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Chapter 17: Problem 17

(a) Calculate the percent ionization of \(0.0075 \mathrm{M}\) butanoic acid \(\left(K_{a}=1.5 \times 10^{-5}\right) .\) (b) Calculate the percent ionization of \(0.0075 \mathrm{M}\) butanoic acid in a solution containing \(0.085 \mathrm{M}\) sodium butanoate.

Short Answer

Expert verified
(a) In pure water, solving the Ka equation gives the equilibrium concentration of H+ ions as \(x = 6.1 \times 10^{-4} \mathrm{M}\). The percent ionization is \( \frac{6.1 \times 10^{-4} \mathrm{M}}{0.0075 \mathrm{M}} \times 100 \approx 8.1 \% \). (b) In the sodium butanoate solution, solving the Ka equation gives the equilibrium concentration of H+ ions as \(x = 3.08 \times 10^{-5} \mathrm{M}\). The percent ionization is \( \frac{3.08 \times 10^{-5} \mathrm{M}}{0.0075 \mathrm{M}} \times 100 \approx 0.41 \% \).

Step by step solution

01

(1) Write the chemical equilibrium

For butanoic acid, the dissociation in water can be represented by the following equation: \( HA\; \leftrightharpoons\; H^{+} + A^{-} \) Where \(HA\) stands for butanoic acid and \(A^-\) represents the conjugate base (butanoate ion).
02

(2) Set up an ICE table

To solve this problem, we will use an ICE (Initial, Change, Equilibrium) table to track the concentrations of each species involved in the reaction. Pure water: | | HA | H+ | A- | | --- | --- | --- | --- | | I | 0.0075 | 0 | 0 | | C | -x | +x | +x | | E | 0.0075-x | x | x | Sodium butanoate solution: | | HA | H+ | A- | | --- | --- | --- | --- | | I | 0.0075 | 0 | 0.085 | | C | -x | +x | +x | | E | 0.0075-x | x | 0.085+x |
03

(3) Solve for x using the Ka expression

The Ka value is given as \(1.5 \times 10^{-5}\) for butanoic acid. We will use this to solve for the equilibrium concentration of hydrogen ions (H+). For pure water: \( K_{a} = \frac{[H^{+}][A^{-}]}{[HA]} \) \( 1.5 \times 10^{-5} = \frac{x \times x}{0.0075 - x} \) For sodium butanoate solution: \(1.5 \times 10^{-5} = \frac{x \times (0.085 + x)}{0.0075 - x}\) Solving these equations will give us the equilibrium concentration of H+ ions.
04

(4) Calculate percent ionization

Percent ionization is the ratio of the concentration of ionized acid to the initial concentration of the acid, multiplied by 100. We will calculate it as follows: Percent ionization in pure water: \( \frac{[H^{+}]}{[HA]} \times 100 \) Percent ionization in sodium butanoate solution: \( \frac{[H^{+}]}{[HA]} \times 100 \) Solve these expressions to find the percent ionization in both cases.

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Most popular questions from this chapter

(a) Consider the equilibrium \(\mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q) .\) Using Le Châtelier's principle, explain the effect of the presence of a salt of \(\mathrm{HB}^{+}\) on the ionization of B. (b) Give an example of a salt that can decrease the ionization of \(\mathrm{NH}_{3}\) in solution.

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A buffer contains a weak acid, HX, and its conjugate base. The weak acid has a \(\mathrm{p} K_{a}\) of \(4.5,\) and the buffer has a \(\mathrm{pH}\) of \(4.3 .\) Without doing a calculation, predict whether \([\mathrm{HX}]=\left[\mathrm{X}^{-}\right]\) \([\mathrm{HX}]>\left[\mathrm{X}^{-}\right],\) or \([\mathrm{HX}]<\left[\mathrm{X}^{-}\right] .\) Explain. \([\) Section 17.2\(]\)
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