Chapter 15 - Applications of Aqueous Equilibria

Neutralization: Strong Acid-Strong Base Chapter 15 - Applications of Aqueous Equilibria Molecular: HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)  SA-SB rxn goes to completion (one-way ) • Write ionic and net ionic rxns H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) H2O(l) + Na+(aq) + Cl-(aq) • Net: H+(aq) + OH-(aq)  H2O(l)  After neutralization, what’s in solution?  Do either of the salt ions react with water?  What is pH of the salt solution? Neutralization: Weak Acid-Strong Base Neutralization: Strong Acid- Weak Base HF (aq) + KOH (aq)  KF (aq) + H2O (l)  Rxn goes to completion due to SB Write ionic and net rxns: HCl (aq) + NH3 (aq)  NH4+ (aq) + Cl- (aq) HF(aq) + K+(aq) + OH-(aq)  K+(aq) + F-(aq) + H2O(l) H+(aq) + Br-(aq) + NH3(aq)  NH4+(aq) + Br-(aq) Net: HF(aq) + OH-(aq)  H2O(l) + F-(aq) Net Ionic: H+(aq) + NH3(aq)  NH4+(aq)  Rxn goes to completion because of strong acid Write ionic and net rxns:  After neutralization, what’s in solution?  After neutralization, what’s in solution?  Do either of the salt ions react with water?  Do either of the salt ions react with water?  What will the pH of the solution be, roughly?  What will the pH of the solution be, roughly? Neutralization: Weak Acid-Weak Base Neutralization Reactions  Reaction does not go to completion since no reactant is completely ionized. • Predict whether the pH after neutralization will be greater than, less than, or equal to 7 for the following combinations: Ionic: CH3COOH(aq) + NH3(aq)  NH4+(aq) + CH3COO-(aq) • HNO2 and KOH Net: same • HCl and LiOH The salt remaining after neutralization contains an acidic and a basic ion, so pH depends on their relative Ka and Kb values. • HBr and NH3 CH3COOH(aq) + NH3(aq)  NH4CH3COO(aq) 1 Common Ion effect The shift in equilibrium caused by the addition of a substance having an ion in common with the equilibrium mixture.  Adding a common ion suppresses the ionization of a weak acid or a weak base. Common Ion Concept Problem Given this reaction: CH3CO2H + H2O  H3O+ + CH3CO2- What happens to the pH of the acetic acid solution if we add NaCH3CO2? [CH3CO2-]  The source of the common ion is typically provided by adding a strong acid, a strong base or a soluble salt to the equilibrium reaction mixture. Eq shifts  [H3O+] , thus pH  Common Ion Effect CH3CO2H + H2O  CH3CO2- + H3O+ 0.100 M CH3CO2H , Ka = 1.810-5 Common Ion Problem • What is the pH of 1.00 M HF solution? Ka = 7.0 x 10- 4 • WA soln: pH = 2.87 Add common ion CH3CO2- , pH  • What is the pH of 1.00 M HF solution after adding 0.500 M NaF? Ka = 7.0  10-4 0.100 M CH3CO2H, add 0.050 M NaCH3CO2 • Mix of WA and conj base: pH = 4.44 16.3 Buffer Solution A solution that resists changes in pH when a small amount of acid or base is added. •Buffers are used to control pH e.g. biological buffers maintain the pH of all body fluids Best buffer systems consist of either a) a weak acid and its conjugate base e.g. HC2H3O2 and NaC2H3O2 b) a weak base and its conjugate acid e.g. NH3 and NH4Cl Which are buffer solutions? • Identify the solutions below that would make good buffer solutions (what criteria need to be met?):        HF and NaF NH3 and NH4Cl KOH and KF CH3COOH and LiCH3COO NaNO3 and HNO3 NaOH and NaCl HCl and NaCH3COO 2 Buffer Capacity Buffer Range Buffers only work within a pH range set by the value of Ka: pH = pKa ± 1 Thus 0.1 < [HA]/[A-] < 10 • Outside this range, we see little buffering effect. • If you know the desired pH of a buffer solution, you can choose an acid with a pKa near the pH. • Also the [acid]:[base] ratio must be less than 10:1 (or 1:10)  Optimum buffer has same amount of HA & A-. • Select an appropriate acid-base pair to get within