Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 17

Tris-hydroxymethyl aminomethane (TRIS) is widely used for the preparation of buffers in biochemical research. Shown here is the structure of TRIS in its protonated form: Its acid dissociation constant, \(K_{\mathrm{a}},\) is \(8.32 \times 10^{-9} M .\) You have available at your lab bench a \(0.1 \mathrm{M}\) solution of TRIS in its protonated form, 0.1 \(M\) solutions of \(\mathrm{HCl}\) and \(\mathrm{NaOH}\), and ample distilled water. Describe the preparation of a 1 L solution of 0.02 M TRIS buffer, pH 7.8.

Short Answer

Expert verified
To create the 1L 0.02M TRIS buffer solution pH 7.8, take 0.2 L of a 0.1M TRIS solution, add 0.02 L of a 0.1M NaOH solution to it, then top off with distilled water up to 1L.

Step by step solution

01

Calculate the \([H^+]\) of the buffer solution

As we have the pH of the buffer solution, we can calculate the concentration of \([H^+]\) ions using the formula \([H^+] = 10^{-pH}\). This gives us \([H^+] = 10^{-7.8}M\).
02

Calculate the ratio of TRIS to TRIS-H+

Now we use the Henderson-Hasselbalch equation, which is \(pH = pK_a + \log(\dfrac{[A^-]}{[HA]})\), where \([A^-]\) is the concentration of deprotonated form (TRIS) and \([HA]\] is the concentration of protonated form (TRIS-H+). Solving for the \(\dfrac{[A^-]}{[HA]}\) ratio gives \(\dfrac{[A^-]}{[HA]} = 10^{(pH - pK_a)} = 10^{(7.8 - (-8.08))} = 10^{15.88}\). Thus, the TRIS to TRIS-H+ ratio should be \(10^{15.88}\), meaning there's practically no TRIS-H+ present compared to TRIS.
03

Decide on the volume of 0.1M TRIS (protonated form) to use

Next, we need to determine how much 0.1M TRIS solution we need to make the 0.02M buffer. We need an end volume of 1 L, so using the equation for dilution \(M1V1 = M2V2\), we get \(V1 = \dfrac{M2V2}{M1} = \dfrac{0.02L \times 1L}{0.1 M} = 0.2 L\). This is the volume of 0.1M TRIS solution we need for our 1L of buffer.
04

Determine the volume of 0.1M NaOH to add to the TRIS solution

Since there is almost no TRIS-H+ present, this means the entire portion of the TRIS initially used (0.02M) will be converted to TRIS by the NaOH. Since the NaOH and the initial TRIS are both 0.1M solutions, the volume (in L) of NaOH to add will be equal to the moles of initial TRIS, which are 0.02M. Therefore, we will add 0.02L of NaOH to the TRIS solution.
05

Finalize the buffer preparation

To finalize the buffer preparation, the TRIS solution with the added NaOH will be made up to a final volume of 1L using distilled water.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

a. If \(50 \mathrm{mL}\) of \(0.01 \mathrm{MHCl}\) is added to \(100 \mathrm{mL}\) of \(0.05 \mathrm{M}\) phosphate buffer at \(\mathrm{pH} 7.2,\) what is the resultant \(\mathrm{pH}\) ? What are the concentrations of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) in the final solution? b. If \(50 \mathrm{mL}\) of \(0.01 \mathrm{MNaOH}\) is added to \(100 \mathrm{mL}\) of \(0.05 \mathrm{M}\) phosphate buffer at \(\mathrm{pH} 7.2,\) what is the resultant \(\mathrm{pH}\) ? What are the concentrations of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) in this final solution?

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.

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

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?

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?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free