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patino_chm2046_chapter16 - Chapter 16 Aqueous Ionic...

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Unformatted text preview: Chapter 16 Aqueous Ionic Equilibrium Review: Chapter 15 Acid-base titration Rxn stoichiometry Solubility rules GOALS pH and buffers pH changes in acid-base titrations Equilibria and solubility …..do not forget the calculations!! The Common Ion Effect Q: What is the effect on the pH of adding Q: 0.10 M NH4Cl (NH4+Cl-) to 0.25 M NH3(aq)? 0.10 Cl (NH to NH3(aq) + H2O ⇄ NH4+(aq) + OH-(aq) NH NH Here we are adding NH4+, an ion COMMON to an COMMON the equilibrium. Le Chatelier predicts that the equilibrium will shift to the left (1), right (2), no change (3). left NH4+ is an acid; will the pH go up (1), down (2), no change (3). no pH of NH3/NH4+ Mixture pH Q: What is the pH of a solution with Q: 0.10 M NH4Cl and 0.25 M NH3(aq)? Add NH4+(aq) NH (aq)? Add (aq) NH3(aq) + H2O ⇄ NH4+(aq) + OH-(aq) NH We expect that the pH will decrease on adding We NH4Cl (adding acid). NH ). The common ion effect is the The common limiting of the ionization of an acid/base by the presence of its acid/ conjugate base/acid. conjugate Buffers • solutions that resist changes in pH when an acid or base is added • they act by neutralizing the added acid or base • but just like everything else, there is a limit to what they can do, eventually the pH changes • made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion (or vice versa) Buffer Solutions Buffer HCl is added to pure HCl water (large dec in pH; 6.29 water large to 1.80). ∆ pH = 4.49 to ). HCl is added to a solution HCl of a weak acid, H2PO4weak and its conjugate base, conjugate HPO42- (slight dec in pH, HPO slight 6.92 to 6.72). ∆ pH = 0.20 6.92 ). Buffer Solutions A buffer solution is a special case of the common ion effect (adding an ion already present). ion ). The function of a buffer is to resist changes in the pH of a solution when acid or base (small amount) is added. is buffers work by applying Le Châtelier’s Principle to weak acid equilibrium Buffer Composition Weak Acid + Conj. Base HOAc + OAcH2PO4+ HPO42NH4+ + NH3 Buffer Solutions Consider HOAc/OAc- to see how buffers work. Consider ACID USES UP ADDED OHWe know that OAc- + H2O ⇄ HOAc + OH- ; Kb = 5.6 × 10-10 OAc base acid Therefore, if we add OH− to the solution, the reverse reaction of the WEAK ACID with added reverse OH- has OH Kreverse = 1/ Kb = 1.8 × 109 Kreverse is VERY LARGE, so HOAc - Buffer Solutions Consider HOAc/OAc- to see how buffers work. Consider CONJUGATE BASE USES UP ADDED H+ HOAc + H2O ⇄ OAc- + H3O+ Ka = 1.8 × 10-5 acid conj base Therefore, the reverse reaction of the Therefore, reverse WEAK BASE with added H+ has WEAK Kreverse = 1/ Ka = 5.6 × 104 Kreverse is VERY LARGE, so OAccompletely gobbles up the added H+!! completely Buffer Solutions Q: What is the pH of a buffer that has [HOAc] Q: = 0.700 M and [OAc-] = 0.600 M? 0.700 HOAc + H2O ⇄ OAc- + H3O+ Ka = 1.8 × 10-5 0.700 M HOAc has pH = 2.45 The pH of the buffer will have 1. pH < 2.45 2. pH > 2.45 3. pH = 2.45 Buffer Solutions Q: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? 