Equilibria. Ch a pt e r Applications of Aqueous. Chemistry 4th Edition McMurry/Fay
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1 16 Ch a pt e r Applications of Aqueous Equilibria Chemistry 4th Edition McMurry/Fay Dr. Paul Charlesworth Michigan Technological University The Common-Ion Effect 01 Chapter 16 Slide 2 The Common-Ion Effect 01 Common Ion: Two dissolved solutes that contain the same ion (cation or anion). The presence of a common ion suppresses the ionization of a weak acid or a weak base. Common-Ion Effect: is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. Chapter 16 Slide 3 1
2 The Common-Ion Effect 02 To determine the ph, we apply I.C.E. and then the Henderson Hasselbalch equation. ph = pka + log pka = logka [ Conjugate Base] [ Acid] When the concentration of HA and salt are high (=0.1 M) we can neglect the ionization of acid and hydrolysis of salt. Chapter 16 Slide 4 The Common-Ion Effect 03 Calculate the ph of a 0.20 M CH 3 COOH solution with no salt added. Calculate the ph of a solution containing 0.20 M CH 3 COOH and 0.30 M CH 3 COONa. What is the ph of a solution containing 0.30 M HCOOH, before and after adding 0.52 M HCOOK? Chapter 16 Slide 5 Buffer Solutions 01 A Buffer Solution: is a solution of (1) a weak acid or a weak base and (2) its salt; both components must be present. A buffer solution has the ability to resist changes in ph upon the addition of small amounts of either acid or base. Buffers are very important to biological systems. Chapter 16 Slide 6 2
3 Buffer Solutions 02 Chapter 16 Slide 7 Buffer Solutions 03 Buffer solutions must contain relatively high acid and base component concentrations, the buffer capacity. Acid and base component concentrations must not react together. The simplest buffer is prepared from equal concentrations of acid and conjugate base. Chapter 16 Slide 8 Buffer Solutions 04 Calculate the ph of a buffer system containing 1.0 M CH 3 COOH and 1.0 M CH 3 COONa. What is the ph of the system after the addition of 0.10 mole of gaseous HCl to 1.0 L of solution? Calculate the ph of 0.30 M NH 3 /0.36 NH 4 Cl buffer system. What is the ph after the addition of 20.0 ml of M NaOH to 80.0 ml of the buffer solution? Chapter 16 Slide 9 3
4 Buffer Solutions 05 Buffer Preparation: Use the Henderson Hasselbalch equation in reverse. 1. Choose weak acid with pk a close to required ph. 2. Substitute into Henderson Hasselbalch equation. 3. Solve for the ratio of [conjugate base]/[acid]. This will give the mole ratio of conjugate base to acid. The acid should always be 1.0. Chapter 16 Slide 10 Buffer Solutions 06 Describe how you would prepare a phosphate buffer with a ph of about How would you prepare a liter of carbonate buffer at a ph of 10.10? You are provided with carbonic acid (H 2 CO 3 ), sodium hydrogen carbonate (NaHCO 3 ), and sodium carbonate (Na 2 CO 3 ). Chapter 16 Slide 11 Acid Base Titrations 01 Chapter 16 Slide 12 4
5 Acid Base Titrations 01 Titration: a procedure for determining the concentration of a solution using another solution of known concentration. Titrations involving only strong acids or bases are straightforward. Titrations involving weak acids or bases are complicated by hydrolysis of salt formed. Chapter 16 Slide 13 Acid Base Titrations 02 Strong Acid Strong Base: The equivalence point is the point at which equimolar amounts of acid and base have reacted. Chapter 16 Slide 14 Acid Base Titrations 03 The ph of a 25 ml 0.10 M HCl sample can be determined after the addition of: 0. No addition of 0.10 M NaOH ml (total) of 0.10 M NaOH ml (total) of 0.10 M NaOH ml (total) of 0.10 M NaOH. Chapter 16 Slide 15 5
6 Acid Base Titrations 04 Weak Acid Strong Base: The conjugate base hydrolyzes to form weak acid and OH. At equivalence point only the conjugate base is present. ph at equivalence point will always be >7. Chapter 16 Slide 16 Acid Base Titrations 05 Chapter 16 Slide 17 Acid Base Titrations 06 The ph of a 25 ml 0.10 M CH 3 COOH sample can be determined after the addition of: 0. No addition of 0.10 M NaOH ml (total) of 0.10 M NaOH ml (total) of 0.10 M NaOH ml (total) of 0.10 M NaOH. Chapter 16 Slide 18 6
7 Acid Base Titrations 07 Exactly 100 ml of 0.10 M nitrous acid are titrated with a 0.10 M NaOH solution. Calculate the ph the ph for: 1. The initial solution. 2. The point at which 80 ml of base have been added. 3. The equivalence point. 4. The point at which 105 ml of base have been added. Chapter 16 Slide 19 Acid Base Titrations 09 Strong Acid Weak Base: The (conjugate) acid hydrolyzes to form weak base and H 3 O +. At equivalence point only the (conjugate) acid is present. ph at equivalence point will always be <7. Chapter 16 Slide 20 Acid Base Titrations 10 Calculate the ph at the equivalence point when 25 ml of 0.10 M NH 3 is titrated with a 0.10 M HCl solution. Calculate the ph at the equivalence point in the titration of 50 ml of 0.10 M methylamine with a 0.20 M HCl solution. Chapter 16 Slide 21 7
