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Chapter 15: Problem 86

Amino acids are the building blocks for all proteins in our bodies. A structure for the amino acid alanine is All amino acids have at least two functional groups with acidic or basic properties. In alanine, the carboxylic acid group has \(K_{\mathrm{a}}=4.5 \times 10^{-3}\) and the amino group has \(K_{\mathrm{b}}=\) \(7.4 \times 10^{-5}\). Because of the two groups with acidic or basic properties, three different charged ions of alanine are possible when alanine is dissolved in water. Which of these ions would predominate in a solution with \(\left[\mathrm{H}^{+}\right]=1.0 M ?\) In a solution with \(\left[\mathrm{OH}^{-}\right]=1.0 M ?\)

Short Answer

Expert verified
In a highly acidic solution with \(\left[\mathrm{H}^{+}\right]=1.0 M\), the predominant ion of alanine is \(\mathrm{NH_3^+CH_3CHCOOH}\) (cationic form). In a highly basic solution with \(\left[\mathrm{OH}^{-}\right]=1.0 M\), the predominant ion of alanine is \(\mathrm{NH_2CH_3CHCOO^-}\) (anionic form).

Step by step solution

01

Identify the ionization state of alanine in water

In an aqueous solution, alanine can exist in three ionization states: cationic (\(\mathrm{NH_3^+CH_3CHCOOH}\)), zwitterionic (\(\mathrm{NH_3^+CH_3CHCOO^-}\)), or anionic (\(\mathrm{NH_2CH_3CHCOO^-}\)) forms. The acid dissociation constant \(K_{\mathrm{a}}\) and base dissociation constant \(K_{\mathrm{b}}\) will help us determine which form predominates under different \(\mathrm{H^+}\) and \(\mathrm{OH^-}\) concentrations. Step 2: Calculate the ionization constants of alanine forms
02

Calculate the ionization constants of alanine forms

Get the ion product of water, \(K_w\), where \(K_w = K_{\mathrm{a}}\times K_{\mathrm{b}}\). Given: \(K_{\mathrm{a}}=4.5 \times 10^{-3}\) and \(K_{\mathrm{b}}=7.4 \times10^{-5}\) Calculate: \(K_w = (4.5 \times 10^{-3}) \times (7.4 \times10^{-5}) = 3.33 \times 10^{-7}\) Step 3: Determine the predominant ion in the acidic solution
03

Determine the predominant ion in acidic solution

In a highly acidic solution with \(\left[\mathrm{H}^{+}\right]=1.0 M\), the concentration of \(\mathrm{H^+}\) ions is much higher than the \(K_{\mathrm{a}}\) value. Hence, the carboxylic acid group will remain protonated, and the amino group will also be protonated, making alanine a cation. Predominant ion in the acidic solution is \(\mathrm{NH_3^+CH_3CHCOOH}\). Step 4: Determine the predominant ion in the basic solution
04

Determine the predominant ion in basic solution

In a highly basic solution with \(\left[\mathrm{OH}^{-}\right]=1.0 M\), we can calculate the concentration of \(\mathrm{H^+}\) using the \(K_w\) value: \(\left[\mathrm{H}^{+}\right]=\frac{K_w}{\left[\mathrm{OH}^{-}\right]}=\frac{3.33 \times 10^{-7}}{1.0} = 3.33 \times 10^{-7} M\) Since \(\left[\mathrm{H}^{+}\right]\) value is much smaller compared to the \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values, it shows that the carboxylic acid group will completely dissociate and lose its proton, while the amino group will remain unprotonated. This means alanine takes the anionic form. The predominant ion in the basic solution is \(\mathrm{NH_2CH_3CHCOO^-}\).

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Most popular questions from this chapter

Malonic acid \(\left(\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)\) is a diprotic acid. In the titration of malonic acid with \(\mathrm{NaOH}\), stoichiometric points occur at \(\mathrm{pH}=3.9\) and \(8.8\). A \(25.00-\mathrm{mL}\) sample of malonic acid of unknown concentration is titrated with \(0.0984 \mathrm{M} \mathrm{NaOH}\), requiring \(31.50 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution. (See Exercise 113.)

What concentration of \(\mathrm{NH}_{4} \mathrm{Cl}\) is necessary to buffer a \(0.52-M\) \(\mathrm{NH}_{3}\) solution at \(\mathrm{pH}=9.00 ?\left(K_{\mathrm{b}}\right.\) for \(\left.\mathrm{NH}_{3}=1.8 \times 10^{-5} .\right)\)

You have a solution of the weak acid HA and add some of the salt NaA to it. What are the major species in the solution? What do you need to know to calculate the \(\mathrm{pH}\) of the solution, and how would you use this information? How does the \(\mathrm{pH}\) of the solution of just the HA compare with that of the final mixture? Explain.

Calculate the number of moles of \(\mathrm{HCl}(g)\) that must be added to \(1.0 \mathrm{~L}\) of \(1.0 \mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) to produce a solution buffered at each pH. a. \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) b. \(\mathrm{pH}=4.20\) c. \(\mathrm{pH}=5.00\)

Calculate the \(\mathrm{pH}\) of a solution that is \(0.40 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) and \(0.80 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3} .\) In order for this buffer to have \(\mathrm{pH}=\) \(\mathrm{p} K_{\mathrm{a}}\), would you add \(\mathrm{HCl}\) or \(\mathrm{NaOH}\) ? What quantity (moles) of which reagent would you add to \(1.0 \mathrm{~L}\) of the original buffer so that the resulting solution has \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) ?
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