0.700 HOAc + H2O ⇄ OAc- + H3O+ Ka = 1.8 × 10-5 [HOAc] [HOAc] initial change equilib 0.700 -x 0.700 - x [OAc-] 0.600 +x 0.600 + x [H3O+] 0 +x x Buffer Solutions Q: What is the pH of a buffer that has [HOAc] = 0.700 M Q: and [OAc-] = 0.600 M? and HOAc + H2O ⇄ OAc- + H3O+ Ka = 1.8 × 10-5 [HOAc] [HOAc] [OAc-] [H3O+] equilib 0.700 - x 0.600 + x x Assuming that x << 0.700 and 0.600, we have [H3O+ ](0.600) Ka = 1.8 x 10-5 = 0.700 [H3O+] = 2.16 × 10-5 and pH = 4.67 > 2.16 2.45 2.45 Buffer Solutions Notice that the expression for calculating the H+ (OH-) conc. of the buffer is [H3O + ] = [H3O + ] = Orig. conc. of HOAc Orig. conc. of OAc- [Acid] • Ka [Conj. base] [OH− ] = • Ka [Base] • Kb [Conj. acid] For a buffer of base B and acid BH+ Notice that the [H+] or [OH-] depend on (1) K depend and (2) the ratio of acid and base concentrations. concentrations. Henderson-Hasselbalch Equation [H3O + ] = [Acid] • Ka [Conj. base] Take the negative log of both sides of this Take negative equation: equation: 1 [Acid] pH = pKa - log [Conj. base] take [acid] [base] pH = pKa + log to remove − ve sign [Conj. base] [Acid] The pH is determined largely by the pKa of the The pK acid and then adjusted by the ratio of acid and acid conjugate base. conjugate pH of Buffer (Ka) Q: What is the pH of a buffer that has [HOAc] = 0.700 Q: M and [OAc-] = 0.600 M? and 0.600 HOAc + H2O ⇄ OAc- + H3O+ HOAc + -5 -5 [H3O ] = 2.1 × 10 and pH = 4.67 2.1 pH K = 1.8 × 10 a [Conj. base] pH = pKa + log [Acid] 0.600 pH = − log (1.8 × 10 ) + log 0.700 −5 pH = 4.74 − 0.067 = 4.67 What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl? find the pKb if given Kb Assume the [B] and [HB+] equilibrium concentrations are the same as the initial Substitute into the HendersonHasselbalch Equation Base Form, find pOH Check the “x is small” approximation Calculate pH from pOH NH3 + H2O ⇄ NH4+ + OH− [HB+ ] pOH = pK b + log [B] ( 0.20 ) pOH = 4.75 + log = 4.35 ( 0.50 ) [OH − ] = 10 -pOH = 10 -4.35 = 4.47 × 10 −5 4.47 × 10 −5 × 100% = 0.0089% < 5% 0.50 pH + pOH = 14.00 pH = 14.00 - 4.35 = 9.65 16 pH from using Kb pH Q: What is the pH of a buffer that has What 0.12 mol of NH 3 and 0.095 mol NH4Cl. Kb = 1.8 × 10-5 0.095 Cl. Equilibrium: NH3 + H2O ⇄ NH4+ + OH− Equilibrium: Solution Solution [acid] pOH = pKb + log [base] 0.095 pOH = 4.74 + log 0.12 pOH = 4.74 +( - 0.10) pOH = 4.64 pH = 14.00 – 4.64 = 9.36 The fundamental equations of buffers The If you have a buffer made of a weak acid, HA, and a salt If containing its conjugate base, A−, such as KA (K+ = containing potassium ion), I recommend you to work with the following equilibrium: following HA + H2O ⇄ H3O+ + A− Ka, K of acid HA. HA Any of the four, [HA], [H3O+], [A−], or Ka may be unknown, whereas the others are given. If you have a buffer made of a weak, B, and a salt If containing its conjugate acid, BH+, such as BHCl (Cl− = containing chloride ion. B is not boron element), I recommend you is ), to work with the following equilibrium: to B + H2O ⇄ BH+ + OH− Kb, K of base B. Any of the four, [B], [BH+], [OH−], or Kb may be unknown Adding an Acid to a Buffer Adding Q: What is the pH when 0.020 mol HCl is added to a solution solution with 0.12 mol of NH3 and 0.095 mol NH4Cl. Kb = 1.8 × 10-5 with Cl. Solution H3O+ from HCl will react with NH3 to produce NH4Cl. [NH3] will dec and [NH4Cl] will increase. The new concns (& K) will Cl] will determine the pH. will NH3 + H3O+ → NH4+ + H2O NH + mol NH3 + mol H3O+ → mol NH4+ mol + 0.12 0.02 0.095 0.12 0.095 -0.02 -0.02 +0.02 -0.02 0.10 0 0.115 K >>1 before rxn before changes after rxn The buffer remains ‘alive’. Now we’ll substitute The concentrations in the Equil. of NH3/NH4+ concentrations Adding an Acid to a Buffer Adding Q: What is the pH when 0.020 mol HCl is added to What 