[理学]Chapter 4 Acid-base titration.ppt

Chapter 4 Acid-base equilibria and titrations,Activity theory 活度理论 Acid base equilibrium a. Acid-base proton theory 酸碱质子理论 b. Autoprotolysis reaction and constant of solvent c. Acid base strength d. Fractional composition equation e. Systematic treatment of equilibrium 3. Calculation of acid-base strength 酸碱强度的计算 4. Logarithm graphic method 对数图解法 5. Buffers 6. Indicators 7. Principle of acid-base titrations 8. End point error 9. Application,Key points,4.1 Activity,4.1.1 hydrated ion水合离子,The Hydrated Radius of an ion in aqueous solution is the effective radius of the ion plus its sheath(外壳）of tightly held water molecules whose electric dipoles are attracted to the ion.,The ion attracts polar water molecules and becomes surrounded or hydrated; thus, ions have a hydration sphere.,4.1 Activity (continued),The greater the charge of the ion, the more it attracts solvent molecules,Hydrated radius of Mg2+ versus Na+ ?,The larger the radius of the naked ion, the more diffuse is its electric charge and the less it attacts solvent molecules,Hydrated radius of K+ versus Na+ ?,4.1 Activity (continued),4.1.2 How do ionized species interact in solution?,a. The charge on the ions attracts or repels other ionic species as well as polar species.(离子中的电荷吸引其它离子或极性物质）,b.The more dense the charge, the stronger the attraction.(电荷密度越大，吸引力越强）,For example: If Ca 2+ present in distilled water, it is surrounded by H2O molecules to form its hydration sphere.,If Ca 2+ is present in water with Cl- present also, the Cl- has a stronger negative attraction than the dipolar negative aspect of H2O. The Cl- will gainer closer access to the Ca 2+ ion.,4.1 Activity (continued),4.1 Activity (continued),离子氛,4.1 Activity (continued),This ionic environment, in a sense, weakens the attractive force compared to the attractive force of the same ion that only has a hydration sphere (in water only), decreases the attraction of the ions ability to attract any additional ions of the opposite charge. （和水化层相比，离子氛的吸引力减弱了,离子氛降低了该离子对其它异性离子的吸引）,So what does it means?,An ion in which the hydration sphere has been changed now has an ionic environment in place of the hydration sphere. （离子氛替代 水化层）,A cation will build up anions around it, and an anion will build up cations around it.（阳离子被阴离子包围，阴离子被阳离子包围）,4.1.3 How do we determine the activity (a)?,ac = C c We know C is the molar concentration (mol/L). c is the activity coefficient for species C. It is dependent on: the ionic strength of the solution (I) the charge of the ion (z) the ionic size ().,4.1 Activity (continued),4.1 Activity (continued),Ionic Strength:,Is a measure of the total concentration of ions in solution,where: ci is molarity of ith ion zi is charge on ith ion Note: effect of square is to remove sign of charge so + and - dont cancel out!,In 1923, P. Debye and E. Hckel used the ionic atmosphere model to derive a theoretical expression that permits the calculation of activity coefficients of ions from their charge and their average size.,4.1 Activity (continued),Debye-Hckel Equation:,1. The equation works fairly well for I 0.1mol/L. 适用于稀溶液 2. As ionic strength increases, the activity coefficient decreases. For all ions, approaches unity as I approaches zero. 离子强度接近零，活度系数接近1 3. As the charge of the ion increases, the departure of its activity coefficient from unity increases. Activity corrections are much more important for an ion with a charge of 3 than for one with a charge of 1.