SPRING REVIEW PART TWO. PRECIPITATION REACTIONS Chapter 19 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to.

SPRING REVIEW PART TWO

PRECIPITATION REACTIONS Chapter 19 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

Analysis of Silver Group All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions.

Analysis of Silver Group Although all salts formed in this experiment are said to be insoluble, they do dissolve to some SLIGHT extent. AgCl(s) Ag + (aq) + Cl - (aq) When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED.

Analysis of Silver Group AgCl(s) Ag + (aq) + Cl - (aq) When solution is SATURATED, expt. shows that [Ag + ] = 1.67 x 10 -5 M. This is equivalent to the SOLUBILITY of AgCl. What is [Cl - ]? This is also equivalent to the AgCl solubility.

Analysis of Silver Group AgCl(s) Ag + (aq) + Cl - (aq) Saturated solution has [Ag + ] = [Cl - ] = 1.67 x 10 -5 M Use this to calculate K c K c = [Ag + ] [Cl - ] = (1.67 x 10 -5 )(1.67 x 10 -5 ) = (1.67 x 10 -5 )(1.67 x 10 -5 ) = 2.79 x 10 -10 = 2.79 x 10 -10

Analysis of Silver Group AgCl(s) Ag + (aq) + Cl - (aq) K c = [Ag + ] [Cl - ] = 2.79 x 10 -10 Because this is the product of “solubilities”, we call it K sp = solubility product constant See Table 19.2 and Appendix J

Precipitating an Insoluble Salt Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 If [Hg 2 2+ ] = 0.010 M, what [Cl - ] is req’d to just begin the precipitation of Hg 2 Cl 2 ? That is, what is the maximum [Cl - ] that can be in solution with 0.010 M Hg 2 2+ without forming Hg 2 Cl 2 ?

Precipitating an Insoluble Salt Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 Recognize that K sp = product of maximum ion concs. Precip. begins when product of ion concs. EXCEEDS the K sp.

Precipitating an Insoluble Salt Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 Solution [Cl - ] that can exist when [Hg 2 2+ ] = 0.010 M,

Precipitating an Insoluble Salt Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 Solution [Cl - ] that can exist when [Hg 2 2+ ] = 0.010 M, If this conc. of Cl - is just exceeded, Hg 2 Cl 2 begins to precipitate.

Precipitating an Insoluble Salt Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 Now raise [Cl - ] to 1.0 M. What is the value of [Hg 2 2+ ] at this point? Solution [Hg 2 2+ ] = K sp / [Cl - ] 2 = K sp / (1.0) 2 = 1.1 x 10 -18 M = K sp / (1.0) 2 = 1.1 x 10 -18 M The concentration of Hg 2 2+ has been reduced by 10 16 !

Separating Metal Ions Cu 2+, Ag +, Pb 2+ K sp Values AgCl1.8 x 10 -10 PbCl 2 1.7 x 10 -5 PbCrO 4 1.8 x 10 -14 K sp Values AgCl1.8 x 10 -10 PbCl 2 1.7 x 10 -5 PbCrO 4 1.8 x 10 -14

Separating Salts by Differences in K sp A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. Which precipitates first? K sp for Ag 2 CrO 4 = 9.0 x 10 -12 K sp for PbCrO 4 = 1.8 x 10 -14 Solution The substance whose K sp is first exceeded precipitates first. The ion requiring the lesser amount of CrO 4 2- ppts. first.

Separating Salts by Differences in K sp A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. Which precipitates first? K sp for Ag 2 CrO 4 = 9.0 x 10 -12 K sp for PbCrO 4 = 1.8 x 10 -14 Solution Calculate [CrO 4 2- ] required by each ion.

Separating Salts by Differences in K sp [CrO 4 2- ] to ppt. PbCrO 4 = K sp / [Pb 2+ ] = 1.8 x 10 -14 / 0.020 = 9.0 x 10 -13 M = 1.8 x 10 -14 / 0.020 = 9.0 x 10 -13 M [CrO 4 2- ] to ppt. Ag 2 CrO 4 = K sp / [Ag + ] 2 = 9.0 x 10 -12 / (0.020) 2 = 2.3 x 10 -8 M = 9.0 x 10 -12 / (0.020) 2 = 2.3 x 10 -8 M PbCrO 4 precipitates first. PbCrO 4 precipitates first. A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. Which precipitates first? K sp for Ag 2 CrO 4 = 9.0 x 10 -12 K sp for PbCrO 4 = 1.8 x 10 -14 Solution

A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. PbCrO 4 ppts. first. K sp (Ag 2 CrO 4 )= 9.0 x 10 -12 K sp (PbCrO 4 ) = 1.8 x 10 -14 How much Pb 2+ remains in solution when Ag + begins to precipitate? Separating Salts by Differences in K sp

