Chapter 15 - Applications of Aqueous Equilibria

Neutralization: Strong Acid-Strong Base Chapter 15 - Applications of Aqueous Equilibria Molecular: HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) SA-SB rxn goes to completion (one-way ) Write ionic and net ionic rxns H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) H 2 O(l) + Na + (aq) + Cl - (aq) GCC CHM152 Net: H + (aq) + OH - (aq) H 2 O(l) After neutralization, what s in solution? Do either of the salt ions react with water? What is ph of the salt solution? Neutralization: Neutralization: Strong Acid- Weak Base HF (aq) + KOH (aq) KF (aq) + H 2 O (l) Rxn goes to completion due to SB Write ionic and net rxns: HF(aq) + K + (aq) + OH - (aq) K + (aq) + F - (aq) + H 2 O(l) Net: HF(aq) + OH - (aq) H 2 O(l) + F - (aq) After neutralization, what s in solution? Do either of the salt ions react with water? What will the ph of the solution be, roughly? HCl (aq) + NH 3 (aq) NH + 4 (aq) + Cl - (aq) Rxn goes to completion because of strong acid Write ionic and net rxns: H + (aq) + Br - (aq) + NH 3 (aq) NH 4+ (aq) + Br - (aq) Net Ionic: H + (aq) + NH 3 (aq) NH 4+ (aq) After neutralization, what s in solution? Do either of the salt ions react with water? What will the ph of the solution be, roughly? Neutralization: Weak Acid-Weak Base Neutralization Reactions CH 3 COOH(aq) + NH 3 (aq) NH 4 CH 3 COO(aq) Reaction does go to completion since no reactant is completely ionized. Ionic: CH 3 COOH(aq) + NH 3 (aq) NH 4 + (aq) + CH 3 COO - (aq) Net: same The salt remaining after neutralization contains an acidic and a basic ion, so ph depends on their relative K a and K b values. Predict whether the ph after neutralization will be greater than, less than, or equal to 7 for the following combinations: HNO 2 and KOH HCl and LiOH HBr and NH 3 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. 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. Common Ion Concept Problem Given this reaction: CH 3 CO 2 H + H 2 O H 3 O + + CH 3 CO 2 - What happens to the ph of the acetic acid solution if we add NaCH 3 CO 2? [CH 3 CO 2- ] Eq shifts [H 3 O + ], thus ph Common Ion Effect Common Ion Problem CH 3 CO 2 H + H 2 O CH 3 CO - 2 + H 3 O + 0.100 M CH 3 CO 2 H WA soln: ph = 2.879 Add common ion CH 3 CO - 2, ph What is the ph of 1.00 M HF solution? K a = 7.0 x 10-4 What is the ph of 1.00 M HF solution after adding 0.500 M NaF? K a = 7.0 x 10-4 0.100 M CH 3 CO 2 H, add 0.050 M NaCH 3 CO 2 Mix of WA and conj base: ph = 4.456 16.3 Buffer Solution Which are buffer solutions? 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. HC 2 H 3 O 2 and NaC 2 H 3 O 2 b) a weak base and its conjugate acid e.g. NH 3 and NH 4 Cl Identify the solutions below that would make good buffer solutions (what criteria need to be met?): HF and NaF NH 3 and NH 4 Cl KOH and KF CH 3 COOH and LiCH 3 COO NaNO 3 and HNO 3 NaOH and NaCl HCl and NaCH 3 COO 2