the range of the desired pH. • Look at table of Ka values (some are shown on the next screen), calculate pKa, and select an appropriate combination. • How would you make a buffer solution with pH = 4.10? What could you use to make this? Table of Ka values Buffer Solutions • Acid Ka HF HNO2 C9H8O4 (aspirin) HCO2H (formic) C6H8O6 (ascorbic) C6H5CO2H (benzoic) CH3CO2H (acetic) HCN C6H5OH (phenol) 3.5 x 10 –4 4.5 x 10 –4 3.0 x 10 –4 1.7 x 10 –4 8.0 x 10 –5 6.5 x 10 –5 1.8 x 10 –5 4.9 x 10 –10 1.3 x 10 –10 Buffer Solutions • Figure 15.3 • HF(aq) + H2O(l)  H3O+(aq) + F-(aq) • Add HCl, which component of the buffer soln will HCl reaction with (acid or base)?  H+ will react with F- (conj. base) • Add NaOH, which component of the buffer soln will NaOH react with (acid or base)?  OH- will react with HF (weak acid) • There will be a pH change in each case, but not as much as if the HCl (or NaOH) were added to water. Henderson-Hasselbalch Equation • When solving for [H3O+] at equilibrium and assuming x << [HA] & [A-] for good buffer action, the equilibrium expression Ka = [H3O+][A-] / [HA] can be rearranged to give a simplified calculation: • pH = pKa + log ( [A-] / [HA])  Can use moles or M – depending on given info  V units cancel since it’s a ratio of [base]/[acid] Buffers shockwave animation 3 Buffer Solutions + HCl • What is the pH of 500.0 mL of 0.10 M formic acid combined with 500.0 mL of 0.20 M sodium formate? Ka = 1.8 x 10-4 • pH = pKa + log ([A-]/[HA]) = 4.04 • Now add 10.0 mL of 0.50 M HCl  calculate moles of substances, then calculate changes  Use a CHANGE Table (initial, change, final) Buffer +NaOH • What will happen when we add a strong base to a buffer system? Buffer Solutions + HCl • To find the new pH, first assume that all the added strong acid reacts with A- to form HA.  H3O+ + A-  H2O + HA  How many moles of HA, A-, and H3O+ are in solution? • Initial mol HA = (0.10 M) (0.5000 L) = 0.050 mol • Initial mol A- = (0.20 M) (0.5000 L) = 0.10 mol • mol HCl = (0.0100 L) (0.50 M) = 0.0050 mol • Final mol A- = 0.10 mol – 0.0050 mol = 0.095 mol A• Final mol HA = 0.050 mol + 0.0050 mol = 0.055 mol HA  Can use final moles in H-H equation. Why? • pH = pKa + log ([A-]/[HA]) = 3.98 (initially 4.04 w/o acid) • pH  slightly when SA added to buffer Buffer Solutions + NaOH • First assume that all OH- reacts with HCO2H: • HCO2H + OH-  H2O + HCO2-  Added base reacts with HA to form A-:  HA and A- • How many moles of base were added?  HA + • final mol HA = 0.050 mol - 0.00500 mol = 0.045 mol OH-  H2O + A- • mol OH- = (0.500 M) (0.0100 L) = 0.00500 mol OH- • Add 10.0 mL of 0.500 M NaOH to the formic acid/sodium formate buffer. What is the new pH? • final mol A- = 0.10 mol + 0.00500 mol = 0.105 mol Practice Ex. Try after class pH Titration Curves 1) Calculate the pH of a 1.25 L buffer solution made up of 0.50 M acetic acid and 0.50 M sodium acetate.Ka = 1.8 x 10-5 • Strong Acid-Strong Base Titrations: 2) What is the pH of the solution when 25.00 mL of 0.40 M NaOH is added to the buffer? • pH = pKa + log ([A-]/[HA]) = 4.11 (initially 4.04 w/o SB) • pH slightly when SB added • In water, pH would change from 7 to 13 if same amount of SB was added. • The equivalence point is the point at which equimolar amounts of acid and base have reacted (review 1st semester notes on calcs). • Graph generated with pH probe  • Figure 15.6 4 pH Titration Curves • 4 key points on a titration curve  At the beginning (before titrant is added)  Before the equivalence point  At the equivalence point  After the equivalence point • Consider titrating 20.00 mL of 0.200 M HCl with 0.100 M NaOH. Find equiv. pt. first! M= mol 1000 mmol mmol = = L 1000 mL mL Strong Acid-Strong Base • Draw 4 beakers. • 1) Draw 