0.12 mol of NH3 and 0.095 mol NH4Cl. Kb = 1.8 × 10-5 0.12 Cl. Solution NH3 + H2O ⇄ NH4+ + OH− NH after rxn 0.10 mol 0.115 mol ? Kb 0.10 mol ———— V [OH−] = 1.8× 10−5 × ————— = 1.57 × 10−5 M [OH ————— 0.115 mol ———— pOH = 4.81 pOH V pH = 9.19 pH Important: pH doesn’t depend on V of buffer solxn Adding an Acid to a Buffer REMOVED!!! REMOVED!!! Q: What is the pH when 0.020 mol HCl is added to Q: What 0.12 mol of NH3 and 0.095 mol NH4Cl. Kb = 1.8 × 10-5 0.12 Cl. Solution Ka = 5.6 × 10-10 ⇒ pKa = 9.25 [base] pH = pKa + log [acid] 0.10 pH = 9.25 + log 0.115 pH = 9.25 +( - 0.061) pH = 9.19 pH decreased from 9.36 to 9.19 (∆ = 0.17) when 0.020 mol HCl was added to buffer containing 0.12 mol NH3 and HCl was 0.095 mol NH4Cl!! 0.095 Preparing a Buffer Preparing We want to buffer a solution at pH = 4.30. We This means [H3O+] = 10-pH = 5.0 × 10-5 M This It is best to choose an acid such that [H3O+] is is about equal to Ka (or pH ≈ pKa). about —then get the exact [H3O+] by adjusting the ratio of by acid to conjugate base. acid [Acid] +] = [H3O • Ka [Conj. base] For a buffer of basic pH (e.g. 10.13) we need to use a For base B with pKb about equal to 3.87 (because pOH =3.87) and a salt containing its conjugate acid BH+ =3.87) Preparing a Buffer We want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 × 10-5 M [H 5.0 POSSIBLE ACIDS/BASES HSO4− / SO42HSO SO HOAc / OAcHOAc HCN / CNHCN -Choose Ka 1.2 x 10-2 -5 √ 1.8 x 10 4.0 x 10-10 an acid such that Ka ≈ [H3O+] (or pKa ≈ pH). (or Best choice is acetic acid / acetate ion. Preparing a Buffer You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 × 10-5 M. With acetic acid, Ka=1.8x10−5 [H 5.0 With + ] = 5.0 x 10-5 = [HOAc] (1.8 x 10 ) -5 [H3O - [OAc ] Solve for Solve [HOAc] - ] ratio = 2.78 [OAc 2.78 = 1 Therefore, if you use 0.100 mol of NaOAc and 0.278 Therefore, mol of HOAc in 1.00 L, the pH = 4.30. mol [HOAc] = 2.78 x [OAc−] = 2.78 x 0.100 = 0.278 M Effectiveness of Buffers The relative concentrations of acid & base should not differ by a more than a factor of 10 to be reasonably effective; [base]/[acid] = 10 or [base]/[acid] = 0.1 • In fact, a buffer will be most effective when the [base]:[acid] = 1 that is, equal concentrations of acid and base a) [acid]=[base] = 0.5 M b) [acid]=[base] = 0.1 M • Also, a buffer will be most effective when the [acid] and the [base] are large 25 Buffering Range for an effective buffer, 0.1 < [base]:[acid] < 10 for • we can use Henderson-Hasselbalch eqn to calculate the maximum and minimum pH over which the buffer will be effective [A - ] pH = pK a + log [HA] Lowest pH pH = pK a + log( 0.10 ) pH = pK a − 1 Highest pH pH = pK a + log(10 ) pH = pK a + 1 therefore, the effective pH range of a buffer is pKa ± 1 when choosing an acid to make a buffer, choose one whose pKa is the closest to the pH of the buffer 26 Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25? Chlorous Acid, HClO2 Nitrous Acid, HNO2 Formic Acid, HCHO2 Hypochlorous Acid, HClO pKa = 1.95 pKa = 3.34 pKa = 3.74 pKa = 7.54 The pKa of HCHO2 is closest to the desired pH of the buffer, so it would give the most effective buffering range. pH – pKa = 0.51, it is in the rank pH - pKa = ± 1 27 All of the following statements concerning buffers are true EXCEPT a) buffers resist pH changes upon addition of small quantities of strong acids or bases. b) buffers are used as indicators in acid-base titrations. c) the pH of a buffer is close to the pKa of the weak acid from which it is made. d) buffers contain appreciable quantities of a weak acid and its conjugate base. e) buffers are resistant to changes in pH when diluted with water. Answer: b Preparing a Buffer Preparing What mass of NH4Cl should be added to 2.55 L of a 0.155 M NH3 to make a buffer having a pH of 9.55? Kb of NH3 = 1.8× 10-5 pKb = -log 1.8× 10−5 = 4.74 Plan: Kb, pOH and [NH3] ⇒[NH4Cl] ⇒mass NH4Cl Cl] pOH =14 –pH =4.45 [OH-]=10−4.45 =3.55x10−5M NH3 +H2O ⇄ NH4+ +OH− The fundamental eq. 0.155 M ? 