Ch. 14/15: Acid-Base Equilibria Sections 14.6, 14.7, 15.1, 15.2

Ch. 14/15: Acid-Base Equilibria Sections 14.6, 14.7, 15.1, 15.2 Creative Commons License Images and tables in this file have been used from the following sources: OpenStax: Creative Commons Attribution License 4.0. ChemWiki (CC BY-NC-SA 3.0): Unless otherwise noted, the StatWiki is licensed under a Creative Commons Attribution- Noncommercial-Share Alike 3.0 United States License. Permissions beyond the scope of this license may be available at copyright@ucdavis.edu. The mineral fluorite (CaF 2 ) is deposited through a precipitation process. Note that pure fluorite is colorless, and that the color in this sample is due to the presence of other metals in the crystal. Chemistry: OpenStax 1 Principles of General Chemistry (CC BY-NC-SA 3.0): http://2012books.lardbucket.org/pdfs/principles-of-generalchemistry-v1.0.pdf 2 Neutralization Reactions Buffers and the Common Ion Effect Write the molecular, ionic, and net ionic equations for the following reactions: 1) HCl(aq) + NaOH(aq) H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) Na + (aq) + Cl - (aq) + H 2 O(l) H + (aq) + OH - (aq) H 2 O(l) or H 3 O + (aq) + OH - (aq) 2H 2 O(l) 2) HF(aq) + NaOH(aq) What happens to HF in solution? HF(aq) + Na + (aq) + OH - (aq) Na + (aq) + F - (aq) + H 2 O(l) HF(aq) + OH - (aq) F - (aq) + H 2 O(l) What will happen when we add sodium acetate to a solution of acetic acid? A system at equilibrium will shift in response to being stressed. The addition of a reactant or a product can be an applied stress. CH 3 COOH(aq) + H 2 O(l) NaCH 3 COO(aq) H 2O Equilibrium is driven toward reactants H 3 O + (aq) + CH 3 COO (aq) Na + (aq) + CH 3 COO - (aq) addition A solution that contains a weak acid and its conjugate base (or a weak base and its conjugate acid) is a buffer. 14.6 (Common Ions) Group Work Buffer solutions resist changes in ph when small amounts of strong acid or base are added. They do this because they already contain both a weak acid and a weak base that don t neutralize each other. When a small amount of strong acid is added, the buffer s base will neutralize the added acid. When a small amount of strong base is added, the buffer s acid will neutralize the base. acid conjugate base CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO (aq) reacts with added base reacts with added acid Identify the solutions below that would make good buffer solutions (what criteria need to be met?): HF and NaF ammonia and ammonium bromide KOH and KF CH 3 COOH and LiCH 3 COO nitric acid and sodium nitrate chlorous acid and potassium chlorite NaOH and NaCl HCl and NaCH 3 COO think about what these would do in solution! 1

Calculate the ph of a 0.50 M CH 3 COOH solution (K a = 1.8x10-5 ). ph = 2.52 What will happen to ph after we add 0.50 M NaCH 3 COO? The initial equation for a buffer solution looks like the dissociation of the weak acid or weak base in water. CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) The ph of a buffer solution can be calculated using two initial concentrations in an ICE table ( ), or we can use the Henderson- Hasselbalch equation which is derived from the K a equation. ph = 4.74 15.3 (Common Ions) 1) Add strong acid (e.g., HCl) HCl (aq) + NaCH 3 COO(aq) CH 3 COOH (aq) + NaCl(aq) Draw the product beaker of: Reactant beaker 1 = 2 moles of buffer solution (can use HA/NaA) Reactant beaker 2 = 1 mole of HCl 2) Add strong base (e.g., NaOH): NaOH(aq) + CH 3 COOH(aq) NaCH 3 COO(aq) + H 2 O(l) Draw the product beaker of: Reactant beaker 1 = 2 moles of buffer solution (can use HA/NaA) Reactant beaker 2 = 1 mole of NaOH Buffer Solution plus Strong Acid Buffer Solution plus Strong Base Calculate the ph of 75.0 ml of a 0.50 M acetic acid (K a = 1.8x10-5 ) and 0.50 M sodium acetate solution. ph = pk a + log (base / acid) = 4.7447 + log (0.50 / 0.50) ph = 4.74 The volume doesn t affect the calculation of the Henderson- Hasselbalch equation. However, we ll need volumes when we calculate ph after adding strong acid or base. Because volumes are often measured in ml, it is helpful to use a new unit, mmol (millimole). Figure 14.18: This diagram shows the buffer action of these reactions. M = mmol / ml mmol = M x ml 2

with Strong Acid Calculate the ph of 75.0 ml of a 0.50 M acetic acid and 0.50 M sodium acetate solution after 5.0 ml of 1.0 M HCl is added. The added acid (H 3 O + ) reacts with the conjugate base (acetate ion). Since a strong acid is added, the reaction goes to completion. Use a change table to determine how many mmoles remain after reaction. Then use Henderson-Hasselbalch equation to calculate ph. NaCH 3 COO(aq) HCl(aq) + NaCl(aq) CH 3 COOH(aq) with Strong Base Calculate the ph of 75.0 ml of a 0.50 M acetic acid and 0.50 M sodium acetate solution after 5.0 ml of 1.0 M NaOH is added. The added base (OH - ) reacts with the acid (acetic). Since a strong base is added, the reaction goes to completion. Use a change table to determine how many mmoles remain after reaction. Then use Henderson-Hasselbalch equation to calculate ph. CH 3 COOH(aq) NaOH(aq) + H 2 O(l) NaCH 3 COO(aq) ph = 4.7447 + log (32.5 / 42.5) = 4.63 ph = 4.7447 + log (42.5 / 32.5) = 4.86 Example 14.20 Practice Calculations Calculate the ph of a 50.0 ml solution made by combining 0.20 M HCOOH with 0.50 M NaHCOO. K a = 1.8x10-4 ph = 4.14 Practice: What is ph when 10.00 ml of 0.50 M HCl is added? ph = 3.87 Work in class: What is ph when 30.00 ml of 2.0 M HCl is added? ph = 0.36 Practice: What is ph when 10.00 ml of 0.50 M NaOH is added? ph = 4.52 Problem 14.20 Practice Calculations The ph of a 50.0 ml buffer solution made by combining 0.20 M HCOOH and 0.50 M NaHCOO is 4.14. K a = 1.8x10-4 What is ph when 30.00 ml of 2.0 M HCl is added? NaHCOO(aq) HCl(aq) + NaCl(aq) HCOOH(aq) 25 mmol 60 mmol - 10. mmol - 25 mmol - 25 mmol - + 25 mmol 0 mol 35 mmol - 35 mmol [H 3 O + ] = 35 mmol / (50 + 30) ml = 0.4375 M ph = 0.36 [H 3 O + ] from HCOOH = 0.00887 M (relatively negligible) What is the ph of a 1.00 L solution that is 0.75 M NH 3 and 0.25 M NH 4 Cl? K b = 1.8 x 10-5 K b K a pk a OR K b pk b pk a ph = pk a + log ([NH 3 ]/[NH 4+ ]) = 9.2552 + log (0.75/0.25) ph = 9.73 To make a buffer with a specific ph: 1) Pick a weak acid whose pk a is close to the desired ph (within one unit of desired ph). 2) Substitute the ph and pk a into the Henderson- Hasselbalch equation to calculate the [base]/[acid] ratio. What is the ph of this solution after 0.01 mol HBr is added to it? ph = 9.2552 + log ((0.75 0.01) / (0.25 + 0.01)) = ph = 9.70 3

