医用基础化学英文版.ppt

The College of Chemistry and Chemical Engineering,Central South University,Chapter 4,Buffer Solution,solutions that resist change in hydronium ion,H,+,and the hydroxide ion,OH,-,concentration(and consequently pH)upon addition of small amounts of acid or base,or upon dilution.,4.1.1,The Essential Feature of a Buffer,Buffer Solution:,4.1,Composition and Action of Buffer Solutions,Buffers are mixtures containing a common ion:either weak acids and their conjugate bases or weak bases and their conjugate acid.,Two common buffers:ammonium-ammonia,carbonate-bicarbonate,NH,4,+,(,aq,),H,+,(,aq,)+NH,3,(,aq,),CO,3,-2,(,aq,)+H,2,O(,l,),H CO,3,-1,(,aq,)+OH,-1,(,aq,),Buffer system,Conjugate acid,Conjugate base,p,K,a,(at 25),HAc,NaAc,HAc,Ac,-,4.76,H,2,CO,3,NaHCO,3,H,2,CO,3,HCO,3,-,6.35,H,3,PO,4,NaH,2,PO,4,H,3,PO,4,H,2,PO,4,-,2.16,TrisHCl,Tris,TrisH,+,Tris,7.85,H,2,C,8,H,4,O,4,KHC,8,H,4,O,4,H,2,C,8,H,4,O,4,HC,8,H,4,O,4,-,2.89,NH,4,Cl,NH,3,NH,4,+,NH,3,9.25,CH,3,NH,3,+,Cl,-,CH,3,NH,2,CH,3,NH,3,+,CH,3,NH,2,10.63,NaH,2,PO,4,Na,2,HPO,4,H,2,PO,4,-,HPO,4,2-,7.21,Buffer systems that are useful at various pH values,Tris:,Tris(Hydroxymethy)methanamin,三羟甲基氨基甲烷,C,CH,2,OH,CH,2,OH,HOH,2,C,NH,2,A Buffered Solution,resists change in its pH when either H,+,or OH,are added.,1.0 L of 0.50 M CH,3,COOH,+0.50 M CH,3,COONa,pH=4.74,Adding 0.010 mol,solid,NaOH,raises the,pH,of the solution to,4.76,a very minor change.,The effect of addition of acid or base to,an unbuffered solution,a buffered solution,acid added,base added,acid added,base added,Buffer with equal concentrations of conjugate base and acid,OH,-,H,3,O,+,Buffer after addition of H,3,O,+,H,2,O+CH,3,COOH H,3,O,+,+CH,3,COO,-,Buffer after addition of OH,-,CH,3,COOH+OH,-,H,2,O+CH,3,COO,-,4.1.2,How a Buffer Works,HAc+H,2,O,H,3,O,+,+Ac,H,+,+,Shift left,+,OH,H,2,O,Shift right,Anti-acid,Anti-base,anti-acid mechanism,anti-base mechanism,The amounts of weak acid and weak base in the buffer must be significantly larger than the amounts of H,3,O,+,or OH,-,that will be added,otherwise the pH cannot remain approximately constant.Thus addition of limited amounts of a strong acid or base is counteracted by the species present in the buffer solution,and the pH changes very little.No solution can keep the pH approximately constant if you add larger amounts of either acid or base that are present in the original buffer.,For a HB-NaB,buffer system,HB+H,2,O,H,3,O,+,+B,NaB,Na,+,+B,K,a,=,H,+,B,HB,H,+,=K,a,HB,B,Apply log on both sides of above equation,pH=pK,a,+lg,B,HB,The Henderson-Hasselbalch Equation,4.2,The Henderson-Hasselbalch Equation,The pH of buffer solution,pH=pK,a,+lg,B,HB,=pK,a,+lg,conjugate base,conjugate acid,pK,a,：,the,log of Ka of the conjugate acid,B,、,HB,：,equilibrium concentration,B,/HB,：,buffer ratio,B,+HB,：,total concentration,HB,=c,HB,c,HB,(dissociated),B,=c,NaB,+,c,HB(dissociated),c,HB,c,NaB,The Henderson-Hasselbalch Equation,pH=pK,a,+,lg,c,B,c,HB,=pK,a,+lg,n,B,/V,n,HB,/V,pH=pK,a,+lg,n,B,n,HB,If the concentrations of conjugate acid and base used are equal,i.e.c,B,=c,HB.,pH=pK,a,+lg,c,B,V,B,c,HB,V,HB,=pK,a,+lg,V,B,V,HB,three different types of of,Henderson-Hasselbalch Equation,Fight dilution,Calculating the pH of a Buffer Solution-1,PROBLEM,Sample Problem 4-1,A buffer is prepared by mixing equal volumes of 0.1 molL,-1,NaAc and 0.2 molL,-1,HAc.What is the pH of the final solution?The p,K,a,of HAc is 4.74.What will the pH be after the addition of 0.005mol of NaOH(s)to 500 mL of the buffer solution described above?,SOLUTION,Before the addition of 0.005mol of NaOH(s)to 500 mL of the buffer solution,Calculating the pH of a Buffer Solution-1,Sample Problem 4-1,SOLUTION,After the addition of 0.005mol of NaOH(s)to 500 mL of the buffer solution,Calculating the pH of a Buffer Solution-1,Sample Problem 4-1,Calculating the pH of a Buffer Solution-2,PROBLEM,Sample Problem 4-2,Calculate the pH of a buffer prepared by mixing