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Chapter 4: Problem 5

For each value of \(K_{\mathrm{a}}\), calculate the corresponding value of \(\mathrm{p} K_{\mathrm{a}}\). Which compound is the stronger acid? (a) Acetic acid, \(K_{\mathrm{a}}=1.74 \times 10^{-5}\) (b) Chloroacetic acid, \(K_{\mathrm{a}}=1.38 \times 10^{-3}\)

Short Answer

Expert verified
Answer: Chloroacetic acid is the stronger acid.

Step by step solution

01

Find pKa for acetic acid

To find the \(\mathrm{p} K_{\mathrm{a}}\) for acetic acid, plug its \(K_{\mathrm{a}}\) value into the \(\mathrm{p} K_{\mathrm{a}}\) formula: \(\mathrm{p} K_{\mathrm{a}} = - \log(1.74 \times 10^{-5})\)
02

Calculate pKa for acetic acid

By performing the calculation, we get the pKa value for acetic acid: \(\mathrm{p} K_{\mathrm{a}} = 4.76\)
03

Find pKa for chloroacetic acid

Now, find the \(\mathrm{p} K_{\mathrm{a}}\) for chloroacetic acid by plugging its \(K_{\mathrm{a}}\) value into the \(\mathrm{p} K_{\mathrm{a}}\) formula: \(\mathrm{p} K_{\mathrm{a}} = - \log(1.38 \times 10^{-3})\)
04

Calculate pKa for chloroacetic acid

Calculate the pKa value for chloroacetic acid: \(\mathrm{p} K_{\mathrm{a}} = 2.86\)
05

Compare pKa values and determine the stronger acid

Comparing the pKa values, we see that chloroacetic acid has the lower pKa (2.86) compared to acetic acid (4.76). Thus, chloroacetic acid is the stronger acid.

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Most popular questions from this chapter

If the \(\Delta G^{\circ}\) for a reaction is \(-4.5 \mathrm{kcal} / \mathrm{mol}\) at \(298 \mathrm{~K}\), what is the \(K_{\text {eq }}\) for this reaction? What is the change in entropy of this reaction if \(\Delta H^{\circ}=-3.2 \mathrm{kcal} / \mathrm{mol}\) ?

Will carbon dioxide be evolved when sodium bicarbonate is added to an aqueous solution of each compound? Explain. (a) Sulfuric acid (b) Ethanol (c) Ammonium chloride

4-Methylphenol, \(\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{OH}\left(\mathrm{p} K_{\mathrm{a}} 10.26\right)\), is only slightly soluble in water, but its sodium salt, \(\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}^{-} \mathrm{Na}^{+}\), is quite soluble in water. In which solution(s) will 4-methylphenol dissolve? (a) Aqueous \(\mathrm{NaOH}\) (b) Aqueous \(\mathrm{NaHCO}_{3}\) (c) Aqueous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\)

Offer an explanation for the following observations. (a) \(\mathrm{H}_{3} \mathrm{O}^{+}\)is a stronger acid than \(\mathrm{NH}_{4}^{+}\). (b) Nitric acid, \(\mathrm{HNO}_{3}\), is a stronger acid than nitrous acid, \(\mathrm{HNO}_{2}\). (c) Ethanol and water have approximately the same acidity. (d) Trifluoroacetic acid, \(\mathrm{CF}_{3} \mathrm{COOH}\), is a stronger acid than trichloroacetic acid, \(\mathrm{CCl}_{g} \mathrm{COOH}\).

One way to determine the predominant species at equilibrium for an acid-base reaction is to say that the reaction arrow points to the acid with the higher value of \(\mathrm{p} K_{a}\). For example, $$ \begin{array}{cr} \mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \longleftarrow \mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+} \\ \mathrm{p} K_{\mathrm{a}} 9.24 & \mathrm{p} K_{\mathrm{a}}-1.74 \\ \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{p} K_{\mathrm{a}} 9.24 & \mathrm{p} K_{\mathrm{a}} 15.7 \end{array} $$ Explain why this rule works.
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