Biochemistry Notes - Lecture 3, Titrations
Biochemistry Notes - Lecture 3, Titrations Bch4053
Popular in General Biochemistry I
Popular in Biochemistry
This 3 page Class Notes was uploaded by Hannah Hartman on Friday September 9, 2016. The Class Notes belongs to Bch4053 at Florida State University taught by Dr. Hong Li in Fall 2016. Since its upload, it has received 18 views. For similar materials see General Biochemistry I in Biochemistry at Florida State University.
Reviews for Biochemistry Notes - Lecture 3, Titrations
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 09/09/16
Titrations viernes, 9 de septiembre de 2016 11:55 AM Titrations: • Titrations are used to determine the amount of acid in a solution by adding bases. • There are two equilibria occurring in solution when an acid is added to water ○ HA = A-+ H+; strong dissociation (large Ka) ○ H+ + OH- = H2O; strong association (small Ka) ○ Titration is forcing the acid to dissociate until it is completely gone. • When a base, typically NaOH, is slowly added to the solution, the hydroxyl ion will neutralize any hydronium ion present by equilibrium (2). More acid will be dissociated and the aid will eventually dissociate completely (to A -). ○ By monitoring the pH value (via a pH meter), you can determine the acid concentration in a solution (quantitatively). • The solution pH is related to the amount of hydroxyl ion according to the HH equation (note that when [-A ] = [HA}, pH = pKa, because the log of the ratio would be O) ○ The HH is a measure of the ratio of [-] and [HA], aka how much is dissociated. ○ You can plot this as the function of pH vs. the concentration of O- Your curve will be the "titration curve" • Titration curves: ○ When [OH-] << 0.5, the solution is in the completely acidic form. ○ When [OH-} = 0.5, the acid form = the conjugated base, or the neutral form = negatively charged form § pKa = pH here. ○ When [OH-] >>> 0.5, the solution is in the completely conjugated base form (negatively charged) How to draw a titration curve: 1. Exam the molecule first. Find out the number of dissociable protons; 2. Obtain the pK aalue(s) for all dissociable group(s); 3. Start the graph by labeling the vertical axis “pH”and the horizontal axis “equivalent OH -”; 4. Find the values of pK an the vertical axis and mark them; 5. Find the value of 0, 0.5, and 1 on the horizontal axis and mark them; (You may need to mark 1.5, 2.0 etc. as well. Depending on the number of dissociable groups) 3. Start the graph by labeling the vertical axis “pH”and the horizontal axis “equivalent OH -”; 4. Find the values of pK an the vertical axis and mark them; 5. Find the value of 0, 0.5, and 1 on the horizontal axis and mark them; (You may need to mark 1.5, 2.0 etc. as well. Depending on the number of dissociable groups) 6. Draw two intercepting lines, one through the first pKa value, the other through the 0.5 equivalent OH- position as shown below (repeat for all pKas): Protonation: • The concept of titration can be applied to any ionizable group on the molecule. The graph shown below are titration curves for acetic acids, imidazole, and phosphoric acid. • Multiple ionizable groups have different pKas, and pose a problem! ○ Note: An acid can have pKa>7! (e.g. HPO4) ○ The lower the pH value, the more protonated the species will be (especially if it has multiple ionizable groups) • Example: ○ Group 1 has a pKa of 5, and group 2 has a pKa of 7. ○ Group 1 is a stronger acid and more likely to lose its proton. Buffers: • Buffers are solutions that resist changes in pH as external acids or bases are added. IT is important to maintain a pH required for the body physiologically; for instance, blood pH is 7.4 , a gastric juice pH is 1; • The buffer chosen such that pH = pKa, where the titration curve is almost flat • Commonly used buffers: ○ Tris (8.1), MES (6.1), PIPES (6.8), MOPS (7.2) and HEPES (7.5) Determining the charge of a titrated molecule: 1. Exam the molecule first. Find out the number of dissociable protons; 2. Calculate the charge of the molecule with ALL DISSOCIABLE PROTONS attached, q0; 3. At equivalent [OH ] = 0, the total charge = 0q ; 4. At equivalent 0< [OH ]<1, (q 01)<total charge< 0 ; - 5. At equivalent [OH ]=1, total charge = 0 -1; 6. If the molecule contains more than one dissociable protons, you will continue to titrate, and the charge of the molecule will further decrease; 7. At equivalent 1<[OH ]<1.5, (q 02)<total charge<q (0-1); - 8. At equivalent [OH ]=2, total charge = 0 -2. Keep in mind how to determine the pH and pKa of dissociable groups. Objectives of Lecture 3 • Textbook: Chap. 2 ○ Describe the process of titration; ○ Draw titration curves for any ionizable group when its pKa is known; know Keep in mind how to determine the pH and pKa of dissociable groups. Objectives of Lecture 3 • Textbook: Chap. 2 ○ Describe the process of titration; ○ Draw titration curves for any ionizable group when its pKa is known; know - how to calculate charge states of the ionizable group when equivalent OH is 0, 0.5, and 1; ○ Draw titration curves for a molecule containing multiple ionizable groups; know how to calculate charges of the molecule at different OH ; ○ Apply HH equation to calculate pH for a buffered solution ○ What is a buffer? What is the function of the buffer for a biological solution? • Homework: Chap. 2:6 (given pH, calculate charge ratio),11 (add strong acid or nd rd base to a buffered solution, calculate new pH) (2 and 3 eds.) • Further reading: Chap. 2, page 53 (3 pp 47), “A Deeper look ”on Bicarbonate buffer system in human blood.
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'