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Chapter 15: Problem 51

When methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2},\) dissolves in water, the resulting solution is slightly basic. Which compound is the Bronsted-Lowry acid and which is the base?

Short Answer

Expert verified
Question: Identify the Bronsted-Lowry acid and base when methylamine (CH3NH2) dissolves in water. Answer: In the reaction, the Bronsted-Lowry base is methylamine (CH3NH2), and the Bronsted-Lowry acid is water (H2O).

Step by step solution

01

Write the reaction equation

First, we'll write the equation for the reaction of methylamine (CH3NH2) with water (H2O), as they act as Bronsted-Lowry base and acid, respectively: \(\mathrm{CH}_{3}\mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CH}_{3}\mathrm{NH}_{3}^{+} + \mathrm{OH}^{-}\) Now, we will identify the Bronsted-Lowry acid and base in this reaction.
02

Identify the Bronsted-Lowry base

In this reaction, the CH3NH2 (methylamine) molecule acts as the Bronsted-Lowry base, as it accepts a proton (H+ ion) from water: \(\mathrm{CH}_{3}\mathrm{NH}_{2} + \mathrm{H}^{+} \rightarrow \mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\)
03

Identify the Bronsted-Lowry acid

In this reaction, the H2O (water) molecule acts as the Bronsted-Lowry acid, as it donates a proton (H+ ion) to methylamine: \(\mathrm{H}_{2}\mathrm{O} - \mathrm{H}^{+} \rightarrow \mathrm{OH}^{-}\) Now that we have identified the acid and base in the reaction, we can answer the original question.
04

Present the answer

In the reaction between methylamine and water, the Bronsted-Lowry base is methylamine (CH3NH2), and the Bronsted-Lowry acid is water (H2O).

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Most popular questions from this chapter

Nicotine Addiction Nicotine is responsible for the addictive properties of tobacco. What is the \(\mathrm{pH}\) of \(1.00 \times 10^{-3} \mathrm{M}\) nicotine?

Predict which solution in each pair below will have the lower pH. a. \(2.56 \times 10^{-2} \mathrm{M} \mathrm{HCl}\) or \(4.09 \times 10^{-2} \mathrm{MHBr}\) b. \(1.00 \times 10^{-5} M\) acctic acid \(\left(K_{\mathrm{a}}=1.76 \times 10^{-5}\right)\) or \(1.00 \times 10^{-5} M\) formic acid \(\left(K_{a}=1.77 \times 10^{-4}\right)\) c. \(22 \mathrm{mM} \mathrm{CH}_{3} \mathrm{NH}_{2}\left(\mathrm{p} K_{\mathrm{b}}=3.36\right)\) or \(22 \mathrm{mM}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\left(K_{\mathrm{b}}=5.9 \times 10^{-4}\right)\) d. \(158 \mathrm{mM} \mathrm{NH}_{3}\left(\mathrm{p} K_{\mathrm{b}}=4.75\right)\) or \(158 \mathrm{m} M\) acetic acid \(\left(p K_{a}=4.75\right)\) e. \(0.00395 M \mathrm{HNO}_{3}\) or \(0.00145 M \mathrm{HClO}_{4}\) f. \(2.05 \times 10^{-1} M\) propionic acid \(\left(K_{2}=1.4 \times 10^{-5}\right)\) or \(2.05 \times 10^{-1} M\) fluoroacetic acid \(\left(K_{2}=2.6 \times 10^{-3}\right)\) g. 375 m \(M\) pyridine \(\left(p K_{b}=8.77\right)\) or \(375 \mathrm{mM}\) aniline \(\left(\mathrm{p} K_{\mathrm{b}}=9.40\right)\) h. \(0.555 M \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\left(K_{2}=3 \times 10^{-3}\right)\) or \(0.355 M \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\left(K_{2}=1 \times 10^{-4}\right)\)

In an aqueous solution of HF, which compound acts as a Bronsted-Lowry acid and which is the Bronsted-Lowry base?

Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of the following solutions: a. \(0.0450 M \mathrm{NaOH}\) b. \(0.160 M \mathrm{Ca}(\mathrm{OH})_{2}\) c. a 1: 1 mixture of \(0.0125 M\) HCl and \(0.0125 M \mathrm{Ca}(\mathrm{OH})_{2}\) d. a 2: 3 mixture of \(0.0125 M \mathrm{HNO}_{3}\) and \(0.0125 M \mathrm{KOH}\)

Hydrogen chloride and water are molecular compounds, yet a solution of HCl dissolved in \(\mathrm{H}_{2} \mathrm{O}\) is an excellent conductor of electricity. Explain why.
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