BCH312 [Practical] 1

Similar documents
Preparation Of Buffer Solutions By Different Laboratory Ways. BCH 312 [Practical]

Preparation of different buffer solutions

Now, the excess strong base will react: HA + OH - A - + H 2 O Start mol End mol

Buffer Solutions. Buffer Solutions

Acids and bases, ph and buffers. Dr. Mamoun Ahram Lecture 2

Volume NaOH Delivered (ml)

Chemistry 222. Start mol End mol

10.1 Acids and Bases in Aqueous Solution

A 95 g/mol B 102 /mol C 117 g/mol D 126 g/mol E 152 g/mol

Chapter 15. Titration Curves for Complex Acid/Base Systems

AP Chemistry: Acid-Base Chemistry Practice Problems

CHEMISTRY 1AA3 Tutorial 2 Answers - WEEK E WEEK OF JANUARY 22, (i) What is the conjugate base of each of the following species?

Questions #4-5 The following two questions refer to the following system: A 1.0L solution contains 0.25M HF and 0.60M NaF (Ka for HF = 7.2 x 10-4 ).

ph Titration of H 3 PO 4 Mixtures Calculation of K 1, K 2, and K 3

Understanding the shapes of acid-base titration curves AP Chemistry

Titration curve of amino acids. BCH 312 [Practical]

Titration Of A Weak Acid With Strong Base. BCH 312 [Practical]

CHEMISTRY - BROWN 13E CH.16 - ACID-BASE EQUILIBRIA - PART 2.

CHEMISTRY - MCQUARRIE 4E CH.21 - BUFFERS & THE TITRATION OF ACIDS & BASES

Part One: Pure Solutions of Weak Acids, Bases (water plus a single electrolyte solute)

Solutions, mixtures, and media

Chemical Equilibria Part 2

Introduction to Acids & Bases. Packet #26

Preparation Of Different Buffer Solutions. BCH 312 [Practical]

Abdullah Zreqat. Laith Abu Shekha. Mamoun Ahram

Full file at Chapter 2 Water: The Solvent for Biochemical Reactions

ph calculations MUDr. Jan Pláteník, PhD Brønsted-Lowry concept of acids and bases Acid is a proton donor Base is a proton acceptor

BUFFERS. RAMESH REDDY.K M.Pharm.,(Ph.D) KRISHNA TEJA PHARMACY COLLEGE

Molarity Revised 2011

Equilibrium constant

Experiment 7 Buffer Capacity & Buffer Preparation

Chemistry 1B Experiment 11 49

1.12 Acid Base Equilibria

Making Buffers v010417

CH3NH2(aq) + H3O + (aq) 3. The reaction of carbon monoxide and diiodine pentoxide is endothermic. The yield could be increased by

Functional Genomics Research Stream. Lecture: February 24, 2009 Buffer & Reagent Production, ph

Nanoscale pictures: Figs. 5.1, 5.4, and 5.5

Buffers and chemicals

Practical Note BIOCHEMICAL CALCULATIONS (BCH 312)

battery acid the most widely used industrial chemical Hydrochloric acid, HCl muriatic acid stomach acid Nitric acid, HNO 3

Analyte: The substance whose concentration is not known in a titration. Usually the analyte is in the flask or beaker beneath the burette.

Homework: 14, 16, 21, 23, 27, 29, 39, 43, 48, 49, 51, 53, 55, 57, 59, 67, 69, 71, 77, 81, 85, 91, 93, 97, 99, 104b, 105, 107

Chapter 10. Acids, Bases, and Salts

CEE March FIRST EXAM (Solutions)

ANALYTICAL CHEMISTRY - CLUTCH 1E CH.8 - MONOPROTIC ACID-BASE EQUILIBRIA.

The texts, tables and images contained in this lecture notes are not my own, they can be found on: References supplied

Chapter 17 Answers. Practice Examples [H3O ] 0.018M, 1a. HF = M. 1b. 30 drops. 2a.

Chapter 17 Additional Aspects of Aqueous Equilibria (Part A)

Solutions. Solution: A solution is homogeneous liquid mixture of two or more substances.

Acid-Base Titration Solution Key

EXPT. 4 DETERMINATION OF pka OF ORTHOPHOSPHORIC ACID

4. Acid Base Equilibria

... so we need to find out the NEW concentrations of each species in the system.