0.0000355 M Kb=[NH4+][OH−]/[NH3] [NH4+] can be calculated [NH4+] [NH4+]= Kb [NH3]/[OH−] [NH [NH +] = 1.8× 10−5× 0.155 / 3.55× 10−5 = 0.0786 M Preparing a Preparing What mass of NH Cl should be added to Buffer Buffer 4 2.55 L of 0.155 M NH3 to make a buffer having a pH of 9.55? Kb = 1.8 × 10-5 Plan: Ka, pH and [NH3] ⇒ [NH4Cl] mass NH4Cl Cl moles of NH4Cl = moles of NH4+ = moles = 0.0786 mol/L x 2.55 L = 0.201 mol Molar mass of NH4Cl = 53.59 g/mol mass NH4Cl = 0.201 mol x 53.49 g/mol= 10.6 g You cannot add isolated NH4+ ions, but a salt such You add as NH4Cl (this example) or NH4NO3. These salts are as These Acid-Base Reactions Acid-Base • Strong acid + strong base HCl + NaOH → NaCl + NaCl • Strong acid + weak base HCl + NH3 → NH4Cl HCl NH • Weak acid + strong base HOAc + NaOH → NaOAc + NaOAc • Weak acid + weak base HOAc + NH3 → NH4OAc HOAc What is the relative What pH before, during, before during & after an acid-base after reaction? reaction? H3CCOOH = HC2H3O2 = HOAc Titrations & titration curves Titrations & titration curves pH pH pH SA-SB, pH = 7 WA-SB, pH > 7 SA-WB, pH < 7 WA-WB, pH ? Solution of known concn added to another Titrant volume, mL Titrant volume, mL QUESTION: Titrate 100. mL of a 0.025 M QUESTION: Titrate 100. mL of a 0.025 M QUESTION: solution of benzoic acid ((Ka= 6.25E-5)) with solution of benzoic acid Ka = 6.25E-5 with solution with solution with 0.100 M NaOH to the equivalence point. 0.100 M NaOH to the equivalence point. 0.100 0.100 pH at pH at pH pH half-way half-way point? point? point? point? Benzoic acid Benzoic acid + NaOH + NaOH pH after pH end pt!! pH at pH at pH pH equivalence equivalence point? point? point? point? pH of solution of pH of solution of pH pH benzoic acid, a benzoic acid, a weak acid? weak acid? weak weak Acid­Base Titration Acid­Base Titration Q: Titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the solution at the equivalence point? Ka of HBz = 6.25x10−5 Ka HBz + NaOH → Na+ + Bz- + H2O HBz + OH− → Bz- + H2O K=1/Kb of BzHBz OH This reactions happens until completion; K>>1 (see next) C6H5CO2H = HBz Benzoate ion = Bz- Acid­Base Titration Acid­Base Titration Q: Titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. the What is the pH of the final solution? At this point, VNaOH = 25 mL Mac Vac = Mbase Vbase At Ka =6.25E-5 HBz + OH− → Bz− + H2O K=1/Kb = ————— ————— Kw = 1E-14 K = 6.25 x 109 too large! Complete reaction! 6.25 Complete The pH of the final solution will be 1. Less than 7 2. Equal to 7 3. Greater than 7 Acid­Base Titrations Acid­Base Titrations The product of the titration of benzoic The acid is the benzoate ion, Bz−. acid Bz- is the conjugate base of a weak acid (HBz). acid Therefore, the final solution is basic: Therefore, Bz− + H2O ⇄ HBz + OH− pH > 7 Bz Kb = 1.6 x 10-10 Acid­Base Reactions Acid­Base Reactions Q: Titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? What Strategy — ffind the conc. of the Strategy — ind the conc. of the conjugate base, Bz−,, iin the solution @ conjugate base, Bz− n the solution @ tthe END of the titration, then calculate he END of the titration, then calculate pH. pH. This is a two-step problem This is a two-step problem 1. stoichiometry off acid-base 1. stoichiometry o acid-base reaction reaction 2. equilibrium calculation 2. equilibrium calculation Acid­Base Reactions Acid­Base Reactions Q:Titrate 100. mL of a 0.025 M solution of benzoic acid with Titrate 0.100 M NaOH to the equivalence point. What is the pH of the final solution? the STOICHIOMETR Y PORTION STOICHIOMETR Y PORTION 1. Calc. moles of NaOH req’d 1. Calc. moles of NaOH req’d (0.100 L HBz)(0.025 M) = 0.0025 mol HBz (0.100 L HBz)(0.025 M) = 0.0025 mol HBz This requires 0.0025 mol NaOH This requires 0.0025 mol NaOH This 0.0025 This 0.0025 2. Calc. volume of NaOH req’d 2. Calc. volume of NaOH req’d 0.0025 mol (1 L // 0.100 mol) = 0.025 L 0.0025 mol (1 L 0.100 mol) = 0.025 L 25 mL of NaOH req’d 25 mL of NaOH req’d Or, from Mac Vac = Mbase Vbase Or, from Mac Vac = Mbase Vbase Or, Or, base base we get Vbase = 25 mL we get Vbase = 25 mL Acid­Base Reactions Acid­Base Reactions Q: Titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? final STOICHIOMETRY PORTION STOICHIOMETRY PORTION HBz + OH− → Bz− + H2O K is too large. HBz + OH− → Bz− + H2O K is too large. HBz HBz 25 mL of NaOH req’d 25 mL of NaOH req’d 25 25 3.. Moles of Bz-- produced = moles HBz = 0.0025 mol 3. Moles of Bz produced = moles HBz = 0.0025 mol 3. 0.0025 3 0.0025 4. Calc. conc. of Bz-4. Calc. conc. of Bz There are 0.0025 mol of Bz-- iin a TOTAL There are 0.0025 mol of Bz n a TOTAL SOLUTION VOLUME of 125 mL = (100. + 25). mL SOLUTION VOLUME of 125 mL = (100. + 25). mL SOLUTION SOLUTION [Bz-] = 0.0025 mol / 0.125 L = 0.020 M 0.0025 0.020 Ac id ­Ba s e R e a c tio ns Ac Q: Titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH at equivalence point? equivalence Eq uiva le nc ee P o int Eq uiva le nc P o int Most important species in solution is benzoate ion, Bz --,, tthe Most important species in solution is benzoate ion, Bz the he the conjugate base of benzoic acid, HBz. conjugate base of benzoic acid, HBz. Bz-- + H2O ⇄ HBz + OH-Bz + H2O ⇄ HBz + OH Bz Bz [Bz--]] [[Bz [Bz Bz initial initial change change Equilib Equilib Equilib Equilib [HBz] [[HBz] [HBz] HBz] 0.020 0.020 -x -x 0.020 - x 0.020 - x 0 0 +x +x x x Kb = 1.6 x 10−10 Kb = 1.6 x 10−10 [[OH--]] OH 0 0 +x +x x x Ac id ­Ba s e R e a c tio ns Ac Q: Titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH at equivalence point? equivalence Eq uiva le nc e P o int Most important species in solution is benzoate ion, Bz -, the conjugate base of benzoic acid, HBz. Bz- + H2O ⇄ HBz + OHBz HBz 2 x Kb = 1.6 x 10-10 Kb = 1.6 x 10 ­10 = 0.020 ­ x x = [OH-] ≈ SQRT(0.020 × 1.6× 10−10)= 1.8 x 106 Q::Titrate 100. mL of a 0.025 M solution of benzoic acid with Q: Titrate 100. mL of a 0.025 M solution of benzoic acid with Q 0.100 M NaOH to the equivalence point. What is the pH at 0.100 M NaOH to the equivalence point. What is the pH at half-way point? (i.e.,12.5 mL added of the 25 mL needed for half-way point? (i.e.,12.5 mL added of the 25 mL needed for the equivalence point) tthe equivalence point) he the pH at half-way point? pH at half-way point? 