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6/12/2015 One Point Quiz  One quiz per table, list everyone’s name  Agree on an answer  You have two minutes.

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Presentation on theme: "6/12/2015 One Point Quiz  One quiz per table, list everyone’s name  Agree on an answer  You have two minutes."— Presentation transcript:

1 6/12/2015 One Point Quiz  One quiz per table, list everyone’s name  Agree on an answer  You have two minutes

2 Theory of Precipitation Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology

3 Problem Set Hydroxide Precipitation Reactions

4 6/12/2015 Ag + (aq) + OH – (aq) Equilibrium Constants and Solubility Product Ag 2 O(s) What is the solubility product for silver oxide? 22+ H 2 O(l) Silver ion reacts with hydroxide ion to form a brown-black oxide. chocolate brown The solubility product equation corresponds to the dissolution of the solid.

5 6/12/2015 Solubility Product 2 Ag + (aq) + 2 OH – (aq) Ag 2 O(s) + H 2 O(l) K c = products reactants [Ag + ] 2 [OH – ] 2 K c · [Ag 2 O(s)] [H 2 O(l)] = [Ag + ] 2 [OH – ] 2 [Ag 2 O(s)] [H 2 O(l)] solids are included in the constant solvent is included in the constant K solubility product K sp

6 6/12/2015 Why is the numeric value of the solubility product important? A precipitate will form from ions when the ion product exceeds the solubility product. 2 Ag + (aq) + 2 OH – (aq) Ag 2 O(s) + H 2 O(l) K sp = [Ag + ] 2 [OH – ] 2

7 6/12/2015 If a room temperature aqueous solution has a pH of 1.0, what is the concentration of the hydrogen ion and the hydroxide ion? more

8 6/12/2015 At pH = 1.0 pH = -log[H + ] [H + ] = 10 –pH = 1 x 10 –1 M How is the hydroxide ion concentration related to the hydrogen ion concentration?

9 6/12/2015 At pH = 1.0 [OH – ] = KwKw [H + ] = 1 x 10 –14 M 2 1 x 10 –1 M = 1 x 10 –13 M H + (aq) + OH – (aq)H 2 O(l) K c = products reactants [H + ] [OH – ] K c · [H 2 O(l)] = [H + ] [OH – ] [H 2 O(l)] solvent is included in the constant K water = 1 x 10 –14 M 2

10 6/12/2015 Iron(III) Hydroxide Write a chemical reaction for the formation of this solid. Fe(OH) 3 has a solubility product (K sp ) of 6.0 x 10 –38 M 4. Will a 0.10 M Fe 3+ solution precipitate as iron(III) hydroxide at a pH of 1.0? Write the solubility product equation for Fe(OH) 3.

11 6/12/2015 < K sp, Fe(OH)3 Will 0.10 M Iron(III) Ion Precipitate as a Hydroxide at pH = 1.0? Q = [Fe 3+ ] [OH – ] 3 Fe(OH) 3 will not precipitate because its ion product is less than the solubility product. = (0.10 M)(1 x 10 –13 M) 3 = 1 x 10 –40 M 4 = 6.0 x 10 –38 M 4

12 6/12/2015 [H + ] = [OH – ] 10 –pH = 1 x 10 –3 M = 1 x 10 –11 M Will 0.10 M Iron(III) Ion Precipitate as a Hydroxide at pH = 3.0?

13 6/12/2015 Iron(III) Hydroxide at pH = 3.0 Fe(OH) 3 will precipitate because its ion product exceeds the solubility product. > K sp, Fe(OH)3 Q = [Fe 3+ ] [OH – ] 3 = (0.10 M)(1 x 10 –11 M) 3 = 1 x 10 –34 M 4 = 6.0 x 10 –38 M 4

14 6/12/2015 Solubility Products of Hydroxides at 25 °C K sp Cd(OH) 2 Co(OH) 2 Cr(OH) 3 Cu(OH) 2 Fe(OH) 3 Mn(OH) 2 Ni(OH) 2 Pb(OH) 2 2.0 x 10 – 14 2.5 x 10 – 16 6.7 x 10 – 31 1.6 x 10 – 19 6.0 x 10 – 38 2.0 x 10 – 13 1.6 x 10 – 16 4.2 x 10 – 15

15 6/12/2015 Selective Precipitation A solution contains 0.10 M Fe 3+ and 0.10 M Ni 2+ If the solution is slowly made basic, what will be the composition of the first solid that forms?

