugate acid and conjugate base of each on the right side. (a) \(\mathrm{HBrO}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{BrO}^{-}(a q)\) (b) \(\mathrm{HSO}_{4}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\) (c) \(\mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{BrO}^{-}\) or \(\mathrm{ClO}^{-},(\mathbf{b}) \mathrm{BrO}^{-}\) or \(\mathrm{BrO}_{2}^{-}\) (c) \(\mathrm{HPO}_{4}^{2-}\) or \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}.\)

(a) Give the conjugate base of the following Bronsted-Lowry acids: (i) HCOOH, (ii) \(\mathrm{HPO}_{4}^{2-} .\) (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) SO \(_{4}^{2-}\) (ii) \(\mathrm{CH}_{3} \mathrm{NH}_{2} .\)

Identify the Lewis acid and Lewis base among the reactants in each of the following reactions: (a) \(\operatorname{Fe}\left(\mathrm{ClO}_{4}\right)_{3}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}(a q)+3 \mathrm{ClO}_{4}^{-}(a q)\) (b) \(\mathrm{CN}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q)\) (c)\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(g)+\mathrm{BF}_{3}(g) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NBF}_{3}(s)\) (d)\(\operatorname{HIO}(l q)+\mathrm{NH}_{2}^{-}(l q) \Longrightarrow \mathrm{NH}_{3}(l q)+\mathrm{IO}^{-}(l q)\) (\(lq\) denotes liquid ammonia as solvent)

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