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Chapter 16: Problem 63

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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Most popular questions from this chapter

\(\mathrm{NH}_{3}(g)\) and \(\mathrm{HCl}(g)\) react to form the ionic solid \(\mathrm{NH}_{4} \mathrm{Cl}(s) .\) Which substance is the Bronsted-Lowry acid in this reaction? Which is the Bronsted-Lowry base?

Label each of the following as being a strong acid, a weak acid, or a species with negligible acidity. In each case write the formula of its conjugate base, and indicate whether the conjugate base is a strong base, a weak base, or a species with negligible basicity: \((\mathbf{a}) \mathrm{HCOOH}\), \((\mathbf{b})\mathrm{H}_{2},(\mathrm{c}) \mathrm{CH}_{4},(\mathbf{d}) \mathrm{HF},(\mathbf{e}) \mathrm{NH}_{4}^{+}\)

Which of the following statements is false? (a) An Arrhenius base increases the concentration of OH \(^{-}\) in water. (b) A Bronsted-Lowry base is a proton acceptor. (c) Water can act as a Bronsted-Lowry acid. (d) Water can act as a Bronsted-Lowry base. (e) Any compound that contains an \(-\)OH group acts as a Bronsted-Lowry base.

(a) Using dissociation constants from Appendix D, determine the value for the equilibrium constant for each of the following reactions. \((\mathrm{i}) \mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{co}_{3^{-}(a q)}+\mathrm{H}_{2} \mathrm{O}(l)\) (ii) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{HCO}_{3}^{-}(a q)\) (b) We usually use single arrows for reactions when the for- ward reaction is appreciable \((K\) much greater than 1\()\) equilibrium is never established. If we follow this convention, which of these equilibria might be written with a single arrow?

Arrange the following 0.10\(M\) solutions in order of increasing acidity: (i) \(\mathrm{NH}_{4} \mathrm{NO}_{3},(\mathrm{ii}) \mathrm{NaNO}_{3},(\mathrm{iii}) \mathrm{CH}_{3} \mathrm{COONH}_{4},(\mathrm{iv})\) \(\mathrm{NaF},(\mathrm{v}) \mathrm{CH}_{3} \mathrm{COONa}.\)
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