Presentation on theme: "Buffers, Titrations, Solubility, and other useless information"— Presentation transcript:
1 Buffers, Titrations, Solubility, and other useless information Chapter 16Buffers, Titrations, Solubility, and other useless informationFrom Kotz, Chemistry & Chemical Reactivity, 5th editionIllustrates the transformation of one insoluble lead compound into an even less soluble compound.
2 16.1 Common Ion EffectAccording to Le Chatelier’s principle: if the concentration of a substance involved in an equilibrium is changed, the system will adjust to accommodate the change and maintain the value of the KCommon ion effect: refers to the equilibrium disturbance involving a weak acid or base and its conjugate partner
3 Common Ion Effect CH3CO2H + OH- H2O + CH3CO2- The addition of more acetate ion suppresses the ionization of the acetic acid and causes the system to shift left.To calculate the pH of a solution containing both a weak acid and its conjugate base, the small changes in initial concentration can (usually) be ignored
4 Common Ion EffectEx. Suppose a solution is 0.10M acetic acid and 0.050M sodium acetate. The Ka of acetic acid is 1.8 x Calculate pH.Ka = [H3O+][CH3CO2-][CH3CO2H]1.8 x 10-5 = [H3O+][0.050] [H3O+] = 3.6x10-5[0.10]pH = What is the pH of the same solution without the common ion added?
5 Common Ion – a special case What is the pH when 25.0mL of M sodium hydroxide is added to 25.0mL of 0.100M lactic acid? (Ka of lactic acid is 1.4x10-4)mol of NaOH is presentmol of lactic acid is presentAll of the base combines with half of the acid mol of lactate ion is produced and mol of lactic acid remainThis is the half equivalence point! Ka = [H+]
6 16.2 Buffer Solutions: Controlling pH A buffer solution is resistant to change in pH when an acid or base is added to the solutionBuffer solutions are simply an example of the common ion effect; they consist of a weak acid and its conjugate base (the common ion) or a weak base and its conjugate acid in solution, in nearly equimolar quantities
7 Buffers There are two requirements for a buffer: Two substances are needed: an acid capable of reacting with added OH- ions and a base that can consume added H3O+ ionsThe acid and base must not react with each other
8 BuffersFrom a list of weak acids (or bases), one is chosen that has a pKa close to the pH value desiredThe relative amounts of the weak acid and its conjugate base are adjusted to achieve exactly the pH required
9 Some Commonly Used Buffer Systems Weak acidConjugate BaseAcid Ka (pKa)Useful pH rangePhthalic acidHydrogen phthalate ion1.3x10-3 (2.89)Acetic acidAcetate ion1.8x10-5 (4.74)Dihydrogen phosphate ionHydrogen phosphate ion6.2x10-8 (7.21)Phosphate ion3.6x10-13 (12.44)
10 BuffersIt is the relative quantities of weak acid and conjugate base that are important; the actual concentrations do not matterDiluting a buffer solution will not change its pHAn increase in the concentration of the buffer components increases the buffer capacity – more acid or base can be added without change in pH
11 Henderson-Hasselbalch Equations (a.k.a. David Hasselhoff) pH = pKa + log [A-][HA]pOH = pKb + log [HB+][B]The Henderson-Hasselbalch equation is generally valid when the ratio of [conj base]/[acid] is less than 10 and greater than 0.1
12 Using Henderson-Hasselbalch Ex. Suppose you dissolve 15.0g of NaHCO3 and 18.0g of Na2CO3 in enough water to make 1.00L of solution. Calculate the pH using Henderson-Hasselbalch.15.0g NaHCO3 = mol NaHCO3/ 1 L18.0g Na2CO3 = mol Na2CO3 / 1 LKa = 4.8x10-11 (from a table)pH = -log(4.8x10-11) + log (.170/.179)pH = 10.3
13 Preparing a Buffer Solution Ex. Prepare a 1.0L buffer solution with a pH of 4.30Select an acid whose pKa is close to the desired pH (from a table)Use either the general equation for a buffer or Henderson-Hasselbalch to calculate the concentration ratio of acid/base needed2.8/1 acid/base concentrationratio
14 How Does a Buffer Maintain pH? If you add 1.0 mL of 1.0M HCl to 1.0L of water, how does the pH change?Water has pH = 70.001 M H+ has pH = 3If you add 1.0mL of 1.0M HCl to 1.0L of 0.7M acetic acid/0.6M sodium acetate buffer, how does the pH change?The pH of the buffer is 4.68 (how do you find this?)