CHEMISTRY - BURDGE-ATOMS FIRST 3E CH.17 - ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA

Acids and Bases Review Worksheet II Date / / Period. Molarity. moles L. Normality [H 3 O +1 ] [OH -1 ] ph poh

AP CHEMISTRY NOTES 10-1 AQUEOUS EQUILIBRIA: BUFFER SYSTEMS

Today. Solubility The easiest of all the equilibria. Polyprotic Acids determining something about an unknown by reacting it with a known solution

-log [H+][OH-] = - log [1 x ] Left hand side ( log H + ) + ( log OH - ) = ph + poh Right hand side = ( log 1) + ( log ) = 14 ph + poh = 14

LESSON ASSIGNMENT. After completing this lesson, you should be able to:

Acids, Bases, and Buffers

Chapter 18 Problem Set Solutions S.E. Van Bramer 2/22/96

12. Acid Base Equilibria

Acids, Bases, and Salts Review for Sections

Chem 112, Fall 05 Exam 3A

REVIEW QUESTIONS Chapter 17

Triprotic H3A, H2A -, HA 2-, A 3-

ph and buffers Dr. Mamoun Ahram Summer, 2018

General Chemistry II CHM 1046 E Exam 2

Molarity, ph, and Buffers

5 Acid Base Reactions

SKKU Physical Pharmacy Laboratory 성균관대학교물리약학연구실

Chapter 2 Water: The Solvent for Biochemical Reactions

Steward Fall 08. Moles of atoms/ions in a substance. Number of atoms/ions in a substance. MgCl 2(aq) + 2 AgNO 3(aq) 2 AgCl (s) + Mg(NO 3 ) 2(aq)

Buffers. How can a solution neutralize both acids and bases? Beaker B: 100 ml of 1.00 M HCl. HCl (aq) + H 2 O H 3 O 1+ (aq) + Cl 1 (aq)

5.1.3 Acids, Bases and Buffers

Monoprotic Acid/Base Equilibria. Monoprotic Acid/Base Equilibria

CHEMICAL EQUILIBRIUM -3

5/10/2017. Chapter 10. Acids, Bases, and Salts

Student Worksheet for Buffers, Ksp, and Titrations

CHEMISTRY 1220 CHAPTER 16 PRACTICE EXAM

General Phenomena: Law of mass action, dissociation of water, ph, buffers

Chapter 17 Additional Aspects of Aqueous Equilibria (Part A)

K w. Acids and bases 8/24/2009. Acids and Bases 9 / 03 / Ionization of water. Proton Jumping Large proton and hydroxide mobility

Ch 7. Systematic Treatment of Equilibrium and Activity

Ionic Equilibria. In the Brönsted Lowry classification, acids and bases may be anions such as HSO 4

EXAM 2 PRACTICE KEY. Leaders: Deborah Course: CHEM 178

Unit The mw of Na 2 CO 3 is : Na=23, O=16, C=12 A) 140 B) 106 C) 96 D) 100 E) 60

ANSWERS Unit 14: Review Acids and Bases

Acids and Bases. Dr. Diala Abu-Hassan, DDS, PhD Lecture 2 Nursing First Semester 014. Dr. Diala Abu-Hassan 1

Proton Transfer Acids - Base. Dr. Fred Omega Garces Chemistry 201. Miramar College

Formation of a salt (ionic compound): Neutralization reaction. molecular. Full ionic. Eliminate spect ions to yield net ionic

Week 6 AB Strength, ph, Kw, Acids

Introduction to Acids & Bases II. Packet #26

Judith Herzfeld 1996,1998. These exercises are provided here for classroom and study use only. All other uses are copyright protected.

Acids, Bases and Buffers

BCH 4053 Spring 2001 Chapter 2 Lecture Notes

PRACTICAL 3 ph AND BUFFERS

Review of Chemistry 11

Acids And Bases. H + (aq) + Cl (aq) ARRHENIUS THEORY


Transcription:

BCH312 [Practical] 1

Triprotic acid is acid that contain three hydrogens ions. It dissociates in solution in three steps, with three Ka values. phosphoric acid is an example of triprotic acid. It dissociates in solution as following: 2

For example if you was asked to prepare sodium phosphate buffer [ NaH 2 PO 4 / Na 2 HPO 4 ]: you can prepare it by.. 1. By mixing NaH 2 PO 4 (conjugate acid ) and Na 2 HPO 4 (conjugate base) in the proper proportions. 2. By starting with H 3 PO 4 and converting it to NaH 2 PO 4 plus Na 2 HPO 4 by adding the proper amount of NaOH. 3. By starting with NaH 2 PO 4 and converting a portion of it to Na 2 HPO 4 by adding NaOH. 4. By starting with Na 2 HPO 4 and converting a portion of it to NaH 2 PO 4 by adding a strong acid such as HCL. 5. By starting with Na 3 PO 4 and converting it to Na 2 HPO 4 plus NaH 2 PO 4 by adding HCL. 6. By mixing Na 3 PO 4 and NaH 2 PO 4 in the proper proportions. 3

