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Chapter 14: Problem 77

A \(0.15-M\) solution of a weak acid is 3.0\(\%\) dissociated. Calculate \(K_{\mathrm{a}}\) .

Short Answer

Expert verified
The acid dissociation constant (\(K_a\)) for the given weak acid solution is calculated as: \[K_a = 1.39 \times 10^{-4}\]

Step by step solution

01

Find the initial concentration of the weak acid

The problem states that we have a 0.15 M solution of the weak acid. This is our initial concentration of the acid \((HA)\). Let it be represented as \(C_{HA}\): \[C_{HA} = 0.15\,\text{M}\]
02

Find the degree of dissociation and calculate equilibrium concentrations

The degree of dissociation is given as 3.0%. That means 3.0% of the weak acid molecules are dissociated into \(H^+\) and \(A^-\) ions at equilibrium. First, we need to convert the percentage to a decimal by dividing by 100: \[\text{Degree of dissociation (decimal)} = \frac{3.0}{100} = 0.03\] Now, multiply the degree of dissociation (as a decimal) by the initial concentration \(C_{HA}\) to determine the change in concentration of all species during the dissociation: \[\text{Change in concentration } = 0.15\,\text{M} \times 0.03 = 0.0045\,\text{M}\] At equilibrium, the concentrations will be: \[[H^+] = [A^-] = 0.0045\,\text{M}\] \[[HA] = 0.15\,\text{M} - 0.0045\,\text{M} = 0.145\,\text{M}\]
03

Calculate the acid dissociation constant, \(K_a\)

The equilibrium expression for the dissociation of weak acid is: \[K_a = \frac{[H^+][A^-]}{[HA]}\] Using the concentrations from Step 2: \[K_a = \frac{(0.0045\,\text{M})(0.0045\,\text{M})}{(0.145\,\text{M})} = \frac{(0.0045)^2\,\text{M}^2}{(0.145)\,\text{M}} = 1.39 \times 10^{-4}\]
04

Write down the final answer

The acid dissociation constant (\(K_a\)) for the given weak acid solution is calculated as: \[K_a = 1.39 \times 10^{-4}\]

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Most popular questions from this chapter

A typical sample of vinegar has a pH of \(3.0 .\) Assuming that vinegar is only an aqueous solution of acetic acid \(\left(K_{\mathrm{a}}=1.8 \times\right.\) \(10^{-5}\) ), calculate the concentration of acetic acid in vinegar.

Quinine $\left(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{N}_{2} \mathrm{O}_{2}\right)$ is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine, $\mathrm{p} K_{\mathrm{b}_{1}}=5.1\( and \)\mathrm{p} K_{\mathrm{b}_{2}}=9.7\left(\mathrm{p} K_{\mathrm{b}}=-\log K_{\mathrm{b}}\right) .$ Only 1 g quinine will dissolve in 1900.0 \(\mathrm{mL}\) of solution. Calculate the pH of a saturated aqueous solution of quinine. Consider only the reaction $\mathrm{Q}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{QH}^{+}+\mathrm{OH}^{-}$ described by \(\mathrm{p} K_{\mathrm{b}_{1}},\) where \(\mathrm{Q}=\) quinine.

Calculate the \(\mathrm{pH}\) and \(\left[\mathrm{S}^{2-}\right]\) in a $0.10-M \mathrm{H}_{2} \mathrm{S}\( solution. Assume \)K_{\mathrm{a}_{1}}=1.0 \times 10^{-7} ; K_{\mathrm{a}_{2}}=1.0 \times 10^{-19}$

Monochloroacetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2},\) is a skin irritant that is used in "chemical peels" intended to remove the top layer of dead skin from the face and ultimately improve the complexion. The value of \(K_{\mathrm{a}}\) for monochloroacetic acid is \(1.35 \times 10^{-3}\) Calculate the pH of a \(0.10-M\) solution of monochloroacetic acid.

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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