15 How Does a Buffer Maintain pH? 0.001 mol of H+ from the HCl combines with mol of acetate ion and produces mol of acetic acidStoich to get new concentrations from the reaction (0.599M acetate ion, 0.701M acetic acid)ICE these concentrations to getequilibrium concentrations and set up buffer equation to solve for [H+]pH = 4.68
16 16.3 Acid-Base TitrationsThe pH at the equivalence point of a strong acid/strong base titration is 7 (neutral)If weak acid is titrated with strong base then pH > 7 at equivalence point due to conj base of weak acidIf weak base is titrated with strong acid then pH < 7 at equivalence point due to conj acid of weak base
18 Titration of a Strong Acid with a Strong Base pH of the initial solution is the pH of the acidAs NaOH is added to the acid solution, amount of HCl declines, volume of solution increases, so H+ concentration decreases and pH slowly increasesThe equivalence point is the midpoint of the vertical portion of the curve; pH is 7 hereAfter all HCl has been used, pH rises slowly as more NaOH is added (and volume increases)The pH at any other point is found using stoichiometry and relationship between pH and [H+]
20 Titration of a Weak Acid with a Strong Base The pH before any titration begins is found from the Ka of the weak acid and the acid concentrationAnywhere between the start and the equivalence point, the Henderson-Hasselbalch equation can be usedAt the half-equivalence point the concentration of the weak acid is equal to the concentration of the conj base, so pH=pKaAt the equivalence point only the conj base remains; the pH is controlled by the conj base Kb and concentrationBeyond the equivalence the pH is found from the volume of the excess base added
22 Titration of a Weak Polyprotic Acid The curve can be divided into three parts:The portion of the curve up to the first equivalence point has a pH determined by the excess of the polyprotic acidThe portion of the curve between the first and second equivalence points has a pH determined by the excess of the amphiprotic substanceThe portion of the curve after the second equivalence point is has a pH determined by the excess of the fully deprotonated conj base
24 Titration of a Weak Base with a Strong Acid At the half-equivalence point, [OH-] = Kb of the weak baseThe pH at the equivalence point is weakly acidic due to the conj acid of the weak base
25 16.4 pH IndicatorsUsually a weak acid or base (treat it as such mathematically)Often a large organic molecule that has different shapes in acid and base solutionThe different structures have different colors that allows for monitoring changing pHChoose an indicator with a Ka near that of the acid being titrated so that the color change occurs at the right stage in the titration
28 Natural Indicators Red rose extract at different pH’s and with Al3+ ions In CH3OHAdd Al3+Add HClAdd NH3Add NH3/NH4+See pages 848–849
29 16.5 Solubility of SaltsSalts are considered to be insoluble if less than 0.01 moles can be dissolved per liter of waterThe equilibrium constant for the solubility of a salt is called the solubility product (Ksp), and from this molar solubility can be calculatedAddition of a common ion depresses the solubility of a salt (Le Chatelier)Direct comparisons of the solubility of two salts on the basis of their Ksp values can only be made for salts having the same ion ratio
30 BaCl2 Ba2+ + 2Cl- x 2x Ksp = [Ba2+][2Cl-]2 Ksp = (x)(4x2) Ksp = 4x3 In solubility problems, s is often substituted for x when solving for molar solubility.NaCl Ksp = s2BaCl2 Ksp = 4s3AlCl3 Ksp = 27s4Al2(SO4)3 Ksp = 108s5
31 Barium Sulfate Ksp = 1.1 x 10-10(b) BaSO4 is opaque to x-rays. Drinking a BaSO4 cocktail enables a physician to exam the intestines.(a) BaSO4 is a common mineral, appearing a white powder or colorless crystals.
32 PbCl2(s) Pb2+(aq) + 2 Cl-(aq) Common Ion EffectPbCl2(s) Pb2+(aq) Cl-(aq)Ksp = 1.9 x 10-5How will the addition of lead(II) ion or chloride ion impact the solubility of lead(II) chloride?If a saturated solution of lead(II) chloride is prepared, what will happen when sodium chloride solution is added to the mixture?