A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. PbCrO 4 ppts. first. K sp (Ag 2 CrO 4 )= 9.0 x 10 -12 K sp (PbCrO 4 ) = 1.8 x 10 -14 How much Pb 2+ remains in solution when Ag + begins to precipitate? Solution We know that [CrO 4 2- ] = 2.3 x 10 -8 M to begin to ppt. Ag 2 CrO 4. What is the Pb 2+ conc. at this point? Separating Salts by Differences in K sp

A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. K sp (Ag 2 CrO 4 )= 9.0 x 10 -12 K sp (PbCrO 4 ) = 1.8 x 10 -14 How much Pb 2+ remains in solution when Ag + begins to precipitate? Solution [Pb 2+ ] = K sp / [CrO 4 2- ] = 1.8 x 10 -14 / 2.3 x 10 -8 M = 7.8 x 10 -7 M = 7.8 x 10 -7 M Separating Salts by Differences in K sp

A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. K sp (Ag 2 CrO 4 )= 9.0 x 10 -12 K sp (PbCrO 4 ) = 1.8 x 10 -14 How much Pb 2+ remains in solution when Ag + begins to precipitate? Solution [Pb 2+ ] = K sp / [CrO 4 2- ] = 1.8 x 10 -14 / 2.3 x 10 -8 M = 7.8 x 10 -7 M = 7.8 x 10 -7 M Lead ion has dropped from 0.020 M to < 10 -6 M Separating Salts by Differences in K sp

Entropy and Free Energy How to predict if a reaction can occur, given enough time? THERMODYNAMICS How to predict if a reaction can occur at a reasonable rate? KINETICS Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

ThermodynamicsThermodynamics Is the state of a chemical system such that a rearrangement of its atoms and molecules would decrease the energy of the system? Is the state of a chemical system such that a rearrangement of its atoms and molecules would decrease the energy of the system? If yes, system is favored to react — a product- favored system. If yes, system is favored to react — a product- favored system. Most product-favored reactions are exothermic. Most product-favored reactions are exothermic. Often referred to as spontaneous reactions. Often referred to as spontaneous reactions. Spontaneous does not imply anything about time for reaction to occur. Spontaneous does not imply anything about time for reaction to occur.

Product-Favored Reactions In general, product-favored reactions are exothermic. Fe 2 O 3 (s) + 2 Al(s) ---> 2 Fe(s) + Al 2 O 3 (s) H = - 848 kJ  H = - 848 kJ

Entropy, S One property common to product- favored processes is that the final state is more DISORDERED or RANDOM than the original. Spontaneity is related to an increase in randomness. The thermodynamic property related to randomness is ENTROPY, S. Reaction of K with water

How probable is it that reactant molecules will react? PROBABILITY suggests that a product- favored reaction will result in the dispersal of energy or of matter or both. Directionality of Reactions

S (gases) > S (liquids) > S (solids) S o (J/Kmol) H 2 O(liq)69.91 H 2 O(gas)188.8 S o (J/Kmol) H 2 O(liq)69.91 H 2 O(gas)188.8 Entropy, S

Increase in molecular complexity generally leads to increase in S. S o (J/Kmol) CH 4 248.2 C 2 H 6 336.1 C 3 H 8 419.4 S o (J/Kmol) CH 4 248.2 C 2 H 6 336.1 C 3 H 8 419.4 Entropy, S

Entropy Changes for Phase Changes For a phase change, S = q/T For a phase change,  S = q/T where q = heat transferred in phase change For H 2 O (liq) ---> H 2 O(g) H = q = +40,700 J/mol  H = q = +40,700 J/mol

Consider 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) S o = 2 S o (H 2 O) - [2 S o (H 2 ) + S o (O 2 )]  S o = 2 S o (H 2 O) - [2 S o (H 2 ) + S o (O 2 )] S o = 2 mol (69.9 J/Kmol) - [2 mol (130.7 J/Kmol) + 1 mol (205.3 J/Kmol)]  S o = 2 mol (69.9 J/Kmol) - [2 mol (130.7 J/Kmol) + 1 mol (205.3 J/Kmol)] S o = -326.9 J/K  S o = -326.9 J/K Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid. Calculating S for a Reaction Calculating  S for a Reaction S o =  S o (products) -  S o (reactants)  S o =  S o (products) -  S o (reactants)

2nd Law of Thermodynamics A reaction is spontaneous (product-favored) if ²S for the universe is positive. S universe = S system + S surroundings  S universe =  S system +  S surroundings S universe > 0 for product-favored process  S universe > 0 for product-favored process First calc. entropy created by matter dispersal (S system ) First calc. entropy created by matter dispersal (  S system ) Next, calc. entropy created by energy dispersal (S surround ) Next, calc. entropy created by energy dispersal (  S surround )

CELL POTENTIAL, E Electrons are “driven” from anode to cathode by an electromotive force or emf. Electrons are “driven” from anode to cathode by an electromotive force or emf. For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M. For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25  C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M. Zn and Zn 2+, anode Cu and Cu 2+, cathode Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

CELL POTENTIAL, E For Zn/Cu cell, voltage is 1.10 V at 25 C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M. For Zn/Cu cell, voltage is 1.10 V at 25  C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M. This is the STANDARD CELL POTENTIAL, E o This is the STANDARD CELL POTENTIAL, E o —a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 C. —a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25  C.