离子的电荷增加，活度系数偏离1越远 4. The smaller the hydrated radius of the ion, the more important activity effects become. 水合离子的半径越小，活度系数越小,4.1 Activity (continued),Points concerning the Debye-Hckel Equation,4.1 Activity (continued),Definition of acid and base酸碱的定义,电离理论 电子理论 质子理论,4.2 Acid-base equilibrium,1. Acid = increases H3O+ or H+ 2. Base = increases OH- 3. Protic = transfer of H+ from one molecule to another 4. Bronsted acid = proton donor (like HCl) 5. Bronsted base = proton acceptor (like NaOH) 6. Salt = ionic solid made from reaction of acid and base (like CaCl2 ：2HCl + Ca(OH)2 = CaCl2 + 2H2O) 7. Conjugate 共轭= acid-base pair (like acetic acid HAc and acetate ion Ac-,4.2 Acid-base equilibrium（continued）,4.2.1 Protic Acids and Bases (Bronsted-Lowry),4.2.2 Autoprotolysis reaction and constant of solvent,Autoprotolysis of water水的质子自递反应,Approximate defination of pH,A useful relation between conc. of H+ and OH- is,4.2 Acid-base equilibrium（continued）,4.2.3 Strengths of acids and base,Strong acids and bases COMPLETELY dissociate and the equilibrium constants are large,HCl H+ + Cl-,Weak acids and bases: PARTIAL dissociation,acid dissociation constant Ka = H+A- / HA,For Base: B + H2O = BH+ + OH-,base hydrolysis constant Kb = BH+ OH- / B,4.2 Acid-base equilibrium,Common Strong Acids HCl Hydrogen chloride HBr Hydrogen bromide HI Hydrogen iodide H2SO4 Sulfuric acid HNO3 Nitric acid HClO4 Perchloric acid Common Strong Bases LiOH Lithium hydroxide NaOH Sodium hydroxide KOH Potassium hydroxide RbOH Rubidium hydroxide CsOH Cesium hydroxide R4NOH Quaternary ammonium hydroxide,Relationship of Ka and Kb The relationship between Ka and Kb for a conjugate pair is: Ka Kb = Kw = 1 10-14 H2A = H+ + HA- Ka (acid) (conjugate base) HA- + H2O = H2 A + OH- Kb (conjugate base) (acid) H2O = H+ + OH- Kw,4.2 Acid-base equilibrium（continued）,For polyprotic acid-base,H3PO4 H2PO4 - + H+ Ka1 Kb3 H2PO4- HPO42- + H+ Ka2 Kb2 HPO42- PO43- + H+ Ka3 Kb1,4.2 Acid-base equilibrium（continued）,多元酸碱在水中逐级离解，强度逐级递减,形成的多元共轭酸碱对中最强酸的解离常数Ka1对 应最弱共轭碱的解离常数Kb3,4.2 Acid-base equilibrium（continued）,一、分析浓度和平衡浓度： 分析浓度：溶液体系达平衡后，各组型体的 平衡浓度之和, 用C表示。 平衡浓度：溶液体系达平衡后，某一型体的 浓度，用 表示。,二、水溶液中酸碱的分布系数 = 某种型体平衡浓度 / 分析浓度,4.2 Acid-base equilibrium（continued）,4.2.4 Fraction of distribution,Monoprotic Systems 一元酸碱体系,4.2.4 Fractional composition equations P216-217,4.2 Acid-base equilibrium（continued）,For a base: B = B / C BH+ = BH+ / C,4.2 Acid-base equilibrium,Ka一定时，HA和A-与pH有关 pH，HA，A- pH pKa，HAc为主 pH = pKa，HAc = Ac- pH pKa时，Ac-为主,4.2 Acid-base equilibrium（continued）,Example,计算pH=5.0时，HAc溶液（0.1000mol/L） 中HAc和Ac-的分布系数及平衡浓度,解：,Diprotic Systems,4.2 Acid-base equilibrium（continued）,pH pKa1，H2A为主 pH = pKa1， H2A = HA- pKa1 pH pKa2 ， HA-为主 pH = pKa2， HA- = A2- pH pKa2，A2-为主,4.2 Acid-base equilibrium（continued）,Conclusion,1）分析浓度和平衡浓度是相互联系却又完全 不同的概念，两者通过联系起来 2）对于任何酸碱性物质，满足 1+ 2 + 3 + - + n = 1 3）取决于Ka，Kb及H+ 的大小，与C无关 4）大小能定量说明某型体在溶液中的分布 ，由可求某型体的平衡浓度,4.2 Acid-base equilibrium（continued）,Example,1. Mass Balance Equation (MBE),The quantity of all species in a solution containing a particular atom (or a group of atoms) must equal the amount of that atom (or a group) delivered to the solution.