16 6/12/2015 Selective Precipitation 6 M NaOH, dropwise Fe 3+, Ni 2+ Method: Determine the hydroxide ion concentration at which each ion will begin to precipitate.

17 6/12/2015 At what [OH – ] will 0.1 M Fe 3+ begin to form iron(III) hydroxide? [Fe 3+ ] [OH – ] 3 = K sp [OH – ] = ( K sp [Fe 3+ ] ) 1/3 = ( 6.0 x 10 –38 M 4 1.0 x 10 –1 M ) 1/3 = 8.4 x 10 –13 M Fe(OH) 3 will begin to form when its ion product exceeds its solubility product. 0.1 M Fe 3+ begins to precipitate when OH – exceeds this

18 6/12/2015 At what [OH – ] will 0.1 M Ni 2+ begin to form nickel(II) hydroxide? [Ni 2+ ] [OH – ] 2 = K sp [OH – ] = ( K sp [Ni 2+ ] ) 1/2 = ( 1.6 x 10 –16 M 3 1.0 x 10 –1 M ) 1/2 Ni(OH) 2 will begin to form when its ion product exceeds its solubility product. = 4.0 x 10 –8 M 0.1 M Ni 2+ begins to precipitate when OH – exceeds this

19 6/12/2015 Analysis  Fe(OH) 3 begins to form at an hydroxide ion concentration of 8.4 x 10 –13 M.  Ni(OH) 2 begins to form at an hydroxide ion concentration of 4.0 x 10 –8 M. At what pH will each solid form?

20 6/12/2015 Selective Precipitation 14 10 6 2 pH  0.1 M Fe 3+ forms Fe(OH) 3 (pH = 1.9)  0.1 M Ni 2+ forms Ni(OH) 2 (pH = 6.6) If the solution is slowly made basic, which solid forms first?

21 6/12/2015 Selective Precipitation 6 M NaOH, dropwise Fe 3+, Ni 2+ dropwise, because we want the ions to precipitate selectively What is the formula of the ion which precipitates first? Fe(OH) 3 red-brown Ni 2+ pale-green

22 6/12/2015 99.9% Complete Separation Determine the hydroxide ion concentration and pH at which 99.9% of the iron(III) ion has precipitated. In what pH range will 99.9% of the iron(III) ion precipitate and the nickel(II) ion remain in solution? Determine the concentration of iron(III) ion when 99.9% of it is precipitated.

23 6/12/2015 99.9% Complete Separation [Fe 3+ ] [OH – ] 3 = K sp [Fe 3+ ] = 0.00010 M [OH – ] = ( 6.0 x 10 –38 M 4 1.0 x 10 –4 M ) 1/3 = 8.4 x 10 –12 M 1.2 x 10 –3 M pH = 2.9 What will be the iron(III) ion concentration if it is 99.9% precipitated? It started 0.10 M [H + ] =

24 6/12/2015 Selective Precipitation 0.0001 M Fe 3+ forms  Fe(OH) 3 (pH = 2.9) 14 10 6 2 pH  0.1 M Fe 3+ forms Fe(OH) 3 (pH = 1.9)  0.1 M Ni 2+ forms Ni(OH) 2 (pH = 6.6) If the solution is adjusted to this pH range, more than 99.9% of the iron(III) ion should precipitate and none of the nickel(II) ion

25 6/12/2015 Solubility Rules  All nitrate (NO 3 – ) salts are soluble and highly conducting. Ti 3+ (aq) + 3 NO 3 – (aq) Ti(NO 3 ) 3 (s)  All Group I (Li +, Na +, K +, Rb +, Cs + ) and ammonium salts are soluble and highly conducting. 2 Na + (aq) + SO 4 2– (aq) Na 2 SO 4 (s)

