# Buffers And Titration Of An Amino Acid

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Practical Exercise 3 – Buffers and Titration of An Amino Acid

Name: Stephanie Tayas Anak Bumphray
Id Number: 100084790
Date Performed: 11 September 2017
Practical Group: Monday 1:30pm – 5:30pm

Introduction:
A buffer solution is which resists large changes in pH when the small amounts of an acid or a base are added to it. The buffer can be calculated based on the equation of ionization pair of acid and conjugate base. Any weak acid can be represented as “HA” and "A-" as the conjugate base. This equation is based known as Henderson-Hasselbalch equation which is used to calculate pH of a solution. Figure 1: Henderson-Hasselbalch equation
Buffer solution can be divided into two which are acidic buffer solutions and alkaline …show more content…

In exercise 1, first, the ratio of [HPO42-] to [H2PO4-] required to produce buffer solutions was calculated with the following pH values (a) pH = 5.9, (b) pH = 7.9. Then, the solutions of 0.1 M H2PO4- and 0.1 M HPO42- was used, these buffer solutions were mixed appropriate volumes to give a total volume of 25 mL of each. The calibrated pH meter was used, the pH of each buffer solution was measured and recorded and the pH of 25mL of distilled water was compared. Next, 2.0 mL of 0.1 NaOH was added to each of the 25mL buffer samples and to the distilled water and each tube was mixed well. The pH of each solution was recorded after the addition of the alkali. Results was tabulated include a column for ∆ …show more content…

= 22.7mL
 Volume of base required = 25mL – 22.7mL = 2.3mL
(b) pH = 6.9 pH = pka + log10 [HPO42-]  6.9 = 6.9 + log10 [HPO42-] [H2PO4-] [H2PO4-] log10 [HPO42-] = 0 [H2PO4-] = 1 1
 Volume of acid required = 25mL (1+1) = 12.7mL
 Volume of base required = 25mL – 12.7mL = 12.5mL
(c) pH = 7.9 pH = pka + log10 [HPO42-]  7.9 = 6.9 + log10 [HPO42-] [H2PO4-] [H2PO4-] log10 [HPO42-] =