化学平衡中，给定物质的分析浓度等于各存在型体平衡浓度之和,4.2 Acid-base equilibrium（continued）,4.2.5 systematic treatment of equilibrium P164,For c(mol/L) H3PO4 solution,For c(mol/L) HAc solution,参见P164-165 例题,2. Charge Balance Equation (CBE),The sum of the positive charges in solution equals the sum of the negative charges in solution. CBE:,4.2 Acid-base equilibrium（continued）,Where C is the conc. of a cation, n is the charge of the cation, A is the conc. of a anion, and m is the magnitude of the charge of the anion.,Example,3. Proton Balance Equation (PBE),The sum of proton gained in acid-base reaction is equal to the sum of proton lost Method 1 PBE obtained from MBE and CBE Consider c (mol/L) NaCN solution MBE: c = HCN+CN- (1) c = Na+ (2) CBE: H+Na+ = OH-+CN- (3) From equation (1),(2) and (3), we can obtain PBE: H+HCN = OH-,4.2 Acid-base equilibrium（continued）,Method 2 Proton reference level PRL 质子参考水准,Choise of PRL a溶液中大量存在的 b参与质子转移反应 Make out PRL 等式左边得质子后产物 等式右边失质子后产物 根据质子得失相等原则列出质子条件式,4.2 Acid-base equilibrium（continued）,4.2 Acid-base equilibrium（continued）,For weak acid HB,1. Chose HB and H2O as proton reference level 2. Gaining proton product: H3O+, 3. Losing proton product: B-, OH- PBE:,For NaHCO3 solution, PRL: PBE: For NaNH4HPO4 solution, PRL: PBE:,4.2 Acid-base equilibrium（continued）,4.3 pH Calculation,4.3.1 Strong acids and bases 强酸强碱溶液P179-181,1. Strong acid HX (c mol/L),PBE: H+ = OH- + X- = OH- + c,OH- =Kw/ H+,Equation 1,H+2 c H+ Kw = 0 accurate equation,Equation 2,H+ = c simplest equation,2. Strong base MOH (c mol/L),Equation 3,OH-2 c OH- Kw = 0 accurate equation,Equation 4,OH- = c simplest equation,4.3 pH Calculation,1. When the concentration is “high” (10-6mol/L), the pH has that value we would calculate by just considering the added H+ or OH-. That is, the pH of 10-5.00 mol/L KOH is 9.00 用最简式,2. When the concentration is “low” (10-8mol/L), the pH is 7.00. we have not added enough acid or base to significantly affect the pH of the water itself.,3. At intermediate concentrations(10-6 to 10-8mol/L), the effects of water ionization and the added acid or base are comparable. Only in this region is it necessary to do a sysmetic equilibrium calculation.-用精确式,4.3 pH Calculation,4.3.2 Monoprotic weak acids and bases,1 Monoprotic weak acids HA (c mol/L),PBE: H+ = OH- + A-,For weak bases, using OH- and Kb substitute H+ and Ka, respectively.,4.3 pH Calculation（continued）,Example,Solution,4.3.3 Polyprotic acids and bases,For c mol/L H2A PBE: H+ = HA- + 2A2- + OH-,4.3 pH Calculation（continued）,So for many protic acids and bases, which have big difference in their Ka1,Ka2 value, we can treat them as monoprotic acid. But if their difference is not significant, then we will use other method that we dont discuss here.,4.3 pH Calculation（continued）,If the dissociation tendency for H2A is small, then H2A = c H+ c,4.3.4 Mixture of strong acid and weak acid,HAc (c mol/L) + HCl (ca mol/L) PBE: H+ = OH- +Ac- + ca Acidic solution, OH- is neglectable,4.3 pH Calculation（continued）,Example,Solution,4.3.5 Mixture of weak acids,cHA mol/L HA + cHB mol/L HB PBE: H+ = OH- +A- + B- Acidic solution, so OH- is neglectable,4.3 pH Calculation（continued）,4.3.6 Amphoterism (acid salt, salt of weak acid and weak base, zwitterion),Acid salt NaHA (c mol/L) PBE: H+ = OH- +A- H2A,4.3 pH Calculation（continued）,4.3 pH Calculation（continued）,Example,For other more complicate system, we can write out PBE first, then simplify the equation,Find the pH of 0.10 mol/L (NH4)2CO3 PBE:,4.3 pH Calculation（continued）,4.3 pH Calculation（continued）,4.4 Logarithm graphic method 对数图解法,横坐标和纵坐标均采用对数表示的一类图解法。 优点： 1.分布分数图适用于讨论各种分布形式浓度较高，分布分数值较大的情况，但若低于0.1，则分布线逐渐与横坐标逼近，无法进行定量分析。 2.代数法求解往往数学处理复杂，若能与浓度对数图配合使用，但很容易判断主次要组分，从而根据允许误差，忽略次要组分，再以代数法求解。 3.绘制浓度对数图时，首先列出表示图形的方程，作出浓度的对数与pH lgc=f(pH) 的关系曲线。它表示在酸碱的分析浓度和离子强度不变的情况下，溶液中各酸碱组分的浓度的对数随pH变化的曲线。