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Chapter 2: Problem 8

Bicine is a compound containing a tertiary amino group whose relevant \(\mathrm{p} K_{\mathrm{a}}\) is 8.3 (Figure 2.17 ). Given \(1 \mathrm{L}\) of \(0.05 \mathrm{M}\) Bicine with its tertiary amino group in the unprotonated form, how much \(0.1 N \mathrm{HCl}\) must be added to have a Bicine buffer solution of \(\mathrm{pH} 7.5 ?\) What is the molarity of Bicine in the final buffer? What is the concentration of the protonated form of Bicine in this final buffer?

Short Answer

Expert verified
To maintain a pH of 7.5, 0.1 L of 0.1 N HCl needs to be added. The molarity of the Bicine in the final buffer will then be 0.045 mol/L, the concentration of the protonated form of Bicine would be 0.00909 mol/L.

Step by step solution

01

Understand The Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is given by the formula: \( \mathrm{pH} = \mathrm{p} K_{\mathrm{a}} + \log \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\) where \(\mathrm{p} K_{\mathrm{a}} = 8.3\), \( \mathrm{A}^{-} \) can be considered as bicine and \(\mathrm{HA}\) as the protonated form of Bicine.
02

Calculate The Ratio Of Bicine To Protonated Bicine

In order to get to a pH balance of 7.5, we need to calculate the ratio of Bicine to protonated Bicine. Substitute the values into the Henderson-Hasselbalch equation: \(7.5 = 8.3 + \log \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\). Solving for the ratio of base to acid, we get \(\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]} = 10^{(7.5-8.3)} = 0.1995 \).
03

Calculate The Amount Of HCl Needed

Use the ratio from step 2 to determine the amount of HCl needed. The moles of Bicine is \(0.05 \mathrm{moles/L} \times 1 \mathrm{L} = 0.05 \mathrm{moles}\). Since the ratio of Bicine to protonated Bicine is 0.1995, then the moles of protonated Bicine is \(0.05 \mathrm{moles} \times 0.1995 = 0.01 \mathrm{moles}\). The amount of HCl necessary to protonate this Bicine is equal to the moles of the protonated form, which is 0.01 moles of HCl. Converting this to liters using the concentration of the HCl gives \(\frac{0.01 \mathrm{moles}}{0.1 \mathrm{N}} = 0.1 \mathrm{L}\). Hence, 0.1 L of 0.1 N HCl needed to be added to maintain a pH of 7.5.
04

Calculate The Molarity of Bicine In The Final Buffer

The total volume of the solution after adding the HCl is \(1L + 0.1L = 1.1L\). Since the molarity is \(moles/volume(mcL)\), the molarity of the Bicine in the final buffer is \(0.5 moles/1.1L = 0.045 mol/L\).
05

Calculate The Concentration Of The Protonated Form Of Bicine

Using the same logic from step 4, the concentration of the protonated form of Bicine is \(0.01 moles/ 1.1L = 0.00909 mol/L\).

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Most popular questions from this chapter

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?

Given \(0.1 M\) solutions of acetic acid and sodium acetate, describe the preparation of \(1 \mathrm{L}\) of \(0.1 \mathrm{M}\) acetate buffer at a pH of 5.4.

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.

Given \(0.1 \mathrm{M}\) solutions of acetic acid and sodium acetate, describe the preparation of \(1 \mathrm{L}\) of \(0.1 \mathrm{M}\) acetate buffer at a pH of 5.4.

Citric acid, a tricarboxylic acid important in intermediary metabolism, can be symbolized as \(\mathrm{H}_{3} \mathrm{A}\). Its dissociation reactions are \\[\begin{array}{ll}\mathrm{H}_{3} \mathrm{A} \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{A}^{-} & \mathrm{p} K_{1}=3.13 \\\\\mathrm{H}_{2} \mathrm{A}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HA}^{2-} & \mathrm{p} K_{2}=4.76 \\\\\mathrm{HA}^{2-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{3-} & \mathrm{p} K_{3}=6.40 \end{array}\\] If the total concentration of the acid and its anion forms is \(0.02 \mathrm{M}\) what are the individual concentrations of \(\mathrm{H}_{3} \mathrm{A}, \mathrm{H}_{2} \mathrm{A}^{-}, \mathrm{HA}^{2-},\) and \(\mathrm{A}^{3-}\) at pH \(5.2 ?\)
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