Buffer Capacity Buffers have a certain range in which they can maintain a constant ph. Outside of this range, we say the buffer s capacity has been exceeded and we see large changes in ph. The best capacity for a buffer is when the conjugate acid and base concentrations are within a factor of 10 (the best capacity is when the acid and base concentrations are equal). The ph of a buffer cannot be more than one ph unit different than the pk a of the weak acid it contains. Example: 1) If we want to make a buffer with a ph of 5.00, which acid solution should we start with? Ascorbic acid: K a = 8.0x10-5 Acetic acid: K a = 1.8x10-5 Hydrofluoric acid: K a = 7.1x10-4 Citric acid: K a = 3.5 x 10-4 2) Acetic acid gives a pk a of 4.7447. ph pk a = log (base/acid) = 0.2553 (base / acid) = 10 0.2553 = 1.80 : 1 (the base should be 1.8x more concentrated than the acid) ICE vs Change Tables 14.7 Strong Acid-Strong Base Titrations Arrows ICE table Change Table The reaction between the strong acid HCl and the strong base NaOH can be represented by: HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) When to use Weak acid or base in water At least one strong reactant Units molarity moles How to find x Labels Solve for x at equilibrium Initial, Change, Equilibrium Find x by determining the L.R. Initial, Change, Final Review terminology: Titrant solution in buret, being added Analyte unknown solution Equivalence point stoichiometric equivalents End point lightest color change Calculate the volume of base needed to reach the equivalence point when 25.00 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Titration Beaker Drawings Draw 4 beakers with strong acid. HCl + NaOH NaCl + H 2 O 1) 2 moles strong acid, 0 moles base 2) 2 moles strong acid, 1 mole base 3) 2 moles strong acid, 2 moles base 4) 2 moles strong acid, 3 moles base Titration Beaker Drawings Draw a WEAK acid reacting with a strong base. CH 3 COOH + NaOH NaCH 3 COO + H 2 O 1) 2 moles weak acid, 0 moles base 2) 2 moles weak acid, 1 mole base 3) 2 moles weak acid, 2 moles base 4) 2 moles weak acid, 3 moles base 4

Acid-Base Titrations Acid-Base Titrations: HCl + NaOH 4 key points on a titration curve At the beginning (before titrant is added) 0.00 ml NaOH Before the equivalence point 10.00 ml NaOH At the equivalence point 25.00 ml NaOH After the equivalence point 30.00 ml NaOH Figure 14.21 1) 0.00 ml NaOH HCl(aq) + NaOH(aq) H 2 O(l) NaCl(aq) 2.50 mmol 0 mmol - 0-0 mmol - 0 mmol - + 0 mmol 2.50 mmol 0 mmol - + 0 mmol Only strong acid in solution [HCl] = [H 3 O + ] ph = -log (0.10) = 1.00 Acid-Base Titrations: HCl + NaOH 2) 10.00 ml NaOH Test Question Example: What is the ph of 25.00 ml of 0.10 M HCl titrated with 10.00 ml of 0.10 M NaOH? HCl(aq) + NaOH(aq) H 2 O(l) NaCl(aq) 2.50 mmol 1.00 mmol - 0-1.00 mmol - 1.00 mmol - +1.00 mmol 1.50 mmol 0 mmol - + 1.00 mmol NaCl is neutral doesn t affect ph [HCl] = [H 3 O + ] = 1.50 mmol / (25+10 ml) = 0.04286 M ph = 1.37 Acid-Base Titrations: HCl + NaOH 3) 25.00 ml NaOH added At the equivalence point, what is in solution? HCl(aq) + NaOH(aq) H 2 O(l) NaCl(aq) 2.50 mmol 2.50 mmol - 0-2.50 mmol - 2.50 mmol - + 2.50 mmol 0 mmol 0 mmol - + 2.50 mmol mmol acid = mmol base; only neutral salt in solution ph = 7.00 (always 7 for strong acid-strong base!) Acid-Base