,4.4.1水及强酸强碱的浓度对数图,1. log H+ = -pH 2. log OH-=log Kw-logH+ =pH-14 3. log H+ pH 是一条斜率为-1，截距为0的直线 4. log OH- pH 是一条斜率为1，截距为-14的直线,4.4 Logarithm graphic method对数图解法,4.4.2 一元弱酸的浓度对数图,在102 mol/L HAc 溶液中，存在的酸碱组分为 HAc, Ac-, H+, OH- log HAc = log(cHAc) = log = log c- pH log (H+ +Ka),4.4 Logarithm graphic method（continued),当H+ Ka，logHAc=logc， 为一条与pH轴平 行，截距为logc的直线； 2) H+ Ka， logHAc=-pH+logc+pKa, 为一条斜率-1, 截距为（logc+pKa），与log H+ 线平行的直线。 3) H+ = Ka， HAc=Ac-=c/2, logHAc= logc-0.3, 故 图上O点的坐标为（pKa, logc-0.3) 1和2两条直线的交点称体系点s，其坐标为（pKa, logc）,4.4 Logarithm graphic method（continued),对于Ac-,按上述同样方法处理。 Conclusion: log HAc pH 与 log Ac- pH 为镜面对称的两条曲线。 绘制步骤： 1 首先确定体系点s 2 过s点作斜率为0，1和1的三条直线 3 确定曲线与两条直线的相切点。与水平线相切于 （ pKa-1, logc), 与斜线相切于（ pKa1, logc-1),4.4 Logarithm graphic method（continued),二元酸的浓度对数图,对数图解法的应用,计算pH值 0.01 mol/L 的HAc 0.01 mol/L 的NaAc 0.01 mol/L 的H2A 0.01 mol/L 的NaHA 计算各种分布形式的浓度和分布分数 （ 0.01 mol/L 的H2A，用强碱滴至pH=9.0,求各种分布形式的浓度分布分数）,A buffer solution resists changes in pH when acids or bases are added or when dilution occurs. The buffer is a mixture of an acid and its conjugate base, or a mixture of a base and its conjugate acid, or strong acid or base.,4.5 Buffers p189,4.5.1 Mixing a weak acid and its conjugate base,cHB mol/L HB + CB mol/L NaB,4.5 Buffers (continued),MBE,CBE,(1)代入(3),(4)代入(2),4.5 Buffers (continued),When the cHB = cB, then pH = pKa,4.5 Buffers (continued),What will be the pH if 0.0100 mol of HA (with pKa=2.00) and 0.0100 mol of A- are dissolved in water to make 1.00 L of solution? Solution: because the solution is acidic, we neglect OH-, using equation (7),Example,Example,A buffer solution,Find the pH of a solution prepared by dissolving 12.43 g of tris (FM 121.135) plus 4.67 g of tris hydrochloride (FM 157.596) in 1.00 L of water.,SOLUTION,Example,Effect of adding acid to a buffer,If we add 12.0mL of 1.00mol/L HCl to the solution used in the previous example, what will be the new pH?,SOLUTION,Moles of HCl added: 0.0120L1.00mol/L=0.0120 mol/L H+,Moles of B=0.1026-0.0120=0.0906 Moles of BH+=0.0296+0.0120=0.0416,We see that the pH of a buffer does not change very much when a limited amount of strong acid or base is added. Addition of 12mL of 1.00mol/L HCl to 1L of unbuffered solution would have lowed the pH to 1.93.,4.5 Buffers (continued),Why does a buffer resist changes in pH?,It does so because the strong acid or base is consumed by B or BH+. If you add HCl to tris, B is convert to BH+, if you add NaOH, BH+ is converted to B. As long as you dont use up the B or BH+ by adding too much HCl or NaOH, the log term of Henderson-Hasselbalch equation does not change very much and the pH does not change very much. The buffer has its maximum capacity to resist changes of pH when pH=pKa,Example,Calculating how to prepare a buffer solution,How many milliliters of 0.500mol/L NaOH should be added to 10.0g of tris hydrochloride to give a pH of 7.60 in a final volume of 250mL?,SOLUTION,The number of moles of tris hydrochloride in 10.0g is,(10.0g)/(157.596g/moL)=0.0635,Example,Preparing a buffer in real life see p 194!,Buffer capacity: is a measure of how well a solution resists changes in pH when strong acid or base is added,4.5 Buffers (continued),Where ca and cb are the number of strong acid and strong base per liter needed to produce a unit change in pH. The greater the buffer capacity, the more resistant the solution is to pH change.,Buffer: cHB mol/L HB + CB mol/L NaB, c= cHB+ CB Adding cMOH strong base MOH,4.5 Buffers (continued),4.5 Buffers (continued),4.5 Buffers (continued),在什么情况下，缓冲溶液的缓冲容量最大呢,So