Calculating Cell Voltage Balanced half-reactions can be added together to get overall, balanced equation. Balanced half-reactions can be added together to get overall, balanced equation. If we know E o for each half-reaction, we could get E o for net reaction. If we know E o for each half-reaction, we could get E o for net reaction. 2 I - ---> I 2 + 2e- 2 H 2 O + 2e- ---> 2 OH - + H 2 ------------------------------------------------- 2 I - + 2 H 2 O --> I 2 + 2 OH - + H 2 2 I - ---> I 2 + 2e- 2 H 2 O + 2e- ---> 2 OH - + H 2 ------------------------------------------------- 2 I - + 2 H 2 O --> I 2 + 2 OH - + H 2

CELL POTENTIALS, E o Can’t measure 1/2 reaction E o directly. Therefore, measure it relative to a STANDARD HALF CELL, SHE. 2 H + (aq, 1 M) + 2e- --> H 2 (g, 1 atm) E o = 0.0 V

Zn/Zn 2+ half-cell hooked to a SHE. E o for the cell = +0.76 V Zn/Zn 2+ half-cell hooked to a SHE. E o for the cell = +0.76 V

Overall reaction is reduction of H + by Zn metal. Zn(s) + 2 H + (aq) --> Zn 2+ + H 2 (g) E o = +0.76 V Therefore, E o for Zn ---> Zn 2+ (aq) + 2e- is +0.76 V. Zn is a better reducing agent than H 2.

TABLE OF STANDARD POTENTIALS

E o and G o E o and  G o E o is related to G o, the free energy change for the reaction. E o is related to  G o, the free energy change for the reaction. G o = - n F E o  G o = - n F E o where F = Faraday constant = 9.6485 x 10 4 J/Vmol and n is the number of moles of electrons transferred Michael Faraday 1791-1867

Michael Faraday 1791-1867 Originated the terms anode, cathode, anion, cation, electrode. Discoverer of electrolysis electrolysis magnetic props. of matter magnetic props. of matter pretty cool guy pretty cool guy electromagnetic induction electromagnetic induction benzene and other organic chemicals benzene and other organic chemicals Was a popular lecturer.

E o and G o E o and  G o G o = - n F E o  G o = - n F E o For a product-favored reaction Reactants ----> Products Reactants ----> Products G o 0  G o 0 E o is positive For a reactant-favored reaction Reactants <---- Products Reactants <---- Products G o > 0 and so E o 0 and so E o < 0 E o is negative

Quantitative Aspects of Electrochemistry Consider electrolysis of aqueous silver ion. Ag + (aq) + e- ---> Ag(s) 1 mol e----> 1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But how to measure moles of e-? Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

Quantitative Aspects of Electrochemistry Consider electrolysis of aqueous silver ion. Ag + (aq) + e- ---> Ag(s) 1 mol e----> 1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But how to measure moles of e-?

But how is charge related to moles of electrons? Charge on 1 mol of e- = (1.60 x 10 -19 C/e-)(6.02 x 10 23 e-/mol) = 96,500 C/mol e- = 1 Faraday Quantitative Aspects of Electrochemistry

1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a)Calc. charge Coulombs = amps x time = (1.5 amps)(15.0 min)(60 s/min) = 1350 C

Quantitative Aspects of Electrochemistry 1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a)Charge = 1350 C (b)Calculate moles of e- used

Quantitative Aspects of Electrochemistry 1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a)Charge = 1350 C (b)Calculate moles of e- used (c)Calc. quantity of Ag

Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Calculate moles of e-

Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Calculate moles of e- c)Calculate charge

Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Calculate moles of e- c)Calculate charge 4.38 mol e- 96,500 C/mol e- = 423,000 C 4.38 mol e- 96,500 C/mol e- = 423,000 C

Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Mol of e- = 4.38 mol c)Charge = 423,000 C

Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Mol of e- = 4.38 mol c)Charge = 423,000 C d)Calculate time

Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Mol of e- = 4.38 mol c)Charge = 423,000 C d)Calculate time About 78 hours

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SPRING REVIEW PART TWO. PRECIPITATION REACTIONS Chapter 19 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to.

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