Buffer Capacity Buffer Range Buffers only work within a ph range set by the value of K a : ph = pk a ± 1 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 pk a near the ph. Select an appropriate acid-base pair to get within the range of the desired ph. Look at table of K a values (some are shown on the next screen), calculate pk a, 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 K a values Buffer Solutions Acid K a HF HNO 2 C 9 H 8 O 4 (aspirin) HCO 2 H (formic) C 6 H 8 O 6 (ascorbic) C 6 H 5 CO 2 H (benzoic) CH 3 CO 2 H (acetic) HCN C 6 H 5 OH (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 HF(aq) + H 2 O(l) H 3 O + (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. Buffer Solutions Henderson-Hasselbalch Equation Figure 15.3 When solving for [H 3 O + ] at equilibrium and assuming x << [HA] & [A - ] for good buffer action, the equilibrium expression K a = [H 3 O + ][A - ] / [HA] can be rearranged to give a simplified calculation: ph = pk a + log ([A - ] / [HA]) Can use moles or M depending on given info 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? K a = 1.8 x 10-4 ph = pk a + log ([A - ]/[HA]) = 4.05 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 Solutions + HCl To find the new ph, first assume that all the added strong acid reacts with A - to form HA. H 3 O + + A - H 2 O + HA How many moles of HA, A -, and H 3 O + 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 Final mol HA = 0.050 mol + 0.0050 mol = 0.055 mol Can use final moles in H-H equation. Why? ph = pk a + log ([A - ]/[HA]) = 3.98 Buffer +NaOH What will happen when we add a strong base to a buffer system? Added base reacts with HA to form A - : HA and A - HA + OH - H 2 O + A - Add 10.0 ml of 0.500 M NaOH to the formic acid/sodium formate buffer. What is the new ph? Buffer Solutions + NaOH First assume that all OH - reacts with HF: HF + OH - H 2 O + F - How many moles of base were added? mol OH - = (0.500 M) (0.0100 L) = 0.00500 mol OH - final mol HA = 0.050 mol - 0.00500 mol = 0.045 mol final mol A - = 0.10 mol + 0.00500 mol = 0.105 mol ph = pk a + log ([A - ]/[HA]) = 4.11 (started at 4.05) In water, ph would change from 7 to 13. Group Quiz #13 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. K a = 1.8 x 10-5 2) What is the ph of the solution when 25.00 ml of 0.40 M NaOH is added to the buffer? Strong Acid-Strong Base Titrations: The equivalence point is the point at which equimolar amounts of acid and base have reacted (review 1 st 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 L 1000 mmol = 1000 ml = mmol 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 HCl + NaOH Titration 1) 0 ml NaOH, ph is calculated from strong acid. [H 3 O + ] = 0.200 M ph = -log (0.200) = 0.70 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) [H 3 O + ] = mmol / total solution volume [H 3 O + ] = 3.50 mmol / 25.00 ml = 0.140 M ph = -log (0.140) = 0.85 3) 40.00 ml NaOH added; at the equivalence point, neutral salt and water in solution. mmol acid = mmol base ph = 7 (always 7 for strong acid-strong base!) 4) 50.00 ml NaOH added, base in excess. By how much? mmol base mmol acid (0.100 M)(50.00 ml) (0.200 M)(20.00 ml) 1.00 mmol base excess / 45.00 ml = 0.01429 M NaOH poh = -log (0.01429) = 1.85, ph = 12.15 Problems 15.13, 15.14 Ascorbic 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 ascorbic acid w/ 0.150 M KOH. Find eq. pt., K a, pk a, and mmols of acid first Draw 4 beakers. 1) Draw 2 moles weak 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

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 Initial ph before adding base. This is a weak acid. How do we find the ph of a weak acid? K a = HA H 2 O H 3 O + A - Initial 0.100 M - 0 0 Change -x - +x +x Equilibrium 0.100 - x - x x 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 = pk a + log ([A - ] / [HA]) Before equiv. pt.: 10.00 ml base Excess acid: 5.00 mmol acid 1.50 mmol base = 3.50 mmol excess acid All strong base is converted to conjugate base: (0.150 M)(10.00 ml) = 1.50 mmol A - divide by total volume = 60.00 ml to get concentration Or leave values in moles for Henderson-Hasselbach ph = pk a + log (base/acid) Key Concept Problem 15.15; Problems 15.16, 15.17 At equivalence point: Find ph of salt (Ch. 14) 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 volume = [A - ] Conjugate base reacts with water (A - + H 2 O HA + OH - ). Set up ICE table (using K b ) and solve for [OH - ] and ph. At equivalence point: 5.00 mmol acid and 5.00 mmol base So 5.00 mmol conj. base (A - ) in flask with 83.33 ml total volume = 0.0600 M A - ICE table: A - + H 2 O HA + OH - K b = K w / K a K b = (x 2 /0.0600-x) x = [OH - ]; then find ph 6