Titrations: HCl + NaOH 4) 30.00 ml NaOH added HCl(aq) + NaOH(aq) H 2 O(l) NaCl(aq) 2.50 mmol 3.00 mmol - 0-2.50 mmol - 2.50 mmol - + 2.50 mmol 0 mmol 0.50 mmol - + 2.50 mmol Base is in excess. [NaOH] = [OH - ] = 0.50 mmol / (25+30 ml) = 0.0090909 M ph = 14 (-log 0.0090909) = 11.96 Problem 14.21 Acid-Base Titrations: HCl + NaOH Table 14.4 Check ph values in titrations Volume of 0.10 M NaOH ph value with 0.10 M HCl 0.00 1.00 5.00 1.18 10.0 1.37 15.0 1.60 20.0 1.96 25.0 7.00 30.0 11.96 35.0 12.22 40.0 12.36 5

Acid-Base Indicators Figure 14.22: Different indicators can be used depending on where the ph at the equivalence point is expected to be. Acid-Base Titrations Comparing Curves SA SB Titration WA SB Titration Weak Acid-Strong Base Titrations Consider the neutralization between acetic acid and sodium hydroxide: CH 3 COOH(aq) + NaOH(aq) NaCH 3 COO (aq) + H 2 O(l) The acetate ion that results from this neutralization undergoes hydrolysis: CH 3 COO (aq) + H 2 O(l)(aq) CH 3 COOH(aq) + OH (aq) Calculate the volume of 0.10 M NaOH needed to reach the equivalence point in a titration of 25.00 ml of 0.10 M acetic acid. K a = 1.8x10-5 4 regions in titration curve (same as SA-SB titration): 1) Before adding base 0.00 ml base 2) Before equivalence point 10.00 ml base 3) At equivalence point: 25.00 ml base 4) After equivalence point 30.00 ml base 1) 0 ml base. HA(aq) + NaOH(aq) H 2 O(l) NaA(aq) 2.50 mmol 0.00 mmol - 0-0.00 mmol - 0.00 mmol - + 0.00 mmol 2.50 mmol 0 mmol - 0.00 mmol This is just weak acid. How do we find the ph of a weak acid? K a = x 2 / (0.10 M x) = 1.8x10-5 x = [H 3 O + ] [H 3 O + ] = 0.001342 M ph = 2.87 2) 10.00 ml base. This is the buffer zone (both weak acid and its conjugate base are present). Set up a 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(aq) + NaOH(aq) H 2 O(l) NaA(aq) 2.50 mmol 1.00 mmol - 0-1.00 mmol -1.00 mmol - +1.00 mmol 1.50 mmol 0 mmol - 1.00 mmol ph = pk a + log ([A - ] / [HA]) = 4.7447 + log (1.00 / 1.50) ph = 4.57 Calculate ph after adding 12.50 ml of NaOH. 6

3) 25.00 ml base. HA(aq) + NaOH(aq) H 2 O(l) NaA(aq) 2.50 mmol 2.50 mmol - 0-2.50 mmol - 2.50 mmol - + 2.50 mmol 0 mmol 0 mmol - 2.50 mmol At equivalence point: only salt in solution Basic salt determines ph NaCH 3 COO Na + (aq) + CH 3 COO - (aq) Acetate ion hydrolyzes to shift ph 3) 25.00 ml base, continued. At equivalence point: Find ph of salt. We ve done this. Need to find concentration of salt: 2.50 mmoles / (25+25 ml) = 0.05 M CH 3 COO (aq) + H 2 O(l)(aq) CH 3 COOH(aq) + OH (aq) Need K b because OH - is produced K b = K w / K a = 1.0x10-14 / 1.8x10-5 = 5.5556x10-10 K b = x 2 / (0.05 x); x = [OH - ] = 5.2705x10-6 M poh = 5.2782 ph = 8.72 4) 30.00 ml base. HA(aq) + NaOH(aq) H 2 O(l) NaA(aq) 2.50 mmol 3.00 mmol - 0-2.50 mmol - 2.50 mmol - + 2.50 mmol 0 mmol 0.50 mmol - 2.50 mmol After the equivalence point: [NaOH] = [OH - ] = 0.50 mmol / (25+30 ml) = 0.0090909 M [A - ] is a minor contributor and doesn t need to be calculated [OH - ] = 5.2705x10-6 M (doesn t change [OH - ]). ph = 11.96 Problem 14.22 Acid-Base Titrations: CH 3 COOH + NaOH Table 14.4 Check ph values in titrations Volume of 0.10 M NaOH ph value with 0.10 M CH 3 COOH 0.00 2.87 5.00 4.14 10.0 4.57 15.0 4.92 20.0 5.35 25.0 8.72 30.0 11.96 35.0 12.22 40.0 12.36 Practice