26 6/12/2015 Solubility Rules  All chloride and bromide (Cl –, Br – ) salts are soluble and highly conducting except salts of Ag +, Pb 2+, Hg 2 2+ and Tl +. Cu 2+ (aq) + 2 Cl – (aq) CuCl 2 (s) Ag + (aq) + Cl – (aq) AgCl(s)  All sulfate (SO 4 2 – ) salts are soluble and highly conducting except salts of Pb 2+, Ba 2+, Ca 2+ and Sr 2+. Fe 2+ (aq) + SO 4 2– (aq) FeSO 4 (s) Ba 2+ (aq) + SO 4 2– (aq) BaSO 4 (s)

27 6/12/2015 Solubility Rules  In almost all cases, when a salt dissolves and dissociates in solution, only one type of cation (positive ion) and one type of anion (negative ion) are formed. N 2 H 5 + (aq) + NO 3 – (aq) N 2 H 5 NO 3 (s) hydrazinium ion

28 Sulfides

29 6/12/2015 Soluble Sulfides Na 2 S15.4 (10 °C)1.97 M (NH 4 ) 2 Svery soluble 5460 m H2SH2S18600 (40 °C) 0.34 (100 °C)0.10 M g/100 g H 2 O molar solubility

30 6/12/2015 Insoluble Sulfides Ag 2 S CdS CoS CuS FeS MnS NiS PbS g/100 g H 2 O 6.4 x 10 -16 1.4 x 10 -13 2.0 x 10 -10 2.8 x 10 16 1.9 x 10 -8 6.2 x 10 -7 5.0 x 10 -10 4.9 x 10 -13 molar solubility 2.6 x 10 -17 1.0 x 10 -14 2.2 x 10 -11 2.9 x 10 -18 2.2 x 10 -9 7.1 x 10 -8 5.5 x 10 -11 2.0 x 10 -14 K sp 6.8 x 10 -50 1.0 x 10 -28 5.0 x 10 -22 8.7 x 10 -36 4.9 x 10 -18 5.1 x 10 -15 3.0 x 10 -21 4.2 x 10 -28

31 6/12/2015 Preparation of Hydrogen Sulfide (acidic solution) 2 H + (aq) + S 2– (aq) H 2 S(aq) H H C H H H C N S      + H 2 O(l) H 2 S(aq) thioacetamide As an aqueous solution H 2 S is called hydrosulfuric acid

32 6/12/2015 Preparation of Hydrogen Sulfide (basic solution) H 2 S(aq) + 2 OH – (aq) S 2– (aq) + 2 H 2 O(l) 0.1 M Na 2 S [S 2– ] = 5.0 x 10 –2 M 0.1 M (NH 4 ) 2 S [S 2– ] = 2.0 x 10 –5 M

33 6/12/2015 Reactions of Hydrosulfuric Acid (polyprotic acid) 2 H + (aq) + S 2– (aq) H 2 S(aq) H + (aq) + HS – (aq) H 2 S(aq) H + (aq) + S 2– (aq) HS – (aq) hydrosulfuric acidbisulfide ion sulfide ion

34 6/12/2015 First Dissociation of H 2 S H + (aq) + HS – (aq) H 2 S(aq) [H + ] [HS – ] [H 2 S] = K 1 = 1.0 x 10 –7 M 1 indicates it is the ionization of the first hydrogen ion Also called K a,1

35 6/12/2015 H + (aq) + S 2– (aq) HS – (aq) Second Dissociation of H 2 S [H + ] [S 2– ] [HS – ] = K 2 = 1.3 x 10 –13 M

36 6/12/2015 2 H + (aq) + S 2– (aq) H 2 S(aq) Double Dissociation of H 2 S [H + ] 2 [S 2– ] [H 2 S] = K 12 = K 1  K 2 =1.3 x 10 –20 M 2 This equilibrium constant can be used for any sulfide ion and pH calculation

37 6/12/2015 Dissociation of Hydrosulfuric Acid H + (aq) + HS – (aq) H 2 S(aq)2 H + (aq) + S 2 – (aq) very acidicneutralvery basic The solubility of H 2 S in water decreases with increasing temperature. The amount of sulfide ion in solution is pH dependent.

38 6/12/2015 What would a graph of the various sulfide containing species versus pH look like? pH  Mole Fraction acidicbasic H2SH2S HS – S2–S2– Excel Graph

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42 The organization of the elements by Moseley is accepted as being superior to that of Meyer and Mendeleev u In which way is it the most superior? prediction of missing elements prediction of elemental properties prediction of compounds formed with hydrogen and oxygen

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