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 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 remaining OH - divide by total volume = 0.0250 M OH - poh = -log (0.09091 M) = 1.60 ph = 14 1.60 = 12.39 Group Quiz #14 Half Equivalence Point Exactly 50.00 ml of 0.20 M hydrazoic acid (K a = 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: 33.34 ml base added (half the volume needed to get to equivalence point): [HA] = (10.00 mmol 5.00 mmol) [A - ] = 5.00 mmol [HA] = [A - ] ph = pk a + log ([A - ]/[HA]) = pk a Animations: http://www.chembio.uoguelph.ca/educmat/chm19104/chemtoons/chemtoons.htm Weak Base-Strong Acid Acid-Base Indicators Figure 15.10 Calculations are similar to weak acid/strong base; opposite values; extra steps to find ph values. See pages 613-615 for outline. 7

Polyprotic Acid Titrations Diprotic Acid Titration One eq. pt. for each proton 2 eq. pts. and 2 half eq. pts. Figure 15.11 - Alanine with NaOH A: acid in water B: pk a1 ; 1 st H removed; 50% H 2 A, 50% HA - C: 1 st equiv. pt. (HA - ) D: pk a2 ; 2 nd H removed; 50% HA -, 50% A 2- E: 2 nd equiv. pt. (A 2- ), all acid reacted F: only excess base Complex Ion Equilibria 16.4 Complex Ion Equilibria Formation of complex ions can increase solubility of an insoluble salt Complex ion: ion containing a central metal cation bonded to one or more molecules or ions which are called ligands. The ligands act as Lewis bases (e.g. NH 3, H 2 O, OH - ) Lewis acid-base reactions also reach a state of equilibrium Metal ion + ligand complex ion K f = [complex ion]/[metal ion][ligand] (formation constant) Hg 2+ + 4I - HgI 4 2- K f = [HgI 4 2- ]/[Hg 2+ ][I - ] 4 Complex Ion Equilibria Complex Ion Equilibria Complex ions are often formed stepwise: Hg 2+ + I - HgI + K f1 = [HgI + ]/[Hg 2+ ][I - ] = 7.9 x 10 12 HgI + + I - HgI 2 K f2 = [HgI 2 ]/[HgI + ][I - ] = 1.0 x 10 11 HgI 2 + I - HgI - 3 K f3 = [HgI 3 - ]/[HgI 2 ][I - ] = 5.0 x 10 3 HgI 3 - + I - HgI 4 2- K f4 = [HgI 4 2- ]/[HgI 3 - ][I - ] = 2.5 x 10 2 Add all four equations together: Hg 2+ + 4I - HgI 4 2- K f = [HgI 4 2- ]/[Hg 2+ ][I - ] 4 K f = K f1 x K f2 x K f3 x K f4 = 1.0x10 30 Values of K f are listed in Table 15.4; they have a wide range of values and thus of stabilities. Unlike Bronsted-Lowrey polyprotic acids, successive K f values may not differ by large factors, so there may be many species coexisting. See Figure 15.11. Demo: orange tornado 8

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 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, K sp. Example. The dissolution of AgBr AgBr(s) Ag + (aq) + Br - (aq) K sp = [Ag + ][Br - ] K sp is the solubility product constant - the equilibrium constant for insoluble salts It is a measure of how soluble a salt is in H 2 O For salts with the same # of ions, the smaller the K sp, the less soluble the salt. Table 15.2 shows some K sp values Write the solubility equilibrium reactions and K sp expressions for a) MgF 2 b) Ca 3 (PO 4 ) 2. A) MgF 2 (s) Mg 2+ (aq) + 2F - (aq) K sp = [Mg 2+ ][F - ] 2 B) Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3- (aq) K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 Solubility/K sp Practice Solubility/K sp Practice K sp for silver bromide is 7.7 x 10-13. Calculate the molar and gram solubility (solve for x). The solubility of calcium sulfate is found experimentally to be 0.67 g/l. Calculate the value of K sp for calcium sulfate. K sp for copper (I) oxide is found to be 2.0 x 10-15. Calculate the molar and gram solubility. The solubility of calcium hydroxide is 0.233 g/l. Calculate K sp. Worked example 15.8 15.10, Problems 15.21 15.24 AgBr: x = 8.8 x 10-7 M; x = 1.6 x 10-4 g/l CaSO 4 : K sp = 2.4 x 10-5 Cu 2 O: x = 7.9 x 10-6 M; x = 1.1 x 10-3 g/l Ca(OH) 2 : K sp = 1.24 x 10-7 9