Calculations 1) Calculate the ph of a 1.0 M solution of acetic acid (K a = 1.8 x 10-5 ). ph = 2.37 2) Calculate the ph of the above solution after 0.50 moles sodium acetate is added to it. Volume = 1.00 L. ph = 4.44 3) Calculate the ph of the above solution (#2) after 0.10 mole of NaOH is added. ph = 4.57 4) Calculate the ph of the solution in #2 after 1.2 moles of NaOH are added. ph = 13.30 What is the ph of a solution made by combining 12.00 ml of 0.10 M HCl with 10.00 ml of 0.15 M NaOH? ph = 12.13 What is the ph of a solution made by combining 15.00 ml of 0.12 M ascorbic acid (K a = 7.9 x 10-5 ) with 10.00 ml of 0.10 M NaOH? ph = 4.20 Determine what substance is in the flask and what substance is the titrant. CC-BY-NA-3.0: http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s20-05- acid-base-titrations.html 7

Determine what substance is in the flask and what substance is the titrant. Determine what substance is in the flask and what substance is the titrant. CC-BY-NA-3.0: http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s20-05- acid-base-titrations.html CC-BY-NA-3.0: http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s20-05- acid-base-titrations.html Determine what substance is in the flask and what substance is the titrant. Acid-Base Titrations Comparing Curves CC-BY-NA-3.0: http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s20-05- acid-base-titrations.html CC-BY-NA-3.0: http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s20-05- acid-base-titrations.html Polyprotic Acid Titration: H 3 PO 4 with NaOH One eq. pt. for each proton. 3 eq. pts. and 3 half eq. pts. 15.1 Insoluble Salts (K sp ) Some combinations of ions in solution form precipitates. Use the solubility rules from Chapter 4/Periodic Table for the following: NaNO 3, PbCl 2, NH 4 Cl, Ag 2 S Even insoluble salts dissolve to a small extent. The solid (undissolved) salt and its dissociated ions establish equilibrium in solution. Soluble salt: > 0.10 M aqueous solution dissolves at room temperature (> 1 g dissolves in 100 ml H 2 O) Insoluble salt: < 0.001 M aqueous solution at room temperature (< 1 g dissolves in 100 ml H 2 O) The value of the equilibrium constant, K sp, (solubility product) is determined at the saturation point. http://www.youtube.com/watch?v=blq5nibwv5g 8

Solubility Equilibria K sp is the solubility product constant. Its value indicates how soluble or insoluble a salt will be in water. See Appendix J (Solubility Products) for K sp values. AgBr(s) Ag + (aq) + Br (aq) K sp = [Ag + ][Br ] = 5.0 x 10-13 PbCl 2 (s) Pb 2+ (aq) + 2Cl (aq) K sp = [Pb 2+ ][Cl ] 2 = 1.6 x 10-5 What does a large K sp value indicate? A small value? Write the dissolution equation and K sp expression for calcium phosphate. Review nomenclature!!!!! (Chapter 2) Figure 15.2: Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of Ag + and Cl ions in equilibrium with undissolved silver chloride. Calculations Using K sp We can solve for concentrations (or gram solubility) given K sp values or solve for K sp given concentrations (or gram solubility). Appendix J (Solubility Products) has a more complete list of K sp values! Substance Ppt color K sp value AgCl white 1.6x10-10 AgBr off-white 5.0x10-13 AgI yellow 1.5x10-16 PbS black 7x10-29 PbCl 2 white 