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.

Short Answer

Expert verified
To prepare this buffer, 0.5 L of the 0.1M triethanolamine solution should be combined with 0.63L of the 0.1 M HCl solution. Then add enough distilled water to make a total volume of 1 L.

Step by step solution

01

Determine the amount of triethanolamine in the free base form.

A solution of 0.05 M triethanolamine in 1L volume is needed. This equates to 0.05 mol of triethanolamine. To calculate the volume of a 0.1 M triethanolamine solution, use the formula \(C1 * V1 = C2 * V2\), where \(C1\) and \(V1\) are the molarity and volume of the available solution, and \(C2\) and \(V2\) are the molarity and volume of the desired solution. Plug in the known quantities to find the volume \(V1 = (C2*V2) / C1\), which results in \(V1 = (0.05*1) / 0.1 = 0.5 L\). This is the volume of the 0.1 M triethanolamine solution that will need to be used.
02

Calculate the amount of acid or base needs to be added to adjust the pH.

To adjust the pH, we use the Henderson-Hasselbalch equation, \([H^+] = 10^(pKa - pH)\) or \([OH^-] = 10^(pH - pKa)\). The given \(pKa\) is 7.8 and the desired pH is 7.6. We therefore add \([H^+] = 10^(7.8 - 7.6) = 0.063\). The molarity of the HCl solution is known, it’s 0.1 M. So, to achieve the desired \([H^+]\), a volume of HCl of \((0.063 * 1L) / 0.1 M = 0.63 L\) should be added to the triethanolamine solution.
03

Combine and dilute.

Combine the 0.5 L of free base triethanolamine and the 0.63 L of HCl solution, and then make up the total volume to 1 L with distilled water. This results in a buffer solution with pH 7.6.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

pKa
Understanding the pKa (acid dissociation constant) of a compound is crucial when preparing buffer solutions. The pKa is a numerical value that represents the strength of an acid in a given substance; it relates to how easily an acid donates its proton(s) to a base. Specifically, a lower pKa value indicates a stronger acid that disassociates more readily in solution.

When dealing with a weak base such as triethanolamine, the pKa refers to its conjugate acid's tendency to release a proton. The pKa value helps us in predicting how the molecule will behave in solution and at what pH level the buffer will effectively stabilise. In the exercise, triethanolamine’s pKa is given as 7.8, which is near neutral pH and ideal for buffers in biological systems, such as cellular environments where pH is typically around 7.
Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is pivotal for calculating the pH of buffer solutions. It relates the pH of a solution to the pKa of the acid and the ratio of the concentrations of its dissociated (A−) and undissociated (HA) forms. The equation is typically written as:
\[ pH = pKa + \text{log} \frac{[A^-]}{[HA]} \]
This critical equation allows us to determine the amounts of acid and conjugate base required to create a buffer with a specific pH. In the exercise, by manipulating the concentrations of triethanolamine and its conjugate acid, the Henderson-Hasselbalch equation guides the preparation of a buffer solution of a desired pH, which is slightly lower than the pKa. This implies that the buffer solution at pH 7.6 will contain slightly more acidic species than its conjugate base.
Molarity
Molarity is a measure of concentration used in chemistry to express the amount of a substance in a given volume of solution. It’s defined as moles of solute per liter of solution (mol/L). It’s fundamental to buffer preparation because accurate molarity allows for precise control over reactant ratios and final pH levels.

Our exercise entails creating a 0.05 M solution of triethanolamine in a 1 L volume. Calculating molarity involves using the formula \(C1V1 = C2V2\) to find out how much of an available higher concentration solution is needed to make a more dilute one. The accuracy of the molarity calculated impacts the effectiveness of the buffer; if too weak or too strong, it cannot effectively resist pH changes.
Buffer solution
A buffer solution is designed to maintain a stable pH despite addition of acids or bases. It manages this by containing a weak acid and its conjugate base, allowing it to neutralize small amounts of added strong acid or base. This characteristic makes buffers particularly essential in biological and chemical processes where pH stability is obligatory.

In the present exercise, a buffer is being created using triethanolamine. By beginning with the base form and adding a specific amount of acid (HCl), we can achieve the desired pH close to the triethanolamine's pKa. The process demonstrated in the solution steps ensures that the final solution can counteract pH changes, adhering to the buffer's primary function.

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Most popular questions from this chapter

The \(K_{\mathrm{a}}\) for formic acid is \(1.78 \times 10^{-4} M.\) a. What is the pH of a \(0.1 \mathrm{M}\) solution of formic acid? b. \(150 \mathrm{mL}\) of \(0.1 \mathrm{MNaOH}\) is added to \(200 \mathrm{mL}\) of \(0.1 \mathrm{M}\) formic acid, and water is added to give a final volume of 1 L. What is the pH of the final solution?

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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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