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Chapter 13: Problem 116

The \(K_{\mathrm{b}}\) values for ammonia and methylamine are \(1.8 \times 10^{-5}\) and \(4.4 \times 10^{-4}\), respectively. Which is the stronger acid, \(\mathrm{NH}_{4}^{+}\) or \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} ?\)

Short Answer

Expert verified
The stronger acid is NH4+ because its Ka value (\(5.5 \times 10^{-10}\)) is higher than the Ka value for CH3NH3+ (\(2.3 \times 10^{-11}\)).

Step by step solution

01

Find the Ka values of NH4+ and CH3NH3+

We will find the Ka values of NH4+ and CH3NH3+ using the Kb values given and the Ka-Kb relationship. For NH4+, we are given the Kb value of its conjugate base, ammonia (NH3), which is 1.8 × 10^{-5}. Using the relation Ka x Kb = Kw, \(K_{a} \times(1.8 \times 10^{-5})=1 \times 10^{-14}\) Solving for Ka: \(K_{a} = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}\) For CH3NH3+, we are given the Kb value of its conjugate base, methylamine (CH3NH2), which is 4.4 × 10^{-4}. Using the relation Ka x Kb = Kw, \(K_{a} \times(4.4 \times 10^{-4})=1 \times 10^{-14}\) Solving for Ka: \(K_{a} = \frac{1 \times 10^{-14}}{4.4 \times 10^{-4}}\)
02

Compare the Ka values

Now we will compare the Ka values of NH4+ and CH3NH3+ to determine which is the stronger acid. Ka for NH4+: \(K_{a} = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}\) \[K_{a} \approx 5.5 \times 10^{-10}\] Ka for CH3NH3+: \(K_{a} = \frac{1 \times 10^{-14}}{4.4 \times 10^{-4}}\) \[K_{a} \approx 2.3 \times 10^{-11}\] Because the Ka value for NH4+ (5.5 × 10^{-10}) is higher than the Ka value for CH3NH3+ (2.3 × 10^{-11}), NH4+ is the stronger acid.

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Most popular questions from this chapter

What are the major species present in 0.250 \(M\) solutions of each of the following acids? Calculate the pH of each of these solutions. a. \(\mathrm{HNO}_{2}\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

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 \(\mathrm{pH}\) 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.

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=378.85 \mathrm{g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=\) \(8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C}\) ). Calculate the \(\mathrm{pH}\) of a \(30.0-\mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at \(35.0^{\circ} \mathrm{C}\) is \(2.1 \times 10^{-14}\).

How would you prepare \(1600 \mathrm{mL}\) of a \(\mathrm{pH}=1.50\) solution using concentrated (12 \(M\) ) HCl?

Calculate the \(\mathrm{pH}\) of a \(0.20-M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.5.6 \times 10^{-4}\right)\)
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