1.6x10-5 Fe(OH) 3 rust red 4x10-38 CaCO 3 white 8.7x10-9 BaSO 4 white 2.3x10-8 The K sp of silver bromide is 5.0 x 10 13. Calculate the molar solubility and solubility (or gram solubility). AgBr(s) Ag + (aq) + Br (aq) Initial concentration (M) 0 0 Change in concentration (M) +x +x Equilibrium concentration (M) x x K sp = [Ag + ][Br ] = 5.0 x 10 13 = x 2 x = 7.1 x 10 7 M (molar sol.) Gram solubility (g/l): use molar mass of salt 7.07107x10-7 mol/l x 187.8 g/mol = 1.3x10-4 g/l (gram sol.) Example 15.1 Solubility Equilibria 1) Lead (II) chloride has a gram solubility of 4.4 g/l. Calculate K sp for this salt. 2) If [Br - ] eq = 0.0236 M, what is K sp for PbBr 2? 3) K sp for iron (III) hydroxide is found to be 4.0 x 10-38. Calculate the molar and gram solubility. 4) The solubility of calcium carbonate is found experimentally to be 9.3x10-3 g/l. Calculate K sp for calcium carbonate. Examples 15.2 15.6 Solubility/K sp Practice 1) PbCl 2 : 4.4 g/l x (1 mol / 278.106 g) = 0.01582 M K sp = 4x 3 = 4 (0.01582) 3 = 1.6x10-5 2) If [Br - ] eq = 0.0236 M, then [Pb 2+ ] = 0.0118 M, K sp = [Pb 2+ ][Br - ] 2 = (0.0118)(0.0236) 2 = 6.57 x 10-6 3) Fe(OH) 3 : K sp = [Fe 3+ ][OH - ] 3 = 27x 4 = 4.0x10-38 x = Solubility/K sp Practice Answers. x = 2.0x10-10 M, 2.1x10-8 g/l = 1.962x10-10 M x (106.875 g/mol) 4) 9.3x10-3 g/l x (1 mol / 100.078 g) = 9.293x10-5 M = x K sp = [Ca 2+ ][CO 3 2- ] = x 2 = (9.293x10-5 ) 2 = 8.6x10-9 Factors Affecting Solubility Common Ion Effect What will happen if we add solid NaF to a solution of saturated CaF 2? CaF 2 (s) Ca 2+ (aq) + 2F - (aq) Common ion effect: add more F - to precipitate more CaF 2 from solution. How does adding a common ion affect the solubility of the solid? A common ion will decrease the solubility of a solid (increase the amount of solid/precipitate that forms). 9

Factors Affecting Solubility Common Ion Effect Calculate the molar solubility of calcium fluoride (in pure water) (K sp = 3.5 x 10-11 ). x = 2.1 x 10-4 M Calculate the molar solubility of calcium fluoride in 0.010 M sodium fluoride. x = 3.5 x 10-7 M Example 15.12 Predicting Precipitation of Ionic Compounds To determine if a precipitate will form as a product of a double replacement reaction or if all of a salt will dissolve in a given amount of water, we calculate Q sp and compare to K sp. Predict products when HCl (aq) is added to Pb(NO 3 ) 2 (aq) 1) Q < K sp ; no precipitate forms (unsaturated) concentration of ions is not high enough to precipitate 2) Q = K sp ; no precipitate forms (saturated) solution has maximum conc. of ions to dissolve. 3) Q > K sp ; precipitate forms (supersaturated; ppt forms) excess ions in solution will cause precipitation Precipitation of Ionic Compounds A solution contains 1.5 x 10-2 M hydrochloric acid and 1.2 x 10-2 M lead(ii) nitrate. Will a precipitate form? K sp (PbCl 2 ) = 1.6 x 10-5 2 HCl (aq) + Pb(NO 3 ) 2 (aq) PbCl 2 (s) + 2 HNO 3 (aq) PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) Q sp = [Pb 2+ ][Cl - ] 2 = (1.2 x 10-2 )(1.5 x 10-2 ) 2 Q sp = 2.7 x 10-6 Q sp < K sp No precipitate will form. Examples 15.7 15.9 Dissolution of Ionic Compounds 245 mg of magnesium carbonate is placed in 1.00 L of water. Will it all dissolve? K sp = 4.0 x 10-5 Two methods to solve: 1) Calculate gram solubility from K sp. Compare to given mass of solid (same as what we ve already worked). 