2 moles HCl. • 2) Draw what happens when 1 mole NaOH is added. • 3) Draw what happens when 2 moles NaOH are added. • 4) Draw what happens when 3 moles NaOH are added. HCl + NaOH Titration • 1) 0 mL NaOH, pH is calculated from strong acid. [H3O+] = 0.200 M  pH = -log (0.200) = 0.700 initial pH of SA • 2) 5.00 mL NaOH, acid is still in excess; but by how much?      mmol acid – mmol base (0.200 M)(20.00 mL) – (0.100 M)(5.00 mL) [H3O+] = mmol / total solution volume [H3O+] = 3.50 mmol / 25.00 mL = 0.140 M pH = -log (0.140) = 0.854 excess SA so pH still quite low! Weak Acid-Strong Base • Acetic acid in flask, KOH is titrant in buret • Find pH at 4 pts. • Figure 15.8 shows strong and weak acids. • Titrate 50.00 mL of 0.100 M acetic w/ 0.150 M KOH.  Find eq. pt., Ka, pKa, and mmols of acid first HCl + NaOH Titration • 3) 40.00 mL NaOH added; at the equivalence point, neutral salt and water in solution.  mmol acid = mmol base = 4.00 mmol  pH = 7 (always 7 for strong acid-strong base!) • 4) 50.00 mL NaOH added, base in excess. Calc moles (or mmol) of excess base remaining     mmol base – mmol acid (0.100 M)(50.00 mL) – (0.200 M)(20.00 mL) 1.00 mmol base excess / 70.00 mL = 0.01429 M NaOH pOH = -log (0.01429) = 1.845, pH = 12.155 excess SB • Problems 15.13, 15.14 Weak Acid-Strong Base • Draw 4 beakers. • 1) Draw 2 moles acetic acid. • 2) Draw what happens when 1 mole KOH is added. • 3) Draw what happens when 2 moles KOH are added. • 4) Draw what happens when 3 moles KOH are added. 5 Weak Acid-Strong Base • • • • • 4 regions in titration curve Before adding base: 0.00 mL base Before equivalence point: 10.00 mL base At equivalence point: ???? mL base After equivalence point: 50.00 mL base *See titration worksheet key for complete solution to this problem! Weak Acid-Strong Base • Initial pH before adding base. This is a weak acid. How do we find the pH of a weak acid? Ka = Initial Change Equilibrium HA H2O  H3O+ A- 0.100 M -x 0.100 - x - 0 +x x 0 +x x x  [H3O  ]  [HA]  K a  0.100  1.8  105  1.34  103 M 0.100 M acetic acid, pH = 2.87 Weak Acid-Strong Base • Before equiv. pt.: this is the buffer zone (both weak acid and its conjugate base are present).  Use change table to find mmoles excess acid and mmoles cong. base produced. All strong base is converted to conjugate acid (strong base is completely consumed).  [HA] = (mmoles acid – mmoles base)/total volume  [A-] = mmoles base / total volume  pH = pKa + log ([A-] / [HA]) Weak Acid-Strong Base • At equivalence point:  5.00 mmol HA and 5.00 mmol SB have reacted  5.00 mmol conj. base (A-) have been formed  Total V = 50.00 mL + 33.33 mL = 83.33 mL  [A-] = 5.00 mmol/83.33 mL = 0.0600 M A WB hydrolysis: A- + H2O  HA + OH Kb = Kw / K a  Set up ICE table!  Kb = (x2/0.0600-x)  x = [OH-]; then find pH  pH = 8.76 (see worksheet key!) Weak Acid-Strong Base • At equivalence point: Find pH of basic salt  We have seen that the product of a weak acid and a strong base is a basic salt and water.  We can expect the pH to be greater than 7.  Since all the weak acid has been consumed and converted to its conjugate base at the equivalence point, the mmoles of A- are now known.  mmoles A- / total mL = [A-]  Where total mL = mL acid + mL base  Conjugate base reacts with water (A- + H2O  HA + OH-).  Set up ICE table (using Kb) and solve for [OH-] and pH. Weak Acid-Strong Base • After the equivalence point: Excess OH- ions  More than enough base is added to react with all the acid in the flask. No HA remains, it is all converted to A-.  pH is determined primarily from excess [OH-].  [A-] is a minor contributor and doesn’t need to be calculated.  [OH-] = (mmoles base – mmoles acid)/ total volume 6 Weak Acid-Strong Base • After the equivalence point:  50.00 mL of 0.100 M acid (5.00 mmol) and 50.00 mL of 0.150 M base (7.50 mmol)  7.50 mmol OH- added – 5.00 mmol acid present = 2.50 mmol OH- remaining  total mL = 50.00 + 50.00 = 100.00 mL  [OH-]= 2.5 mmol/100.00 mL = 0.0250 M OH pOH = -log (0.0250 M) = 1.602  pH = 14 – 1.602 = 12.398 Half Equivalence Point • 33.34 mL base added (half the volume needed to get to eq pt for quiz #14 problem) Group Quiz #14 • Exactly 50.00 mL of 0.20 M hydrazoic acid (Ka = 1.9 x 10-5) are titrated with a 0.15 M KOH solution. Calculate the pH when 66.67 mL of base have been added Weak Base-Strong Acid • Figure 15.10 • At this point half the acid has been neutralized mmol substances at half eq pt: mmol HA = (10.00 mmol – 5.00 mmol) = 5.00 mmol • mmol A- = 5.00 mmol • Thus, [HA] = [A-] • pH = pKa + log ([A-]/[HA]) = pKa Acid-Base Indicators • Calculations are similar to weak acid/strong base; opposite values; extra steps to find pH values. See pages 613 - 615 for outline. Polyprotic Acid Titrations • One eq. pt. for each proton • 2 eq. pts. and 2 half eq. pts. • Figure 15.11 Alanine with NaOH 7 Solubility equilibria Dissolution of slightly soluble salts in water important biological examples:  tooth decay - tooth enamel dissolves in acidic soln  formation of kidney stones - salts precipitate in kidney Example. The dissolution of AgBr AgBr(s)  Ag+(aq) + Br -(aq) Ksp = [Ag+][Br -] Ksp is the solubility product constant - the equilibrium constant for insoluble salts • It is a measure of how soluble a salt is in H2O For salts with the same # of ions, the smaller the Ksp, the less soluble the salt. 15.10 Solubility Equilibria Many ions combine to form solid precipitates in aqueous solutions. (Solubility rules) These “insoluble” salts dissolve to a small extent and form a saturated solution. The undissolved solid and the dissociated ions in solution establish an equilibrium system that is characterized by the solubility product constant, Ksp. Write the solubility equilibrium reactions and Ksp expressions for a) MgF2 b) Ca3(PO4)2. A) MgF2(s)  Mg2+(aq) + 2F-(aq) Ksp = [Mg2+][F-]2 B) Ca3(PO4)2 (s)  3Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO43-]2 Table 15.2 shows some Ksp values Solubility Solubility/Ksp Practice • Ksp can be used to calculate the solubility, which can be compared for different salts. • Solubility (molar solubility) = molar concentration of dissolved salt; ion concentrations are related to this by their coefficients. • Solubility can also be expressed in g/L • Examples: • AgCl: [Ag+] = x [Cl-] = x Ksp = (x)(x) • Ag2S: [Ag+] = 2x [S2-] = x Ksp = (2x)2(x) 3 + • Fe(OH)3: [Fe ] = x [OH ] = 3x Ksp = (x)(3x)3 • Ksp for AgBr is 7.7 x 10-13. Calculate the molar and gram solubility (solve for x). • If a saturated solution prepared by dissolving CaF2 in water has [Ca2+] = 3.310-4 M, what is the value of Ksp? • Ksp for Cu2O is 2.0 x 10-15. Calculate the molar solubility. • The solubility of Ca(OH)2 is 0.233 g/L. Calculate Ksp. • Worked example 15.8 – 15.10, Problems 15.21 – 15.24 8 Solubility/Ksp Practice • Solubility = x in ICE table! • Ksp = eq constant for dissolution of salt! • • • • AgBr: x = 8.8 x 10-7 M; x = 1.6 x 10-4 g/L CaF2: Ksp = 1.4 x 10-10 Cu2O: x = 7.9 x 10-6 M Ca(OH)2: Ksp = 1.24 x 10-7 Common Ion Effect • Solubility is decreased when a common ion is added AgCl(s)  Ag+(aq) + Cl-(aq) • Common ion effect: add more Cl- to precipitate Ag+ from solution. • Saturated solution of AgCl: Find (x)  Ksp = 1.70 x 10-10 • Add 0.100 M Cl 0.100 >> 1.30 x 10-5, so [Cl-]= 0.100 M • What is the new [Ag+]? Note this is x in ICE! Common Ion Effect What will happen if I add solid NaF to a solution of saturated SrF2? SrF2(s)  Sr2+(aq) + 2F-(aq) a) Calculate the molar solubility