8 Acid Base Titrations 11 Polyprotic Acids: Chapter 16 Slide 22 Solubility Equilibria 01 Solubility Product: is the product of the molar concentrations of constituent ions and provides a measure of a compound s solubility. MX 2 (s) æ M 2+ (aq) + 2 X (aq) K sp = [M 2+ ][X ] 2 Chapter 16 Slide 23 Solubility Equilibria 02 Al(OH) x BaCO x 10 9 BaF x 10 6 BaSO x Bi 2 S x CdS 8.0 x CaCO x 10 9 CaF x Ca(OH) x 10 6 Ca 3 (PO 4 ) x Cr(OH) x CoS 4.0 x CuBr 4.2 x 10 8 CuI 5.1 x Cu(OH) x CuS 6.0 x Fe(OH) x Fe(OH) x FeS 6.0 x PbCO x PbCl x 10 4 PbCrO x PbF x 10 8 PbI x 10 8 PbS 3.4 x MgCO x 10 5 Mg(OH) x MnS 3.0 x Hg 2 Cl x HgS 4.0 x NiS 1.4 x AgBr 7.7 x Ag 2 CO x AgCl 1.6 x Ag 2 SO x 10 5 Ag 2 S 6.0 x SrCO x 10 9 SrSO x 10 7 SnS 1.0 x Zn(OH) x ZnS 3.0 x Chapter 16 Slide 24 8
9 Solubility Equilibria 03 The solubility of calcium sulfate (CaSO 4 ) is found experimentally to be 0.67 g/l. Calculate the value of K sp for calcium sulfate. The solubility of lead chromate (PbCrO 4 ) is 4.5 x 10 5 g/l. Calculate the solubility product of this compound. Calculate the solubility of copper(ii) hydroxide, Cu(OH) 2, in g/l. Chapter 16 Slide 25 Solubility Equilibria 04 Ion Product (Q): solubility equivalent of the reaction quotient. It is used to determine whether a precipitate will form. Q < K sp Q = K sp Q > K sp Unsaturated Saturated Supersaturated; precipitate forms. Chapter 16 Slide 26 Solubility Equilibria 05 Exactly 200 ml of M BaCl 2 are added to exactly 600 ml of M K 2 SO 4. Will a precipitate form? If 2.00 ml of M NaOH are added to 1.00 L of M CaCl 2, will precipitation occur? Chapter 16 Slide 27 9
10 Fractional Precipitation 01 Chapter 16 Slide 28 Fractional Precipitation 01 Fractional precipitation is a method of removing one ion type while leaving others in solution. Ions are added that will form an insoluble product with one ion and a soluble one with others. When both products are insoluble, their relative K sp values can be used for separation. Chapter 16 Slide 29 Fractional Precipitation 02 Silver nitrate is slowly added to a solution that is M in Cl ions & M in Br ions. Calculate the concentration of Ag + ions (in mol/l) required to initiate (a) the pptnof AgBr, and (b) the pptn of AgCl. K sp s of AgCl and Ag 3 PO 4 are 1.6 x and 1.8 x 10 18, respectively. If Ag + is added to 1.00 L of a solution containing 0.10 mol Cl and 0.10 mol PO 4 3, calculate the concentration of Ag + ions required to initiate (a) the pptn of AgCl, and (b) the pptn of Ag 3 PO 4. Chapter 16 Slide 30 10
11 The Common-Ion Effect and Solubility 01 Chapter 16 Slide 31 The Common-Ion Effect and Solubility 01 The solubility product (K sp ) is an equilibrium constant; precipitation will occur when the ion product exceeds the K sp for a compound. If AgNO 3 is added to saturated AgCl, the increase in [Ag + ] will cause AgCl to precipitate. Q = [Ag + ] 0 [Cl ] 0 > K sp Chapter 16 Slide 32 The Common-Ion Effect and Solubility 02 Chapter 16 Slide 33 11
12 The Common-Ion Effect and Solubility 03 Chapter 16 Slide 34 The Common-Ion Effect and Solubility 04 Calculate the solubility of silver chloride (in g/l) in a 6.5 x 10 3 M silver chloride solution. Calculate the solubility of AgBr (in g/l) in: (a) pure water (b) M NaBr Chapter 16 Slide 35 Complex Ion Equilibria and Solubility 01 A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Most metal cations are transition metals because they have more than one oxidation state. The formation constant (K f ) is the equilibrium constant for the complex ion formation. Chapter 16 Slide 36 12
13 Complex Ion Equilibria and Solubility 02 Chapter 16 Slide 37 Complex Ion Equilibria and Solubility 03 Chapter 16 Slide 38 Complex Ion Equilibria and Solubility 04 ION K f Ag(NH 3 ) x 10 7 Ag(CN) x Cu(CN) x Cu(NH 3 ) x Cd(CN) x CdI x 10 6 ION K f HgCl x HgI x Hg(CN) x Co(NH 3 ) x Zn(NH 3 ) x 10 9 Chapter 16 Slide 39 13
14 Complex Ion Equilibria and Solubility 05 A 0.20 mole quantity of CuSO 4 is added to a liter of 1.20 M NH 3 solution. What is the concentration of Cu 2+ ions at equilibrium? If 2.50 g of CuSO 4 are dissolved in 9.0 x 10 2 ml of 0.30 M NH 3, what are the concentrations of Cu 2+, Cu(NH 3 ) 4 2+, and NH 3 at equilibrium? Chapter 16 Slide 40 Complex Ion Equilibria and Solubility 06 Calculate the molar solubility of AgCl in a 1.0 M NH 3 solution. Calculate the molar solubility of AgBr in a 1.0 M NH 3 solution. Chapter 16 Slide 41 Complex Ion Equilibria and Solubility 07 Chapter 16 Slide 42 14
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