H 3 PO 4 HCl donate H + H 2 PO 4 HPO 4 PO 4 3 NaOH accept H +

Prepare 0.1 liters of 0.045 M sodium phosphate buffer, ph=7.5, [pka1= 2.12, pka2 = 7.21 and pka3 = 12.30]: a) From concentrated (15M) H 3 PO 4 and solution of 1.5 M NaOH. b) From solid NaH 2 PO 4 and solid NaOH. Calculations: 1 st Write the equations of phosphoric acid dissociation and the pka of corresponding ones: Because phosphoric acid [H 3 PO 4 ] is triprotic acid it has 3 dissociation phases so: Regardless of which method is used, the first step involves determine the buffer ionic species, calculating number of moles and amounts of the two ionic species in the buffer. 2 nd Choose the pka value which is near the ph value of the required buffer, to be able to know the ionic species involved in your buffer: The ph of the required buffer [ph =7.5] is near the value of pka2, consequently, the two major ionic species present are H 2 PO 4 ( conjugate acid ) and HPO 4 (conjugate base), with the HPO 4 predominating {since the ph of the buffer is slightly basic}. 5

Calculations cont : 3 rd calculate No. of moles for the two ionic species in the buffer: ph =pka2 + log [ HPO 4 ] / [H 2 PO 4 ] Note that : [A ] = HPO 4, [HA] = H 2 PO 4 Since the buffer concentration is 0.045M, so assume [A ] = y, [HA]= 0.045 y : 7.5 = 7.2 + log ( y / 0.045y ) 7.57.2 = log ( y / 0.045y ) 0.3= log( y / 0.045y ) antilog for both sides 2=( y / 0.045y) y= 0.09 2 y 3 y = 0.09 y= 0.9/3 = 0.03M conc. of [ HPO 4 ] = [A] = y So, conc. of [H 2 PO 4 ] =[HA] = 0.045 y = 0.045 0.03 = 0.015 M Now find the number of mole for the two ionic species in the buffer: No. of moles of = HPO 4 (A ) = M x V = 0.03 x 0.1 = 0.003 moles. No. of moles of H 2 PO 4 (HA)= M x V = 0.015 x 0.1 = 0.0015 moles. 6 Note that Total no. of moles of phosphate buffer = M x V= 0.045 x 0.1 = 0.0045 moles.

Now, to prepare the required buffer: a) From concentrated (15M) H 3 PO 4 and solution of 1.5 M NaOH. Remember that the two ionic species involved in the buffer are: Calculations: Start with 0.0045 mole of H 3 PO 4 add 0.0045 moles of NaOH to convert H 3 PO 4 completely to H 2 PO 4 (HA), then add 0.003 moles of NaOH to convert H 2 PO 4 to give HPO 4 (A ): No. of moles needed of NaOH= 0.0045+0.003= 0.0075 moles Volume of NaOH needed= no.of moles / M = 0.0075/ 1.5 = 0.005 L =5 ml Volume of H 3 PO 4 needed =no.of moles / M = 0.0045/ 15 =0.0003 L = 0.3 ml 0.0045 moles 0.0045 moles HPO 4 H 3 PO 4 H 2 PO 4 H 2 PO 4 0.003 moles 0.0015 moles NaOH 0.0045 moles NaOH 0.003 moles So: Add 5ml of NaOH to the 0.3 ml of concentrate H 3 PO 4, mix ; then add sufficient water to bring the final volume to 0.1 liters (100 ml), and check the ph. 7

b) From solid NaH 2 PO 4 and solid NaOH. Remember that the two ionic species involved in the buffer are: Calculations: Start with 0.0045 mole of NaH 2 PO 4 (HA) and add 0.003 moles of NaOH to convert NaH 2 PO 4 to give Na 2 HPO 4 (A ): Weight in grams of NaH 2 PO 4 needed = no.of moles x MW =0.0045 x 141.98 = 0.638 g Weight in grams of NaOH needed = no. of moles x MW = 0.003 x 40 =0.12 g So: Dissolve the 0.638g of NaH 2 PO 4 and 0.12g of NaOH in some water, mix ; then add sufficient water to bring the final volume to 0.1 liters (100 ml), and check the ph. 0.0045 moles HPO 4 H 2 PO 4 H 2 PO 4 0.003 moles 0.0015 moles NaOH 0.003 moles 8

9

To learn how to prepare a buffer by different laboratory ways. 10

Prepare 0.1 liters of 0.045 M sodium phosphate buffer, ph=7.5, [pka1= 2.12, pka2 = 7.21 and pka3 = 12.30]: a) From concentrated (15M) H 3 PO 4 and solution of 1.5 M NaOH : Add 5ml of NaOH to the 0.3 ml of concentrate H 3 PO 4, mix ; then add sufficient water to bring the final volume to 0.1 liters (100 ml), and check the ph. b) From solid NaH 2 PO 4 and solid NaOH : Dissolve the 0.638g of NaH 2 PO 4 and 0.12g of NaOH in some water, mix ; then add sufficient water to bring the final volume to 0.1 liters (100 ml), and check the ph. 11

Prepare 0.1 liters of 0.045 M sodium phosphate buffer, ph=7.5, [pka1= 2.12, pka2 = 7.21 and pka3 = 12.30]: c) You are provided with solid Na 2 HPO 4 and 2M HCl. d) You are provided with solid Na 3 PO 4 and 2 M HCL. 12