2) Find molarity of solid. Calculate Q sp. Compare to K sp to determine if precipitate will occur. If Q sp > K sp, supersaturated and salt won t dissolve If Q sp < K sp, unsaturated and salt will dissolve Dissolution of Ionic Compounds Dissolution of Ionic Compounds 245 mg of magnesium carbonate is placed in 1.00 L of water. Will it all dissolve? K sp = 4.0 x 10-5 Method 1: Calculate gram solubility from K sp. Compare to given mass of solid (same as what we ve already worked). K sp = [Mg 2+ ][CO 3 2- ] = x 2 = 4.0 x 10-5 x = (4.0 x 10-5 ) = 6.325x10-3 M x = 6.325x10-3 M x (84.32 g/1 mol) = 0.53 g/l 0.245 g/l < 0.53 g/l unsat., salt will dissolve 245 mg of magnesium carbonate is placed in 1.00 L of water. Will it all dissolve? K sp = 4.0 x 10-5 Method 2: Find molarity of solid. Calculate Q sp. Compare to K sp to determine if precipitate will occur. 245 mg = 0.0245 g/l x (1 mol/84.32 g) = 2.906 x 10-3 M K sp = [Mg 2+ ][CO 3 2- ] = x 2 = 4.0 x 10-5 Q sp = (2.906 x 10-3 ) 2 = 8.442 x 10-6 Q sp < K sp, unsaturated and salt will dissolve 10

Factors Affecting Solubility - ph Mg(OH) 2 (s) Mg 2+ (aq) + 2OH (aq) K sp = [Mg 2+ ][OH ] 2 = 1.2 x 10 11 ; weak base because insoluble x = 1.4 x 10 4 M At equilibrium: [OH ] = 2(1.4 x 10 4 M ) = 2.8 x 10 4 M ph = 14.00 3.55 = 10.45 Qualitative Analysis Lab: involves the principle of selective precipitation and can be used to identify the types of ions present in a solution. This is the qualitative analysis lab you ll be doing. In a solution with a ph of less than 10.45, the solubility of Mg(OH) 2 increases. Why? Acid added will react with OH - and remove it from solution. This will shift the equilibrium to the right and increase solubility. Example 15.10 Selective Precipitation Some compounds can be separated based on selective precipitation. This is the separation of precipitates based on different K sp values. For example, AgCl (K sp = 1.6x10-10 ), AgBr (K sp = 7.7x10-13 ), and AgI (K sp = 8.3x10-17 ) all precipitate. Because silver iodide s K sp value is so low (it is less soluble than the other salts), it would precipitate before the other two. Almost all of the AgI would precipitate before AgCl and AgBr would. Complex Ions Complex ion formation: A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Because AgCl is an acidic salt, adding base will dissolve it. AgCl (s) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) + Cl - (aq) Ammonia is called a ligand in this compound. Ligands are small neutral molecules or ions that often act as Lewis bases. Adding acid to the product will break the coordinate covalent bond between silver and ammonia to reprecipitate AgCl. Ag + (aq) + Cl - (aq) AgCl (s) Formation Constant (K f ) values Table 15.2: Common Complex Ions and Their Formation Constants Substance K f at 25 o C [Cd(CN) 4 ] 2-1.3x10 7 Ag(NH 3 ) + 2 1.7x10 7 [AlF 6 ] 3-7.0x10 19 15.2 Lewis Acids and Bases A Lewis base is a substance that can donate a pair of electrons. A Lewis acid is a substance that can accept a pair of electrons to form a coordinate covalent bond. The larger the formation constant, the more stable the complex ion is. More values can be found in Appendix K (Formation Constants for Complex Ions). 66 11