Presentation on theme: "Chapter 19 – Ionic Equilibria in Aqueous Solutions"— Presentation transcript:
1Chapter 19 – Ionic Equilibria in Aqueous Solutions 19.1 – Equilibria of Acid-Base Buffer Systems19.2 – Acid-base Titration Curves19.3 – Equilibria of Slightly Soluble Ionic Compounds19.4 – Equilibria Involving Complex Ions19.5 – Application of Ionic Equilibria to Chemical Analysis
219.1 - Equilibria of Acid-Base Buffer Systems The effect of addition of acid or base to …an unbuffered solutionacid addedbase addedor a buffered solutionacid addedbase added
319.1 - Equilibria of Acid-Base Buffer Systems Table 19.1The Effect of Added Acetate Ion on the Dissociation of Acetic Acid[CH3COOH]initial[CH3COO-]added% Dissociation*pH0.100.000.0500.151.32.890.0364.440.0184.740.0124.92* % Dissociation =[CH3COOH]dissoc[CH3COOH]initialx 100
419.1 - Equilibria of Acid-Base Buffer Systems Figure 19.3How a buffer works.Buffer after addition of H3O+H2O + CH3COOH H3O+ + CH3COO-Buffer with equal concentrations of conjugate base and acidBuffer after addition of OH-CH3COOH + OH H2O + CH3COO-H3O+OH-
519.1 - Equilibria of Acid-Base Buffer Systems Sample Problem 19.1Calculating the Effect of Added H3O+ or OH-on Buffer pHPROBLEM:Calculate the pH:(a) of a buffer solution consisting of 0.50M CH3COOH and 0.50M CH3COONa(b) after adding 0.020mol of solid NaOH to 1.0L of the buffer solution in part (a)(c) after adding 0.020mol of HCl to 1.0L of the buffer solution in part (a)Ka of CH3COOH = 1.8x10-5. (Assume the additions cause negligible volume changes.PLAN:We know Ka and can find initial concentrations of conjugate acid and base. Make assumptions about the amount of acid dissociating relative to its initial concentration. Proceed step-wise through changes in the system.SOLUTION:CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)Concentration (M)(a)InitialChangeEquilibrium0.50-0.50- x-+ x+ x0.50-x-0.50 +xx
619.1 - Equilibria of Acid-Base Buffer Systems Sample Problem 19.1Calculating the Effect of Added H3O+ and OH-on Buffer pHcontinued (2 of 4)[H3O+] = x[CH3COOH]equil ≈ 0.50M[CH3COO-]initial ≈ 0.50M[H3O+][CH3COO-][CH3COOH][H3O+] = x = Ka[CH3COO-][CH3COOH]Ka == 1.8x10-5MCheck the assumption:1.8x10-5/0.50 X 100 = 3.6x10-3 %(b)[OH-]added =0.020 mol1.0L soln= 0.020M NaOHCH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O (l)Concentration (M)Before addition0.50-0.50-Addition-0.020--After addition0.480.52-
719.1 - Equilibria of Acid-Base Buffer Systems Sample Problem 19.1Calculating the Effect of Added H3O+ and OH-on Buffer pHcontinued (3 of 4)Set up a reaction table with the new values.CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)Concentration (M)Initial0.48-0.52Change- x-+ x+ xEquilibrium0.48 -x-0.52 +xx[H3O+] = 1.8x10-50.480.52= 1.7x10-5pH = 4.770.020 mol1.0L soln(c)[H3O+]added == 0.020M H3O+Before addition0.50-0.50-Addition-0.020--After addition0.480.52-
819.1 - Equilibria of Acid-Base Buffer Systems Sample Problem 19.1Calculating the Effect of Added H3O+ and OH-on Buffer pHcontinued (4 of 4)Set up a reaction table with the new values.CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)Concentration (M)Initial0.52-0.48Change- x-+ x+ xEquilibrium0.52 -x-0.48 +xx[H3O+] = 1.8x10-50.480.52= 2.0x10-5pH = 4.70
919.1 - Equilibria of Acid-Base Buffer Systems The Henderson-Hasselbalch EquationHA + H2O H3O+ + A-Ka =[H3O+] [A-][HA]Ka [HA][A-][H3O+] =[A-][HA]- log[H3O+] = - log Ka + logpH = pKa + log[base][acid]
1019.1 - Equilibria of Acid-Base Buffer Systems Buffer capacity is the ability to resist pH change.The more concentrated the components of a buffer, the greaterthe buffer capacity.The pH of a buffer is distinct from its buffer capacity.A buffer has the highest capacity when the componentconcentrations are equal.Buffer range is the pH range over which the buffer acts effectively.Buffers have a usable range within ± 1 pH unit of the pKa ofits acid component.