HA with NaA.The buffer contains HA at a concentration of 0.25 molL,-1,.After 0.20g NaOH is added to 100mL of this buffer,the pH of the resulting solution is 5.60.The acid-ionization constant of HA is 5.010,-6,(pK,a,=5.3).,SOLUTION,HA(aq)+OH,-,(aq),A-(aq)+H,2,O(,l,),Initial(mol)0.025 0.005(,added,)0.1,change(mol)-0.005 -0.005 +0.005,Equil-(mol)0.02 0 0.1 +0.005,Step 1:,calculate the concentration()of conjugate base,MA,in the buffer before the addition of NaOH,Calculating the pH of a Buffer Solution-2,Sample Problem 4-2,SOLUTION,Step 2:,calculate the pH of the original buffer,Calculating the pH of a Buffer Solution-2,Sample Problem 4-2,4.3.1,The Concepts of Buffer Capacity,Buffer capacity,is defined as the amount of strong acid or base needed to change the pH of one liter of buffer by 1 unit.,Buffer capacity,is the ability to resist pH change.,Or,more specifically,4.3,Buffer Capacity and Buffer Range,where,is the buffer capacity and has units of moles per liter per pH(molL,-1,pH,-1,);d,n,a,(or b)stands for moles of strong acid or strong base which are added to a buffer solution to cause the change in pH,dpH.,The following can be derived from above one,：,=2.303,HBB,/,c,total,unit,：,mol,L,1,pH,1,The magnitude of,indicates the relative strength of buffer capacity.,The larger the value of,the greater the capacity of the buffer to resist changes in pH.,4.3.2,Factors Affecting Buffer Capacity,Buffer capacity depends on two factor:,Relative one:,buffer ratio,B,/,HB.,Absolute one:,total conc.of buffer,B,+,HB,When C,total,if fixed,：,c,B,c,HB,=,1,1,(,max,),c,B,c,HB,=,10,1,min,c,B,c,HB,=,1,10,decrease,min,For a given buffer pair,the more the buffer ratio approach 1,the stronger the capacity;when the buffer ratio equals 1,the capacity reaches its maximum.,decrease,max,=0.576c,total,When the buffer ratio,c(B,-,)/c(HB),is fixed,the more concentrated the components of a buffer,the greater the buffer capacity.,The pH of a buffer is distinct from its buffer capacity.,When the total concentration of buffer is fixed,the more the ratio of c(B,-,),/c(HB)approaches 1,the more the buffer capacity.When,c(B,-,)/c(HB)=1,the,buffer has the highest capacity.,Conclusion:,Buffer range,the pH range over which,the buffer acts effectively.,Buffers have a usable range within 1 pH unit of the pK,a,of its acid component.,4.3.3,Buffer Range,1,c,B,c,HB,=,1,(,max,),c,B,c,HB,=,10,1,min,c,B,c,HB,=,1,10,decrease,min,decrease,pH=,pK,a,1,buffer effective range,1.Choose the conjugate acid-base pair.,Calculate the ratio of buffer component concentrations.,3.Determine the buffer concentration.,4.Mix the solution and adjust the pH.,4.4,Preparing a Buffer,General steps for a buffer preparation:,An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of the acid rain on limestone-rich soils.How many grams of Na,2,CO,3,must be added to 1.5L of freshly prepared 0.20M NaHCO,3,to make the buffer?K,a,of HCO,3,-,is 4.7x10,-11,.,PLAN,We know the K,a,and the conjugate acid-base pair.Convert pH to H,3,O,+,find the number of moles of carbonate and convert to mass.,Preparing a buffer-1,PROBLEM,Sample Problem 4-3,SOLUTION,HCO,3,-,(,aq,)+H,2,O(,l,)CO,3,2-,(,aq,)+H,3,O,+,(,aq,),K,a,=,CO,3,2-,H,3,O,+,HCO,3,-,pH=10.00;H,3,O,+,=1.0 x10,-10,4.7x10,-11,=,CO,3,2-,(1.0,10,-10,),0.2,CO,3,2-,=0.094M,moles of Na,2,CO,3,=(1.5L)(0.094mols/L),=0.14,=15 g Na,2,CO,3,0.14 moles,105.99g,mol,Preparing a buffer-1,Sample Problem 4-3,Method 1:,Method 2:,We know the pH of the buffer is 10.0.The conc.of,NaHCO,3,is,0.20M,Using Henderson-Hasselbalch Equation we can find out the conc.of CO,3,2-,in the buffer.,K,a,of HCO,3,-,is 4.7x10,-11,.,CO,3,2-,=0.094M,moles of Na,2,CO,3,=(1.5L)(0.094mols/L),=0.14,=15 g Na,2,CO,3,0.14 moles,105.99g,mol,PROBLEM,Preparing a buffer-2,Sample Problem 4-4,There is 2 liter of 0.50molL,-1,NH,3,H,2,O and 2 liter of 0.50molL,-1,HCl(hydrochloric acid)in a laboratory.A technician wants to use them to prepare a