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Chapter 16: Problem 63

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

Short Answer

Expert verified
For a 0.400 M solution of hydrazoic acid (HN3), the percent ionization is calculated as follows: 1. x = \(\sqrt{(1.9 \times 10^{-5})(0.400)} \approx 8.68 \times 10^{-4}\) 2. Percent Ionization = \(\frac{8.68 \times 10^{-4}}{0.400}\) × 100% ≈ 0.217% For a 0.100 M solution of hydrazoic acid (HN3), the percent ionization is calculated as follows: 1. x = \(\sqrt{(1.9 \times 10^{-5})(0.100)} \approx 4.36 \times 10^{-4}\) 2. Percent Ionization = \(\frac{4.36 \times 10^{-4}}{0.100}\) × 100% ≈ 0.436%

Step by step solution

01

(General Formula for Percent Ionization)

(The formula for percent ionization is given by: Percent Ionization = \(\frac{[\text{H}^+]}{[\text{HA}]_\text{initial}}\) x 100%)
02

(Acid Ionization Reaction)

(Express the ionization of hydrazoic acid as an equilibrium reaction: \[ \mathrm{HN}_{3} \rightleftharpoons \mathrm{H}^{+} + \mathrm{N}_{3}^{-} \])
03

(Setting up the ICE Table)

(Establish an ICE table to keep track of the initial, change, and equilibrium concentrations of all species: ``` HN3 --> H+ + N3- Initial [C] 0 0 Change -x +x +x Equilib [C]-x x x ```)
04

(Expression for Ka)

(Write the expression for the equilibrium constant, Ka: \[K_a = \frac{[\mathrm{H}^+][\mathrm{N}_{3}^-]}{[\mathrm{HN}_3]}\])
05

(Substitution of Equilibrium Concentrations in Ka expression)

(Substitute the equilibrium concentrations from the ICE table into the Ka expression: \[K_a = \frac{x^2}{([C] - x})\]) Now, we will proceed to solve the given problems using the above methods.
06

(Percent Ionization for 0.400 M HN3)

(Given concentration, [HA] = 0.400 M, and Ka = 1.9 x 10^-5) 1. First, we will assume that x is very small compared to [C], hence we can approximate the denominator by [C]: \[K_a \approx \frac{x^2}{[C] }\] 2. Solve for x (the concentration of H+ ions at equilibrium): \[x^2 = K_a \cdot [C] \Rightarrow x = \sqrt{K_a \cdot [C]}\] 3. Calculate the percent ionization: Percent Ionization = \(\frac{x}{[HA]}\) × 100%
07

(Percent Ionization for 0.100 M HN3)

(Given concentration, [HA] = 0.100 M, and Ka = 1.9 x 10^-5) 1. First, we will assume that x is very small compared to [C], hence we can approximate the denominator by [C]: \[K_a \approx \frac{x^2}{[C] }\] 2. Solve for x (the concentration of H+ ions at equilibrium): \[x^2 = K_a \cdot [C] \Rightarrow x = \sqrt{K_a \cdot [C]}\] 3. Calculate the percent ionization: Percent Ionization = \(\frac{x}{[HA]}\) × 100%

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Most popular questions from this chapter

At \(50^{\circ} \mathrm{C}\), the ion-product constant for $\mathrm{H}_{2} \mathrm{O}\( has the value \)K_{w}=5.48 \times 10^{-14} \cdot(\mathbf{a})$ What is the \(\mathrm{pH}\) of pure water at \(50^{\circ} \mathrm{C} ?\) (b) Based on the change in \(K_{w}\) with temperature, predict whether \(\Delta H\) is positive, negative, or zero for the autoionization reaction of water: $$ 2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) $$

A particular sample of vinegar has a pH of 2.20 . If acetic acid is the only acid that vinegar contains \(\left(K_{a}=1.8 \times 10^{-5}\right)\), calculate the concentration of acetic acid in the vinegar.

The following observations are made about a diprotic acid $\mathrm{H}_{2} \mathrm{~A}:\( (i) \)\mathrm{A} 0.10 \mathrm{M}\( solution of \)\mathrm{H}_{2} \mathrm{~A}\( has \)\mathrm{pH}=3.30\(. (ii) \)\mathrm{A} 0.10 \mathrm{M}$ solution of the salt NaHA is acidic. Which of the following could be the value of \(\mathrm{p} K_{a 2}\) for \(\mathrm{H}_{2} \mathrm{~A}\) : (i) 3.22 , (ii) 5.30 , (iii) \(7.47,\) or (iv) \(9.82 ?\)

Calculate the pH of each of the following strong acid solutions: $(\mathbf{a}) 0.0178 \mathrm{M} \mathrm{HNO}_{3},(\mathbf{b}) 0.500 \mathrm{~g}$ of \(\mathrm{HClO}_{3}\) in \(5.00 \mathrm{~L}\) of solution, \((\mathbf{c}) 5.00 \mathrm{~mL}\) of $2.00 \mathrm{M} \mathrm{HCl}\( diluted to \)0.500 \mathrm{~L}$, (d) a mixture formed by adding \(75.0 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HCl}\) to \(125 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{HBr}\).

A \(0.25 M\) solution of a salt NaA has \(\mathrm{pH}=9.29 .\) What is the value of \(K_{a}\) for the parent acid HA?
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