1119.1 - Equilibria of Acid-Base Buffer Systems Figure 19.4The relation between buffer capacity and pH change.
1219.1 - Equilibria of Acid-Base Buffer Systems Sample Problem 19.2Preparing a BufferPROBLEM:An environmental chemist needs a carbonate buffer of pH to study the effects of the acid rain on limsetone-rich soils. How many grams of Na2CO3 must she add to 1.5L of freshly prepared 0.20M NaHCO3 to make the buffer? Ka of HCO3- is 4.7x10-11.PLAN:We know the Ka and the conjugate acid-base pair. Convert pH to [H3O+], find the number of moles of carbonate and convert to mass.SOLUTION:Ka =[CO32-][H3O+][HCO3-]HCO3-(aq) + H2O(l) CO32-(aq) + H3O+(aq)4.7x10-11 =[CO32-](0.20)1.0x10-10pH = 10.00; [H3O+] = 1.0x10-10[CO32-] = 0.094Mmoles of Na2CO3 = (1.5L)(0.094mols/L)= 0.140.14 moles105.99gmol= 15 g Na2CO3
1319.2 – Acid-base Titration Curves Colors and approximate pH range of some common acid-base indicators.Figure 19.5pH
14Curve for a strong acid-strong base titration Figure 19.7Curve for a strong acid-strong base titration
1519.2 – Acid-base Titration Curves Titration of 40.00mL of M HPr with M NaOHFigure 19.8Curve for a weak acid-strong base titrationpH = 8.80 at equivalence pointpKa of HPr = 4.89methyl red[HPr] = [Pr-]
1619.2 – Acid-base Titration Curves Sample Problem 19.3Calculating the pH During a Weak Acid-Strong Base TitrationPROBLEM:Calculate the pH during the titration of mL of M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of M NaOH:(a) 0.00mL(b) mL(c) mL(d) mLPLAN:The amounts of HPr and Pr- will be changing during the titration. Remember to adjust the total volume of solution after each addition.SOLUTION:(a) Find the starting pH using the methods of Chapter 18.Ka = [Pr-][H3O+]/[HPr][Pr-] = x = [H3O+][Pr-] = x = [H3O+]x = 1.1x10-3 ; pH = 2.96(b)HPr(aq) + OH-(aq) Pr-(aq) + H2O (l)Amount (mol)Before addition--Addition---After addition-
1719.2 – Acid-base Titration Curves Sample Problem 19.3Calculating the pH During a Weak Acid-Strong Base Titrationcontinued[H3O+] = 1.3x10-5molmol= 4.3x10-6MpH = 5.37(c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr -] will be( mol)( L) + ( L)= MKa x Kb = KwKb = Kw/Ka = 1.0x10-14/1.3x10-5 = 7.7x10-10[H3O+] = Kw / = 1.6x10-9MpH = 8.80(d) 50.00mL of NaOH will produce an excess of OH-.mol XS base = (0.1000M)( L L) = mol[H3O+] = 1.0x10-14/ = 9.0x10-11MM =pH = 12.05
18Curve for a weak base-strong acid titration Titration of 40.00mL of M NH3 with M HClFigure 19.9pKa of NH4+ = 9.25Curve for a weak base-strong acid titrationpH = 5.27 at equivalence point
19Titration of 40.00mL of 0.1000M H2SO3 with 0.1000M NaOH Figure 19.10Curve for the titration of a weak polyprotic acid.pKa = 7.19pKa = 1.85Titration of 40.00mL of M H2SO3 with M NaOH
2019.3 – Equilibria of Slightly Soluble Ionic Compounds Sample Problem 19.4Writing Ion-Product Expressions for SlightlySoluble Ionic CompoundsPROBLEM:Write the ion-product expression for each of the following:(a) Magnesium carbonate(b) Iron (II) hydroxide(c) Calcium phosphate(d) Silver sulfidePLAN:Write an equation which describes a saturated solution. Take note of the sulfide ion produced in part (d).SOLUTION:(a) MgCO3(s) Mg2+(aq) + CO32-(aq)Ksp = [Mg2+][CO32-](b) Fe(OH)2(s) Fe2+(aq) + 2OH- (aq)Ksp = [Fe2+][OH-] 2(c) Ca3(PO4)2(s) Ca2+(aq) + 2PO43-(aq)Ksp = [Ca2+]3[PO43-]2(d) Ag2S(s) Ag+(aq) + S2-(aq)S2-(aq) + H2O(l) HS-(aq) + OH-(aq)Ksp = [Ag+]2[HS-][OH-]Ag2S(s) + H2O(l) Ag+(aq) + HS-(aq) + OH-(aq)