1. <7 1. <7 2. =7 2. =7 3. >7 3. >7 @ equivalence @ equivalence equivalence equivalence point, pH = 8.25 point, pH = 8.25 point, point, pH at half-way pH at half-way pH pH mL) mL) mL) mL) point (12.5 point (12.5 Ac id ­Ba s e R e a c tio ns Ac Titrate 100. mL of a 0.025 M solution of benzoic acid Titrate with 0.100 M NaOH. with What is the pH at the half-way point? What HBz + H2O ⇄ H3O+ + Bz- Both HBz and Bz-- are Both HBz and Bz are present at same [[ ]= M present at same ]= M This is a BUFFER! This is a BUFFER! Ka = 6.3 x 10-5 + [HBz] [H 3 O ] = - • Ka [Bz ] At the half-way point, [HBz] = [Bz-], because half of At [HBz], initial moles of HBz has been converted to Bz− initial Therefore, [H3O+] = Ka = 6.3 x 10-5 pH = 4.20 = pKa of the acid A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH Write an equation for the rxn of KOH with HNO2. Determine the mol of acidbefore & mol of added KOH Make a stoichiometry table and determine the moles of HNO2 in excess and moles NO2− made HNO2 + OH− → NO2− + H2O 40.0 mL × 0.001 L 0.100 mol HNO 2 × = 0.00400 mol HNO 2 1 mL 1L 0.001 L 0.200 mol KOH 5.00 mL × × = 0.00100 mol KOH 1 mL 1L HNO2 mols before 0.00400 mols added -0.00100 mol reacted mols after NO2- OH− 0 ≈0 - 0.00100 +0.0010 -0.00100 0 0.00300 0.00100 ≈0 44 A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH. HNO2 + H2O ⇄ NO2− + H3O+ ( ) pK a = − log K a = − log 4.6 × 10 −4 = 3.15 Ka = 4.6 x 10-4 Write an equation for the reaction of HNO2 with H2O Determine Ka and pKa for HNO2 Use the HendersonHasselbalch eqn to determine the pH NO 2 − pH = pK a + log HNO 2 0.00100 pH = 3.15 + log = 2.67 0.00300 HNO2 NO2- OH− 0 ≈0 mols Before 0.00400 mols added 0.00100 mols After 0.00300 0.00100 ≈ 0 45 Titra tio n C urve s : Ac e tic a c id titra te d with Na O H T itra p H = 8 .7 Weak acid titrated with a strong Weak base base Strong acid (HCl) titrated with Strong acid (HCl) titrated with Strong a strong base (KOH) a strong base (KOH) Same for HCl + NaOH, and Same for HCl + NaOH, and HNO3 + KOH or NaOH HNO3 + KOH or NaOH Weak diprotic acid (H2C2O4) titrated with a (H titrated strong base (NaOH) strong Figure 18.7 Figure Weak base (NH3) Weak base (NH3 ) ttitrated with a strong itrated with a strong acid (HCl) or HNO3 acid (HCl) or HNO3 pH indicators pH An acid-base titration requires an acid-base indicator to determine the equivalence point. indicator This is usually an organic compound that is a weak acid or weak base (similar to dyes of flowers). (similar The acid form (HInd) and the conjugate The and base (Ind-) have different colors. Thanks to base have that we can ‘see’ the equivalence point. that HInd(aq) + H2O(liq) ⇄ H3O+ (aq) + Ind-((aq) (aq) Ind aq) HInd phenolphthalein Ac id ­Ba ss e A c id ­Ba e Ind ic a to rs Ind ic a to rs (we a k (we a k a c id ss /b a ss e ss )) a c id /b a e PRECIPITATION REACTIONS PRECIPITATION Solubility of Salts Apply equilibrium principles to Apply (i) acid-base and now……… acid-base (ii) precipitation reactions. precipitation Precipitation rxn: AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq) Ag+ + NO3− + K+ + Cl− → AgCl(s) + K+ + NO3− Ag+(aq) + Cl−(aq) ⇄ AgCl(s) the net ionic eqxn Ksp of AgCl Ag+ Pb2+ Hg22+ Ag+ Pb2+ Hg22+ ClAgCl PbCl2 Hg2Cl2 AgCl PbCl2 Hg2Cl2 Solid at the bottom Solid of beaker of All salts formed in this All experiment are said to be INSOLUBLE and INSOLUBLE form when mixing moderately concentrated solutions of the metal ion with chloride ions. ion Review solubility rules, chapter 4 of tetxbook. Ag+ Pb2+ Hg22+ Ag+ Pb2+ Hg22+ Cl AgCl PbCl22 Hg22Cl2 AgCl PbCl Hg Cl2 Ksp of AgCl Although AgCl is said to be insoluble, it insoluble it dissolves to a SLIGHT extent. dissolves AgCl(s) ⇄ Ag+(aq) + Cl−(aq) AgCl(s) Ag