the bigger the conc. of the buffer solution, and the ratio of HB/B is more approach 1:1, the higher the buffer index,4.5 Buffers (continued),4.5 Buffers (continued),In choosing a buffer, seek one whose pKa is as close as possible to the desired pH. The useful pH range of a buffer is usually considered to be pKa1 pH unit. Outside this range, there is not enough of either the weak acid or the weak base to react with added base or acid. Buffer capacity can be increased by increasing the concentration of the buffer.,4.6 Indicator p239,For example: methyl orange (MO) 甲基橙,Indicator：A weak acid or weak base whose various protonated species have different colors.,4.6 Indicator- Transition range (变色范围),Factors affecting the transition interval,Concentration of the indicator Mixing indicator Ionic strength Temperature,4.6 Indicator(continued),The concentration of two-color indicator has no effect on the transition interval, but for monocolor indicator, its conc. does affect the transition interval.,4.6 Indicator(continued),Two indicators which has similar Ka values (two color complemented) One indicator + one dye (use as a color background),4.6 Indicator(continued),Mixing indicator,4.7 Principle of acid-base titration,4.7.1 Titrating a strong acid with a strong base Titration reaction Equilibrium constant,For each type of titration studied, our goal is to construct a graph showing how the pH changes as titrant is added. The first step in each case is to write the chemical reaction, then use that reaction to calculate composition and pH after each addition of titrant.,Because the equilibrium constant for this reaction is 1014, it is fair to say that “it goes to completion. Any amount of H+ added will consume a stoichiometric amount of OH-,There are four regions of the titration curve that represent different kinds of calculations Before titration, pH is determined by the initial concentration of HCl Before reaching the SP, the pH is determined by excess H+ in the solution At the SP, H+ is just sufficient to react with all the OH- to make H2O. The pH is determined by the dissociation of water After the SP, pH is determined by the excess OH- in the solution.,4.7 Principle of acid-base titration,Equation governing a strong-acid(HX) titration,4.7 Principle of acid-base titration,Region 1 Before the titration,H+ = 0.10mol/L, pH = 1.00,Region 2 Before the SP (suppose add NaOH 18.00 mL),Region 3 At the SP,Region 4 After the SP (suppose add NaOH 20.02 mL),4.7 Principle of acid-base titration,8.0-9.6, ttp:9.1,3.1-4.4, ttp:3.4,4.4-6.2, ttp:5.2,计量点,计量点： 1. The slope (dpH/dV) is greatest,2. The second derivative is 0 (d2y/dx2=0) 这一点也称拐点,4.7 Principle of acid-base titration,C增大10倍，突跃范围增大2个pH单位,4.7.2 Titration of a weak acid with a strong base,4.7 Principle of acid-base titration,The titration curve has two easily identified points. One is the stoichiometric point which is the steepest part of the curve; the other landmark is the point where pH=pKa. This latter point has minimum slope.,The titration curve depends on the acid dissociation constant of HA and on the concentrations of the reactants.,Titration reaction,Equilibrium constant,4.7 Principle of acid-base titration,CBE,4.7 Principle of acid-base titration,Which you can rearrange to the form,改变从02，求相应的pH，即可绘制滴定曲线,Equation governing a weak-acid (HA) titration,4.7 Principle of acid-base titration,4.7 Principle of acid-base titration,Stoichiometric point,4.7 Principle of acid-base titration,As HA becomes