buffer with pH=9.00 without the addition of water.How many liters of buffer can the technician prepare at most?What are the concentrations of NH,3,H,2,O and NH,4,+,in the buffer?The pk,b,(NH,3,H,2,O)=4.74.,SOLUTION,To prepare the buffer with the volume as more as possible,2 liter of 0.500molL,-1,NH,3,H,2,O must be utilized completely,while only a part of the HCl can be used.,Let the volume of HCl used be x L,so,the total volume of the buffer prepared is(2.00+x)L.After the neutralization,Preparing a buffer-2,Sample Problem 4-4,NH,3,(aq)+HCl(aq),NH,4,+,(aq)+Cl,-,(,l,),Initial(mol)1.0 0.5V 0,change(mol)-0.5V -0.5V +0.5V,Equil-(mol)1.0-0.5V 0 0.5V,V=1.3L,so the biggest volume of buffer is 3.3L.,Preparing a buffer-2,Sample Problem 4-4,4.5,Buffering Action in Human Blood,The pH of the blood in a healthy individual remains remarkably constant at 7.35 to 7.45.This is because the blood contains a number of buffers that protect against pH change due to the presence of acidic or basic metabolites.From a physiological viewpoint,a change of 0.3 pH unit is extreme.Acid metabolites are ordinarily produced in greater quantities than basic metabolites,and carbon dioxide is the principal one.,The buffer capacity of blood for handling CO,2,is estimated to be distributed among various buffer systems as follows:hemoglobin and oxyhemoglobin,62%;H,2,PO,4,-,/HPO,4,2-,22%;plasma protein,11%;bicarbonate,5%.Proteins contain carboxylic(,羧基,)and amino groups,which are weak acids and bases,respectively.They are,therefore,effective buffer agents.The combined buffer capacity of blood to neutralize acids is designated by clinicians as the alkali reserve,and this is frequently determined in the clinical laboratory.Certain diseases cause disturbances in the acid balance of the body.,For example,diabetes may give rise to:acidosis,which can be fatal.,An important diagnostic analysis is the CO,2,/HCO,3,-,balance in blood.This ratio is related to the pH of the blood by the Henderson-Hasselbalch equation:,where H,2,CO,3,can be considered equal to the concentration of dissolved CO,2,in the blood;6.10 is p,K,a1,of carbonic acid in blood at body temperature(37).Normally,the bicarbonate concentration in blood is about 26.0 mmolL,-1,while the concentration of carbon dioxide is 1.3 mmolL,-1,.Accordingly,for the blood,If the pH of the blood drops below 7.35,a potentially fatal condition called,acidosis,results.If the pH of the blood increases to over 7.45,another serious condition known as,alkalosis,results.,The HCO,3,-,/H,2,CO,3,buffer system is the most important one in buffering blood in the lung(alveolar blood).As oxygen from inhaled air combines with hemoglobin,the oxygenated hemoglobin ionizes,releasing a proton.This excess acid is removed by reacting with HCO,3,-,:,You know that carbonic acid also breaks down into water and gaseous carbon dioxide.The extra carbon dioxide is released in the lungs and exhaled.When extra OH,-,finds its way into the blood system,it reacts with the H,2,CO,3,.The extra bicarbonate ions formed are eventually expelled by the kidneys.,But note that the HCO,3,-,/H,2,CO,3,ratio at pH 7.4 is 26 mmolL,-1,/1.3 mmolL,-1,=20:1.This is not a very effective buffer ratio;and as significant amounts of HCO,3,-,are converted to H,2,CO,3,the pH would have to decrease to maintain the new ratio.But,fortunately,the H,2,CO,3,produced is rapidly decomposed to CO,2,and H,2,O by the enzyme decarboxylase(,脱羧酶,),and the CO,2,is exhaled by lungs.Hence,the ratio of HCO,3,-,/H,2,CO,3,remains constant at 20:1.,Study Goals,Master the composition and the mechanism of buffer solution,2.Know well how to calculate the pH of buffer solution using,The Henderson-Hasselbalch Equation,3.Master the concepts such as buffer capacity,buffer range,the factors influencing buffer capacity,4.Master how to prepare a buffer,5.Understand the application of buffer action in medicine,Exercises,3,、,5,、,7,、,8,