undissolved solid in equilibrium with its ions in solution. When equilibrium has been established, no more When AgCl dissolves and the solution is SATURATED. SATURATED Ag+ Pb2+ Hg22+ Ag+ Pb2+ Hg22+ Cl Cl AgCl PbCl22 Hg22Cl2 AgCl PbCl Hg Cl2 Ksp of AgCl AgCl(s) ⇄ Ag+(aq) + Cl-(aq) When the soln is SATURATED, expt. shows that When SATURATED expt. [Ag+] = 1.67 x 10-5 M. [Ag This is equivalent to the SOLUBILITY of AgCl. This SOLUBILITY of What is [Cl-]? [Cl-] = [Ag+] = 1.67 x 10-5 M, [Cl the same. -based on the stoichiometry, one Cl− per one Ag+ Ag+ Pb2+ Hg22+ Ag+ Pb2+ Hg22+ ClCl AgCl PbCl22 Hg22Cl2 AgCl PbCl Hg Cl2 Ksp of AgCl AgCl(s) ⇄ Ag+(aq) + Cl-(aq) The saturated solution has The [Ag+] = [Cl-] = 1.67 x 10-5 M Use this to calculate Kc Kc = Ksp = [Ag+] [Cl-] AgCl is solid AgCl = (1.67 x 10-5)(1.67 x 10-5) = (1.67 x 10-5)2 = 2.79 x 10-10 Ksp = solubility product constant Lead(II) chloride Lead(II) PbCl2(s) ⇄ Pb2+(aq) + 2 Cl-(aq) (s) (aq) Ksp = 1.9 x 10-5 = [Pb2+][Cl–]2 Silver Silver Agcarbonate CO (s) ⇄ 2Ag (aq) (s) carbonate 2 3 + + CO32-(aq) (aq) Ksp = 8.5 x 10-12 = [Ag+]2[CO32-] Ksp from solubility Consider PbI2 dissolving in water. Consider PbI2(s) ⇄ Pb2+(aq) + 2 I-(aq) (s) Calculate Ksp if the molar solubility Calculate PbI2 = 0.00130 M PbI Solution Solution 1. Solubility = [Pb2+] = 1.30 × 10-3 M 1.30 [I-] = ? [I [I-] = 2 x [Pb2+] = 2.60 × 10-3 M 2.60 Ksp = (1.30 × 10-3 )(2.60 × 10-3 )2 = 8.79 × 10-9 Ksp = solubility × (2× solubility)2 = 4 S3 Ksp from solubility Consider MgF2 dissolving in water. MgF2(s) ⇄ Mg2+(aq) + 2 F-(aq) (s) Calculate Ksp Calculate 4 if molar its solubility = 2.65 × 10--4 M solubility Solution (alternatively) Solution Ksp = [Mg2+] [F-]2 = [Mg2+] {2 • [Mg2+]}2 [Mg Ksp = 4 [Mg2+]3 = 4 (solubility)33 = 4 (solubility) Ksp = 4 ((2.65 × 10-4))33 = 7.44 × 10-11 Ksp = 4 2.65 × 10-4 = 7.44 × 10-11 Solubility from Ksp Q: Ksp of PbBr2= 6.6 × 10-6. Calculate the SOLUBILITY of the salt in mol/L & g/L. PbBr2(s) ⇄ Pb2+(aq) + 2Br-(aq) x→ x SOLUTION I C E 2x Step 1 Pb2+ 0 +x x 2Br0 +2x 2x Solubility from Ksp Q: Ksp of PbBr2= 6.6 × 10-6. Calculate the of the salt in mol/L & g/L. solubility Step 2 Substitute into Ksp expression & solving for x: Ksp = [Pb2+][Br-]2 = (x)(2x)2 = 4 x3 x=3 K sp 4 6.6 × 10 x=3 4 −6 mol = 0.012 L Solubility in g/L: 0.012 mol/L × 367 g/mol Solubility = 4.4 g/L . 4.4 Precipitating an Insoluble Salt Precipitating Hg2Cl2(s) ⇄ Hg22+(aq) + 2 Cl-(aq) (s) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2 If [Hg22+] = 0.010 M, what [Cl-] is required to is just begin the precipitation of Hg2Cl2? just That is, what is the maximum [Cl -] that that can be in solution with 0.010 M Hg22+ can without forming Hg2Cl2(s)? without Precipitating an Insoluble Salt Precipitating Hg2Cl2(s) ⇄ Hg22+(aq) + 2 Cl-(aq) (s) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2 Solution [Cl-] that can exist when [Hg22+] = 0.010 M, [Cl-]2 = Ksp / [Hg22+] [Cl− ] = K sp 0.010 = 1.1 x 10-8 M If this conc. of Cl- is just exceeded, Hg2Cl2 begins to precipitate! begins Precipitating an Insoluble Salt Precipitating Hg2Cl2(s) ⇄ Hg22+(aq) + 2 Cl-(aq) (s) Ksp = 1.1 x 10-18 (Now raise [Cl-]) If we continue adding Cl− until [Cl-] = 1.0 M, What is the value of [Hg22+] at this at point? point? Solution [Hg22+] = Ksp / [Cl-]2 [Cl sp = Ksp / (1.0)2 = 1.1 x 10-18 M The concentration of Hg22+ has been reduced (from 0.010) by 1016 ! The Effect of pH on Solubility • for insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide – low pH, the higher the solubility (H+ reacts with OH-) – higher pH = increased [OH−] = lowering solubility High [OH−] high pH left M(OH)n(s) ⇄ Mn+(aq) + nOH−(aq) High [H+] low pH right • for insoluble ionic compounds that contain anions of weak acids (HA), the lower the pH (higher [H3O+], lower [A−]), the higher the solubility M2(CO3)n(s) ⇄ 2 Mn+(aq) + nCO32−(aq) H3O+(aq) + CO32− (aq)⇄ HCO3− (aq) + H2O(l) high K65 Barium Sulfate Barium Ksp = 1.1 x 10-10 (a) BaSO4 is a common mineral, appearing as a white powder or colorless crystals. (b) BaSO4 is opaque to x-rays. Drinking a BaSO4 cocktail enables a physician to exam the intestines. The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure Calculate water and (b) in 0.010 M Ba(NO3)2. water Ksp for BaSO4 = 1.1 x 10-10 1.1 BaSO4(s) ⇄ Ba2+(aq) + SO42-(aq) (s) Solution Solubility in pure water = [Ba2+] = [SO42-] = x Ksp = [Ba2+] [SO42-] = x2 x = (Ksp)1/2 = 1.1 x 10-5 M Solubility in pure water = 1.1 x 10-5 mol/L The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. (b) Ksp for BaSO4 = 1.1 x 10-10 1.1 BaSO4(s) ⇄ Ba2+(aq) + SO42-(aq) (s) add some Ba2+ Solution Solution Solubility in pure water = 1.1 x 10-5 mol/L. Now add BaSO4 to water already containing 0.010 M Ba2+ (one common ion. The other is SO42−) 0.010 (one Which way will the “common ion” shift the equilibrium? Will solubility of BaSO4 be less than or greater than in pure water?___ The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. (b) Ksp for BaSO4 = 1.1 x 10-10 1.1 BaSO4(s) ⇄ Ba2+(aq) + SO42-(aq) (s) Solution Solution initial change equilib. equilib. [Ba2+] [Ba 0.010 +y 0.010 + y [SO42-] 0 +y y = the molar solubility the of BaSO4 of The Common Ion Effect The Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. (b) Ksp for BaSO4 = 1.1 x 10-10 1.1 BaSO4(s) ⇄ Ba2+(aq) + SO42-(aq) (s) Solution Solution Ksp = [Ba2+] [SO42-] = (0.010 + y) (y) Because y < 1.1 x 10-5 M (= x, the solubility in pure water), this means 0.010 + y is about equal to 0.010 0.010. Therefore, 0.010 Ksp = 1.1 x 10-10 = (0.010)(y) y = 1.1 x 10-8 M = solubility in presence of 0.010 M 1.1 added Ba2+ ion. y is indeed < 1.1 x 10-5 M added 1.1 The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. Ba(NO Ksp for BaSO4 = 1.1 x 10-10 1.1 BaSO4(s) ⇄ Ba2+(aq) + SO42-(aq) (s) SUMMARY SUMMARY Solubility BaSO4 in pure water = x = 1.1 x 10-5 M Solubility BaSO4 in presence of added (0.010 M) Ba2+ = 1.1 x 10-8 M Solubility = [SO42−] Ba Le Chatelier’s Principle is followed! (solubilty Le dec; or, more BaSO4 ppts. as equil shifts left) dec; Dissolving Precipitates by forming Dissolving Complex Ions Ions Insoluble ionic compounds can dissolve in the presence of a Lewis base. AgCl(s) ⇄ Ag+(aq) + Cl-(aq) Ksp = 1.8 × 10-10 Ag+(aq) + 2NH3(aq)⇄ [Ag(NH3)2]+(aq) Kf = 1.6 × 107 (aq) 2NH (aq) Complex Ion ---------------------------------------------------------------------formation formation + AgCl(s) + 2NH3(aq) ⇄ [Ag(NH3)2] (aq) + Cl (aq) equilibrium (aq) (aq) Cl constant constant Ch. 14: add two equations, Knet = product of the two Ch. net equilibrium constants for the two equations. equilibrium Knet = Ksp • Kf = (1.8× 10-10) (1.6× 107) = 2.9 × 10-3 2.9 this is a moderate value equilibrium constant this ...
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This note was uploaded on 11/07/2011 for the course CHM 2045 taught by Professor Geiger during the Fall '08 term at University of Central Florida.

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