of SrF2 in pure water (Ksp = 4.3 x 10-9). b)Calculate the molar solubility of SrF2 in 0.010 M NaF. Common Ion Effect • • • • SrF2(s)  Sr2+(aq) + 2F-(aq) (in water) Equil: x 2x Ksp = [Sr2+][F-]2 = 4.3 x 10-9 x = 1.0 x 10-3 M • • • • SrF2(s)  Sr2+(aq) + 2F-(aq) (in NaF) Initial: 0 0.010 Equil: x 0.010 + 2x Ksp = [Sr2+][F-]2 = 4.3 x 10-9 • x = 4.3 x 10-5 M • Worked ex 15.11, Problem 15.25 Group Quiz #15 • Calculate the molar solubility of Cu(OH)2. Ksp = 1.6 x 10-19 • Calculate the molar solubility of Cu(OH)2 when 0.10 M Cu(NO3)2 is added. Effect of pH on Solubility Addition of an acid can increase the solubility of an insoluble basic salt. E.g. CaF2(s)  Ca2+(aq) + 2F-(aq) Add strong acid (e.g. HCl) provides H3O+ ions: H3O+ + F-  HF + H2O  Adding H+ causes [F-] , equilibrium shifts right to form more F- & solubility of CaF2. 9 pH and solubility • E.g. For the following salts, predict whether the salt will dissolve in an acidic solution. • A. AgBr _____________ • B. CdCO3 _____________ • C. PbCl2 _____________ • D. BaS _____________ Precipitation of Ionic Cmpds • Worked Example 15.14: Will a precipitate form when 0.150 L of 0.10 M lead (II) nitrate and 0.100 L of 0.20 M sodium chloride are mixed? • What is the precipitate that will form? What are Ksp and the equilibrium expression? • Find total concentration of each ion once mixed. • Problem 15.29 Precipitation of Ionic Cmpds • • • • Q > Ksp Supersaturated; ppt forms Q = Ksp Saturated Q < Ksp Unsaturated; no ppt forms 2.45 mg of magnesium carbonate is placed in 1.00 L of water. Will it all dissolve? Ksp = Precipitation of Ionic Cmpds • Note: book refers to IP, same idea as Q (value of concentration ratios NOT necessarily at equilibrium) • Compare Q and K. System will shift until Qsp = Ksp (a saturated solution) • Example: We want to know if precipitate will form when 0.150 L of 0.10 M lead (II) nitrate and 0.100 L of 0.20 M sodium chloride are mixed? • Find Qsp and relate it to Ksp. Precipitation • PbCl2(s)  Pb2+(aq) + 2Cl-(aq) • Calculate moles of each ion. • Convert moles to concentration using total volume. • Plug concentrations into Qsp equation. • If Qsp > Ksp, rxn shifts left (toward solid) and precipitate will form. • If Qsp < Ksp, rxn shifts right (toward ions) and precipitate will not form. Group quiz 16 Will a solution containing 4.0 x 10-5 M Cland 2.0 x 10-4 M Ag+ form a precipitate of AgCl? Ksp = 1.70 x 10-10 (Hint: think Q) 4.0 x 10-5 • Find molarity of solid. Write equilibrium expression for solubility. Calculate Qsp. 10 Complex Ion Equilibria Formation of complex ions can increase solubility of an insoluble salt Complex ion: ion containing a metal cation bonded to one or more molecules or ions which are called ligands. The ligands act as Lewis bases (e.g. NH3, H2O, OH-) • E.g. Al(H2O)63+, CuBr4- Complex Ion Equilibria • Lewis acid-base reactions also reach a state of equilibrium • Metal ion + ligand  complex ion • Kf = [complex ion]/[metal ion][ligand] • Kf = formation constant • Hg2+ + 4I-  HgI42• Kf = [HgI42-]/[Hg2+][I-]4 • Kf values are usually pretty large Adding NH3 to AgCl solution increases the solubility of AgCl: 1) AgCl(s)  Ag+(aq) + Cl-(aq) K1 = 1.7 x10-10 + + 2) Ag (aq) + 2NH3(aq)  Ag(NH3)2 (aq) K2 = 1.6 x 107 3) AgCl(s) + 2NH3(aq)  Ag(NH3)2+(aq) + Cl-(aq) Knet = K1 x K2 = 2.7 x 10-3 •In qual lab we add ammonia to dissolve our AgCl precipitate so it can be separated from Hg. Qualitative Analysis • Skipping section 15.14: Selective Precipitation • 15.15 Qualitative analysis is used to identify unknown ions in a solution.  152LL should read this section before qual lab! • Each ion can be precipitated out by addition of selective reagents. • Purely qualitative research, like solving a puzzle. 11