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Chapter 2: Problem 18

a. Draw the titration curve for Bicine, assuming the \(\mathrm{p} K_{\mathrm{a}}\) for its free COOH group is 2.3 and the \(\mathrm{p} K_{\mathrm{a}}\) for its tertiary amino group is 8.3 b. Draw the structure of the fully deprotonated form (completely dissociated form) of bicine. c. You have available a \(0.1 ~ M\) solution of Bicine at its isoelectric point \(\left(\mathrm{pH}_{\mathrm{I}}\right), 0.1 \mathrm{M}\) solutions of \(\mathrm{HCl}\) and \(\mathrm{NaOH}\), and ample distilled \(\mathrm{H}_{2} \mathrm{O}\) Describe the preparation of 1 L of 0.04 M Bicine buffer, pH 7.5 d. What is the concentration of fully protonated form of Bicine in your final buffer solution?

Short Answer

Expert verified
a. The titration curve starts at pH 2.3 and reaches a mid-point at pH 2.3, then ascends until it reaches pH 8.3 which represents the mid-point of the second buffering region. b. The dissociated form of Bicine has a COO- group and a free amino group. c. Dissolve 0.04 mol of Bicine in water and adjust the pH to 7.5 using NaOH, then adjust the volume to 1 L with water. d. The concentration of fully protonated form of bicine in the final solution can be obtained using Henderson-Hasselbalch equation and the total buffer concentration.

Step by step solution

01

- Titration Curve Drawing

Plot the pKa values on the y-axis and the volume of teritiary base added on the x-axis. Initially the curve starts at pH 2.3, the pKa of COOH group. As the titration proceeds, the pH increases. When half of the COOH group has been titrated, the pH equals the pKa of the COOH group. Further titration triggers the reaction with the tertiary amine group. The pH then increases until it equals pKa of the amine group (pH 8.3) when half of the tertiary amine group has been titrated.
02

- Drawing Structure of Deprontonated Form

The fully deprotonated form of Bicine has lost all of its protons. Since Bicine has a carboxyl group and a tertiary amine group, both have lost their protons - COOH group becomes COO-, and the amino group becomes free of H+.
03

- Buffer Preparation

Firstly, calculate the amount of bicine needed using the formula: amount (mol) = volume(L) x concentration(M). So the amount is 1L * 0.04M = 0.04 mol. Dissolve this amount of Bicine in 750 mL of water. Secondly, we need to adjust the pH to 7.5 using HCl or NaOH. Since the pI of Bicine (the average of its two pKa values) is less than 7.5, NaOH will be needed to raise the pH. Add it slowly until you reach the desired pH. Finally, adjust the final volume to 1 L with distilled water.
04

- Concentration Calculation

Use Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]). Since we know the pH (7.5), pKa (8.3), we can calculate the ratio [A-]/[HA], which represents the ratio of deprotonated to protonated forms of Bicine. Solve for [HA], knowing that [A-] + [HA] equals the total concentration of the buffer (0.04 M).

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Most popular questions from this chapter

Shown here is the structure of triethanolamine in its fully protonated form: Its \(\mathrm{p} K_{\mathrm{a}}\) is \(7.8 .\) You have available at your lab bench \(0.1 \mathrm{M}\) solutions of \(\mathrm{HCl}, \mathrm{NaOH}\), and the uncharged (free base) form of triethanolamine, as well as ample distilled water. Describe the preparation of a 1 L solution of 0.05 M triethanolamine buffer, pH 7.6.

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Given a solution of \(0.1 \mathrm{M}\) HEPES in its fully protonated form, and ready access to \(0.1 \mathrm{M} \mathrm{HCl}, 0.1 \mathrm{M} \mathrm{NaOH}\) and distilled water, describe the preparation of 1 L of 0.025 M HEPES buffer solution, \(\mathrm{pH} 7.8\)

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Calculate the \(\mathrm{pH}\) of the following. a. \(5 \times 10^{-4} \mathrm{MHCl}\) d. \(3 \times 10^{-2}\) M KOH b. \(7 \times 10^{-5} M\) NaOH e. \(0.04 \mathrm{m} M \mathrm{HCl}\) c. \(2 \mu M\) HCl f. \(6 \times 10^{-9}\) M HCl
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