Solubility Common Ion Effect K sp can be used to calculate the solubility, which can be compared for any salts. Solubility (S) = molar concentration of dissolved salt; ion concentrations are related to this by their coefficients. Examples: AgCl: [Ag + ] = S [Cl - ] = S K sp = (S)(S) Ag 2 S: [Ag + ] = 2S [S 2- ] = S K sp = (2S) 2 (S) Fe(OH) 3 : [Fe 3+ ] = S [OH - ] = 3S K sp = (S)(3S) 3 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) K sp = 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 Common Ion Effect What will happen if I add solid NaF to a solution of saturated SrF 2? SrF 2 (s) Sr 2+ (aq) + 2F - (aq) Problem 15.101: a) Calculate the molar solubility of strontium fluoride in pure water (K sp = 4.3 x 10-9 ). Calculate the molar solubility of strontium fluoride in 0.010 M sodium fluoride. SrF 2 (s) Sr 2+ (aq) + 2F - (aq) (in water) Equil: x 2x K sp = [Sr 2+ ][F - ] 2 = 4.3 x 10-9 x = 1.0 x 10-3 M SrF 2 (s) Sr 2+ (aq) + 2F - (aq) (in NaF) Initial: 0 0.010 Equil: x 0.010 + 2x K sp = [Sr 2+ ][F - ] 2 = 4.3 x 10-9 x = 4.3 x 10-5 M Worked ex 15.11, Problem 15.25 Effect of ph on Solubility Group Quiz #15 Addition of an acid can increase the solubility of an insoluble basic salt. E.g. CaF 2 (s) Ca 2+ (aq) + 2F - (aq) Add strong acid (e.g. HCl) provides H 3 O + ions: H 3 O + + F - HF + H 2 O Adding H + causes [F - ], equilibrium shifts right to form more F - & solubility of CaF 2. Calculate the solubility, both molar and gram, of copper(ii) hydroxide. K sp = 1.6 x 10-19 Calculate the molar solubility of copper(ii) hydroxide when 0.10 M copper(ii) nitrate is added. 10

ph and solubility E.g. For the following salts, predict whether the salt will dissolve in an acidic solution. A. AgBr B. CdCO 3 C. PbCl 2 D. BaS Adding NH 3 to AgCl solution increases the solubility of AgCl: 1) AgCl(s) Ag + (aq) + Cl - (aq) K 1 = 1.7 x10-10 2) Ag + (aq) + 2NH 3 (aq) Ag(NH 3 ) 2 + (aq) K 2 = 1.6 x 10 7 3) AgCl(s) + 2NH 3 (aq) Ag(NH 3 ) 2 + (aq) + Cl - (aq) K net = K 1 x K 2 = 2.7 x 10-3 Mixture of AgCl and AgBr can be separated: 5 M NH 3 dissolves AgCl, but not AgBr. Precipitation of Ionic Cmpds 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 Q sp = K sp (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 Q sp and relate it to K sp. 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 K sp and the equilibrium expression? Find total concentration of each ion once mixed. Problem 15.29 Precipitation Precipitation of Ionic Cmpds PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) Calculate moles of each ion. Convert moles to concentration using total volume. Plug concentrations into Q sp equation. If Q sp > K sp, rxn shifts left (toward solid) and precipitate will form. If Q sp < K sp, rxn shifts right (toward ions) and precipitate will not form. Q > K sp Supersaturated; ppt forms Q = K sp Saturated Q < K sp Unsaturated; no ppt forms 2.45 mg of magnesium carbonate is placed in 1.00 L of water. Will it all dissolve? K sp = 4.0 x 10-5 Find molarity of solid. Write equilibrium expression for solubility. Calculate Q sp. 11

Precipitation of Ionic Cmpds Group quiz 16 What are the silver and bromide ion concentrations if 0.0244 g of AgBr and 0.0111 g of NaBr are put into 1.00 L of water? Will AgBr dissolve completely? Write solubility equilibrium expressions for AgBr and NaBr. Calculate Q sp. Q sp = [Ag + ][Br - ] [Ag + ] = 1.30 x 10-4 M, [Br - ] = 1.30 x 10-4 + 1.08 x 10-4 M K sp = 7.7 x 10-13 Q sp > K sp, what does this mean? Not all of the AgBr can dissolve Worked example 15.14, Problem 15.29 Will a solution containing 4.0 x 10-5 M Cl - and 2.0 x 10-4 M Ag + form a precipitate of AgCl? K sp = 1.70 x 10-10 (Hint: think Q) 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. 12