33 Common Ion Effect and Salt Solubility Ex. If solid AgCl is placed in 1.00L of 0.55M NaCl, what mass of AgCl will dissolve?AgCl Ag+ + Cl- Ksp = 1.8x10-10s = 1.3x10-5 mol/LAgClAg+Cl-I0.55C+xExx
34 Common Ion Effect and Salt Solubility Assume x is very small compared to 0.55 (because Ksp is so small)Ksp = 1.8x10-10 = (x)(0.55)X = 4.4x10-10 mol/L (which is less than 1.3x10-5 mol/L, as predicted by Le Chatelier’s principle
35 Effect of Basic Anions on Salt Solubility Any salt containing an anion that is the conjugate base of a weak acid will dissolve in water to a greater extent than given by Ksp PbS Pb2+ + S2-The conj base can hydrolyze water, which lowers the [conj base] causing more of the salt to dissolve to reestablish equilibrium S2- + H2O HS- + OH-Salts of phosphate, acetate, carbonate, cyanide, and sulfide can be affected
36 Effect of Basic Anions on Salt Solubility Insoluble salts in which the anion is the conjugate base of a weak acid will dissolve in strong acidsAnions such as acetate, carbonate, hydroxide, phosphate, and sulfide dissolve in strong acids Ex. Mg(OH)2 + 2 H3O+ Mg H2OSalts are not soluble in strong acid if the anion is the conj base of a strong acid Ex. AgCl is not soluble in strong acid because Cl- is a very weak base of a very strong acid
37 16.6 Precipitation Reactions Precipitation is the reverse process of dissolvingIf you write a dissolving reaction, its equilibrium constant expression, and its Ksp, you can write the reverse reaction and its equilbrium constant expression; notice that it has 1/Ksp!
38 Ksp and the Reaction Quotient, Q If Q = Ksp the solution is saturatedThe ion concentrations are at equilibrium valuesIf Q<Ksp the solution is not saturatedIf more of the solid is present it will continue to dissolve until equilibrium is reached; if there is no solid present, more can be addedIf Q>Ksp the solution is supersaturatedThe ion concentrations are too high and precipitation will occur until equilibrium is reached
39 Solubility and the Reaction Quotient Solid PbI2 (Ksp = 9.8 x 10-9) is placed in a beaker of water. After a period of time, the lead(II) concentration is measured and found to be 1.1 x 10-3 M. Has the system yet reached equilibrium?Q = [Pb2+][2x I-]2Q = 5.3 x This is less than Ksp, more PbI2 can dissolve.Can you figure out how much more can be added?
40 16.7 Solubility and Complex Ions Metal ions form complex ions with Lewis bases, such as ammonia and waterThe formation of complex ions increases the solubility of metal ions as predicted by the Ksp
41 16.8 Solubility, Ion Separations, and Qualitative Analysis Separating Salts by Differences in KspAdd CrO42- to solid PbCl2. The less soluble salt, PbCrO4, precipitatesPbCl2(s) + CrO42- PbCrO4 + 2 Cl-Salt KspPbCl x 10-5PbCrO x 10-14
42 Separating Salts by Differences in Ksp PbCl2(s) + CrO42- PbCrO4 + 2 Cl-Salt KspPbCl x 10-5PbCrO x 10-14PbCl2(s) Pb Cl- K1 = KspPb2+ + CrO42- PbCrO4 K2 = 1/KspKnet = K1 • K2 = 9.4 x 108Net reaction is product-favored
43 Separating Salts by Differences in Ksp A solution contains M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first?Ksp for Ag2CrO4 = 9.0 x 10-12Ksp for PbCrO4 = 1.8 x 10-14SolutionThe substance whose Ksp is first exceeded precipitates first.The ion requiring the lesser amount of CrO42- ppts. first. You MUST use molar solubility to determine this!
44 Separating Salts by Differences in Ksp A solution contains M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first?Ksp for Ag2CrO4 = 9.0 x 10-12Ksp for PbCrO4 = 1.8 x 10-14SolutionCalculate [CrO42-] required by each ion.[CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+]= 1.8 x / = 9.0 x M[CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2= 9.0 x / (0.020)2 = 2.3 x 10-8 MPbCrO4 precipitates first
45 Dissolving Precipitates by forming Complex Ions Examine the solubility of AgCl in ammonia.AgCl(s) Ag+ + Cl- Ksp = 1.8 x 10-10Ag NH3 --> Ag(NH3) Kform = 1.6 x 107AgCl(s) + 2 NH3 Ag(NH3) Cl-Knet = Ksp • Kform = 2.9 x 10-3By adding excess NH3, the equilibrium shifts to the right.