2119.3 – Equilibria of Slightly Soluble Ionic Compounds Sample Problem 19.5Determining Ksp from SolubilityPROBLEM:(a) Lead (II) sulfate is a key component in lead-acid car batteries. Its solubility in water at 250C is 4.25x10-3g/100mL solution. What is the Ksp of PbSO4?(b) When lead (II) fluoride (PbF2) is shaken with pure water at 250C, the solubility is found to be 0.64g/L. Calculate the Ksp of PbF2.PLAN:Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression.SOLUTION:PbSO4(s) Pb2+(aq) + SO42-(aq)(a)Ksp = [Pb2+][SO42-]1000mLL4.25x10-3g100mL soln303.3g PbSO4mol PbSO4= 1.40x10-4M PbSO4Ksp = [Pb2+][SO42-] = (1.40x10-4)2 =1.96x10-8
2219.3 – Equilibria of Slightly Soluble Ionic Compounds Sample Problem 19.5Determining Ksp from Solubilitycontinued(b) PbF2(s) Pb2+(aq) + 2F-(aq) Ksp = [Pb2+][F-]20.64gL soln245.2g PbF2mol PbF2= 2.6x10-3 MKsp = (2.6x10-3)(5.2x10-3)2 =7.0x10-8
2319.3 – Equilibria of Slightly Soluble Ionic Compounds Table Solubility-Product Constants (Ksp)of Selected Ionic Compounds at 250CName, FormulaKspAluminum hydroxide, Al(OH)33 x 10-34Cobalt (II) carbonate, CoCO31.0 x 10-10Iron (II) hydroxide, Fe(OH)24.1 x 10-15Lead (II) fluoride, PbF23.6 x 10-8Lead (II) sulfate, PbSO41.6 x 10-8Mercury (I) iodide, Hg2I24.7 x 10-29Silver sulfide, Ag2S8 x 10-48Zinc iodate, Zn(IO3)23.9 x 10-6
2419.3 – Equilibria of Slightly Soluble Ionic Compounds Sample Problem 19.6Determining Solubility from KspPROBLEM:Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH)2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH)2 in water if the Ksp is 6.5x10-6.PLAN:Write out a dissociation equation and Ksp expression; Find the molar solubility (S) using a table.SOLUTION:Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Ksp = [Ca2+][OH-]2Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration (M)Initial-Change-+S+ 2SEquilibrium-S2SKsp = (S)(2S)2S == 1.2x10x-2M
2519.3 – Equilibria of Slightly Soluble Ionic Compounds Table Relationship Between Ksp and Solubility at 250CNo. of IonsFormulaCation:AnionKspSolubility (M)2MgCO31:13.5 x 10-81.9 x 10-42PbSO41:11.6 x 10-81.3 x 10-42BaCrO41:12.1 x 10-101.4 x 10-53Ca(OH)21:25.5 x 10-61.2 x 10-23BaF21:21.5 x 10-67.2 x 10-33CaF21:23.2 x 10-112.0 x 10-43Ag2CrO42:12.6 x 10-128.7 x 10-5
2619.3 – Equilibria of Slightly Soluble Ionic Compounds Figure 19.12The effect of a common ion on solubilityPbCrO4(s) Pb2+(aq) + CrO42-(aq)CrO42- addedPbCrO4(s) Pb2+(aq) + CrO42-(aq)
2719.3 – Equilibria of Slightly Soluble Ionic Compounds Sample Problem 19.7Calculating the Effect of a Common Ion onSolubilityPROBLEM:In Sample Problem 19.6, we calculated the solubility of Ca(OH)2 in water. What is its solubility in 0.10M Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.PLAN:Set up a reaction equation and table for the dissolution of Ca(OH)2. The Ca(NO3)2 will supply extra [Ca2+] and will relate to the molar solubility of the ions involved.SOLUTION:Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration(M)Initial-0.10Change-+S+2SEquilibrium-S2SKsp = 6.5x10-6 = ( S)(2S)2 = (0.10)(2S)2S << 0.10Check the assumption:S == 4.0x10-30.10M4.0x10-3x 100 =4.0%
