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Chapter 14: Problem 69

For propanoic acid $\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right),$ determine the concentration of all species present, the \(\mathrm{pH},\) and the percent dissociation of a \(0.100-M\) solution.

Short Answer

Expert verified
The concentration of all species in the $0.100$ M propanoic acid solution are as follows: $[HC_3H_5O_2] = 0.0996$ M, $[C_3H_5O_2^-] = 3.6 \times 10^{-4}$ M, and $[H^+] = 3.6 \times 10^{-4}$ M. The pH of the solution is $3.44$, and the percent dissociation of propanoic acid is $0.36 \%$.

Step by step solution

01

Write the dissociation equation for propanoic acid

Propanoic acid dissociates into propanoate ions and hydrogen ions as follows: \[HC_3H_5O_2 \rightleftharpoons C_3H_5O_2^- + H^+\]
02

Use the ICE table to determine the change in concentration

Let's set up the ICE table for propanoic acid dissociation: | | HC3H5O2 | C3H5O2- | H+ | |-------|---------|---------|-----| |Initial| 0.100 | 0 | 0 | |Change | -x | +x | +x | |Equil. | 0.100-x | x | x | Here, x represents the change in concentration of each species during the dissociation process.
03

Write the expression for Ka and solve for x

The expression for Ka is: \[K_a = \frac{[C_3H_5O_2^-][H^+]}{[HC_3H_5O_2]}\] Substitute the equilibrium concentration values from the ICE table and the given Ka value: \[1.3 \times 10^{-5} = \frac{x \times x}{0.100 - x}\] We can assume that x is small compared to 0.100, so 0.100 - x is approximately 0.100. Therefore, the equation becomes: \[1.3 \times 10^{-5} = \frac{x^2}{0.100}\] Solve for x: \[x = \sqrt{1.3 \times 10^{-5} \times 0.100} = 3.6 \times 10^{-4}\]
04

Calculate the concentration of all species present

Now we can find the equilibrium concentration of each species by plugging x back into the ICE table: | | HC3H5O2 | C3H5O2- | H+ | |-------|---------|---------|-----| |Initial| 0.100 | 0 | 0 | |Change | -x | +x | +x | |Equil. | 0.100-x | x | x | | | HC3H5O2 | C3H5O2- | H+ | |-------|---------|---------|-----| |Equil. | 0.100-3.6×10⁻⁴ | 3.6×10⁻⁴ | 3.6×10⁻⁴ | Equilibrium concentrations: HC3H5O2: 0.100 - 3.6 × 10⁻⁴ = 0.0996 M C3H5O2-: 3.6 × 10⁻⁴ M H+: 3.6 × 10⁻⁴ M
05

Determine the pH of the solution

The pH of the solution can be calculated using the following formula: \[pH = -\log[H^+]\] \[pH = -\log(3.6 \times 10^{-4}) = 3.44\]
06

Calculate the percent dissociation

Percent dissociation is the ratio of the dissociated acid concentration to the initial acid concentration, multiplied by 100: \[% \thinspace dissociation = \frac{[H^+]}{[HC_3H_5O_2]_{initial}} \times 100\] \[% \thinspace dissociation = \frac{3.6 \times 10^{-4}}{0.100} \times 100 = 0.36\%\] The concentration of all species in the solution are as follows: [HC3H5O2] = 0.0996 M, [C3H5O2-] = 3.6 × 10⁻⁴ M, and [H+] = 3.6 × 10⁻⁴ M. The pH of the solution is 3.44, and the percent dissociation of propanoic acid is 0.36%.

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Most popular questions from this chapter

Place the species in each of the following groups in order of increasing acid strength. a. $\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{S}, \mathrm{H}_{2} \mathrm{Se}\( (bond energies: \)\mathrm{H}-\mathrm{O}, 467 \mathrm{kJ} / \mathrm{mol}$ $\mathrm{H}-\mathrm{S}, 363 \mathrm{kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{kJ} / \mathrm{mol} )$ b. $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}$ c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}^{+}\) d. \(\mathrm{NH}_{4}^{+}, \mathrm{PH}_{4}^{+}\) (bond energies: $\mathrm{N}-\mathrm{H}, 391 \mathrm{kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H},$ 322 \(\mathrm{kJ} / \mathrm{mol} )\) Give reasons for the orders you chose.

A typical vitamin \(\mathrm{C}\) tablet (containing pure ascorbic acid, \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}\) ) weighs $500 . \mathrm{mg}\( . One vitamin \)\mathrm{C}$ tablet is dissolved in enough water to make 200.0 \(\mathrm{mL}\) of solution. Calculate the \(\mathrm{pH}\) of this solution. Ascorbic acid is a diprotic acid.

Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced chemical equations for the reactions causing the solution to be acidic or basic. The relevant \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values are found in Tables 14.2 and \(14.3 .\) $\begin{array}{ll}{\text { a. NaNO, }} & {\text { d. } \mathrm{NH}_{4} \mathrm{NO}_{2}} \\ {\text { b. NaNO_ }_{2}} & {\text { e. } \mathrm{KOCl}} \\\ {\text { c. } \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHClO}_{4}} & {\text { f. } \mathrm{NH}_{4} \mathrm{OCl}}\end{array}$

Place the species in each of the following groups in order of increasing acid strength. Explain the order you chose for each group. a. \(\mathrm{HIO}_{3}, \mathrm{HBrO}_{3}\) b. \(\mathrm{HNO}_{2}, \mathrm{HNO}_{3}\) c. HOCl, HOI d. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{3} \mathrm{PO}_{3}\)

Use Table 14.3 to help order the following bases from strongest to weakest. $$ \mathrm{NO}_{3}^{-}, \quad \mathrm{H}_{2} \mathrm{O}, \quad \mathrm{NH}_{3}, \quad \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} $$ Also order the following acids from strongest to weakest. $$ \mathrm{HNO}_{3}, \quad \mathrm{H}_{2} \mathrm{O}, \quad \mathrm{NH}_{4}^{+}, \quad \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} $$
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