2819.3 – Equilibria of Slightly Soluble Ionic Compounds Sample Problem 19.8Predicting the Effect on Solubility of AddingStrong AcidPROBLEM:Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of these ionic compounds:(a) Lead (II) bromide(b) Copper (II) hydroxide(c) Iron (II) sulfidePLAN:Write dissolution equations and consider how strong acid would affect the anion component.SOLUTION:(a) PbBr2(s) Pb2+(aq) + 2Br-(aq)No effect.(b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)OH- is the anion of water, which is a weak acid. Therefore it will shift the solubility equation to the right and increase solubility.(c) FeS(s) Fe2+(aq) + S2-(aq)S2- is the anion of a weak acid and will react with water to produce OH-.FeS(s) + H2O(l) Fe2+(aq) + HS-(aq) + OH-(aq)Both weak acids serve to increase the solubility of FeS.
2919.3 – Equilibria of Slightly Soluble Ionic Compounds Sample Problem 19.9Predicting Whether a Precipitate Will FormPROBLEM:A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO3)2 is mixed with 0.200L of 0.060M NaF?PLAN:Write out a reaction equation to see which salt would be formed. Look up the Ksp valus in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < Ksp. Remember to consider the final diluted solution when calculating concentrations.SOLUTION:CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = 3.2x10-11mol Ca2+ = L(0.30mol/L) = 0.030mol[Ca2+] = 0.030mol/0.300L = 0.10Mmol F- = L(0.060mol/L) = 0.012mol[F-] = 0.012mol/0.300L = 0.040MQ = [Ca2+][F-]2 =(0.10)(0.040)2 = 1.6x10-4Q is >> Ksp and the CaF2 WILL precipitate.
3019.5 – Ionic Equilibria in Chemical Analysis Sample Problem 19.12Separating Ions by Selective PrecipitationPROBLEM:A solution consists of 0.20M MgCl2 and 0.10M CuCl2. Calculate the [OH-] that would separate the metal ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of Cu(OH)2 is 2.2x10-20.PLAN:Both precipitates are of the same ion ratio, 1:2, so we can compare their Ksp values to determine which has the greater solubility.It is obvious that Cu(OH)2 will precipitate first so we calculate the [OH-] needed for a saturated solution of Mg(OH)2. This should ensure that we do not precipitate Mg(OH)2. Then we can check how much Cu2+ remains in solution.SOLUTION:Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = 6.3x10-10Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = 2.2x10-20[OH-] needed for a saturated Mg(OH)2 solution == 5.6x10-5M
3119.5 – Ionic Equilibria in Chemical Analysis Sample Problem 19.12Separating Ions by Selective PrecipitationcontinuedUse the Ksp for Cu(OH)2 to find the amount of Cu remaining.[Cu2+] = Ksp/[OH-]2 = 2.2x10-20/(5.6x10-5)2 =7.0x10-12MSince the solution was 0.10M CuCl2, virtually none of the Cu2+ remains in solution.
3219.5 – Ionic Equilibria in Chemical Analysis Figure 19.17The general procedure for separating ions in qualitative analysis.Add precipitating ionAdd precipitating ionCentrifugeCentrifuge
3319.5 – Ionic Equilibria in Chemical Analysis A qualitative analysis scheme for separating cations into five ion groups.Figure 19.18Acidify to pH 0.5; add H2SCentrifugeAdd NH3/NH4+ buffer(pH 8)CentrifugeAdd (NH4)2HPO4CentrifugeAdd6M HClCentrifuge