Catalyst

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Catalyst 1. Calculate the K sp for AgC 2 H 3 O 2 whose mass solubility is 1.02 g/100 mL 2. A solution containing 0.010 M Ba 2+ and 0.10 M Ag + , which solid will precipitate first when Na 2 SO 4 is added to the solution? Justify your reasoning with calculations. NOTE the K sp (BaSO 4 ) = 1.1 x 10 -10 and K sp (AgSO 4 ) = 1.1 x 10 -5 . En d

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Catalyst. 1. Calculate the K sp  for AgC 2 H 3 O 2  whose mass solubility is 1.02 g/100 mL  - PowerPoint PPT Presentation

Transcript of Catalyst

Page 1: Catalyst

Catalyst1. Calculate the Ksp for AgC2H3O2 whose

mass solubility is 1.02 g/100 mL  2. A solution containing 0.010 M Ba2+ and

0.10 M Ag+, which solid will precipitate first when Na2SO4 is added to the solution? Justify your reasoning with calculations. NOTE the Ksp (BaSO4) = 1.1 x 10-10 andKsp (AgSO4) = 1.1 x 10-5.

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Cobalt Complexes

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Lecture 8.3 – Complex Ion Formation, Equilibrium Constants, and ΔG

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Today’s Learning Targets• LT 8.6 – I can discuss how coordination

complexes form between metal ions and ligands. Furthermore, I can determine the coordination number for a coordination complex

• LT 8.7 – I can calculate the formation constant for complex ions and relate that to the Ksp for a slightly soluble compound.

• LT 8.8 – I can calculate the free energy of a chemical reaction by utilizing my knowledge the value of K.

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Transition Metal Complexes

•When transition metals bond they are the central atom and they bond extensively with numerous substances.

•The complexes that form are known as metal complexes

•The molecule that bond with metals are known as ligands▫A ligand is always a Lewis Base

•The number of ligands bound to the metal is the coordination number

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Common Ligands

•The most common ligands are H2O, NH3, Cl-, and CN- due to their free lone pair of electrons that can be donated to a metal.

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The Metal Ligand Bond

•The bond between a metal and a ligand is a Lewis acid-base interaction

•Metal = electron pair acceptor (acid)•Ligand = electron pair donor (base)

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Formation of Complex Ions•The formation of complex ions can

drastically impact the solubility of a metal.

•Ligands act to pull metal ions into solution and displace solvating H2O, thus, making the solution more soluble

AgCl (s) ← → ⏐ Ag+ (aq) + Cl- (aq)

Ag+(aq) + 2NH3+ (aq) ← → ⏐ Ag(NH 3)2

+ (aq)

Overall : AgCl (s) + 2NH3+ (aq) ← → ⏐ Ag(NH 3)2

+ +Cl−(aq)

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Complex Ions

•The previously mentioned coordination complexes or metal complexes are referred to here as complex ions.

•For the equilibrium

•The formation constant, Kf, is

Ag+(aq) + 2NH3(aq)← → ⏐ Ag(NH3)2+(aq)

K f =[Ag(NH3)2

+]

[Ag+][NH3]

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Class Example•Calculate the concentration of Ag+

present in solution at equilibrium when concentrated ammonia is added to a 0.010 M solution of AgNO3 to give an equilibrium concentration of [NH3] = 0.20 M. Neglect the small volume change that occurs when NH3 is added

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Table Talk•Calculate [Cr3+] in equilibrium with

Cr(OH)4- when 0.010 mol of Cr(NO3)3 is

dissolved in 1 L of solution buffered at pH = 10.0. The Kf for Cr(OH)4

- is 8 x 1029

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Class Example•The Ksp for AgI is 1 x 10-16 and Kf for Ag(CN)2

- is 1 x 1021. Using these values, (a) calculate the molar solubility of AgI in pure water and (b) calculate the equilibrium constant for the reaction:

•and, (c) determine the molar solubility of AgI in a 0.100 M NaCN solution.

AgI(s) + 2CN −(aq)← → ⏐ Ag(CN)2−(aq) + I−(aq)

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Table Talk•The Kf of Ag(NH3)2 is 1.7 x 107 and the Ksp of

AgCl (s) is 1.6 x 10-10. Calculate (a) the molar

solubility of AgCl in pure water, (b) the equilibrium constant for the reaction:

•and, (c) the solubility of AgCl (s) in 0.50 M NH3

AgCl(s) + 2NH3+(aq)← → ⏐ Ag(NH3)2

+(aq) +Cl−(aq)

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Collaborative Poster•Solve the two problems

that are on the handout that you and your group received.

•Explain all steps that you do so that someone who was not here could learn how to do these problems just from your poster.

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Free Energy

•Recall that we calculated standard free energies (ΔGo), which were when all concentrations were 1 M and temperature was 298 K.

•Most reactions occur at non-standard condition and we can calculate ΔG at this point with:

ΔG = ΔGo + RT lnQ

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Free Energy and K

•At equilibrium ΔG=0 and Q = K. Therefore:

ΔG = ΔGo + RT lnQ

0 = ΔGo + RT lnK

−ΔGo = RT lnK

ΔGo = −RT lnK

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Class Example•The equilibrium for the Haber process at 25 oC

is

•What is the Kp for this reaction?

N2(g) + 3H2(g)← → ⏐ 2NH3(g)

ΔHfo

(kJ/molSo (J/ (mol x K)

N2 0 191.50

H2 0 130.58

NH3

-80.29 192.5

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Table Talk•Calculate the value of the equilibrium constant

for the following reaction:

ΔHfo

(kJ/molSo (J/ (mol x K)

H2 0 130.58

Br2 0 174.9

HBr

-36.23 198.48

H2(g) + Br2(l)← → ⏐ 2HBr(g)

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Relay Races1.If K is calculated to be a value of 0.5 for a reaction at 25 oC,

then is the reaction spontaneous?2. If ΔG = 52 kJ/mol at 52 oC, then what is the value of K?3. If K is determined to be 523 at 30 oC, then is the reaction

spontaneous at these conditions?4. Use the following standard free energy of formation for

formic acid (HCO2). Calculate the Ka for the reaction:

5. The Kf of Ag(NH3)2 is 1.7 x 107 and the Ksp of AgCl (s) is 1.6 x 10-10

. Calculate the equilibrium constant and ΔG for the reaction:

ΔGfo

(kJ/mol

HCO2 -372.3

H+ 0

HCO2-

-351.0

AgCl(s) + 2NH3+(aq)← → ⏐ Ag(NH3)2

+(aq) +Cl−(aq)

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•Read 17.6, 19.7, and 20.2•Homework: 17.60 17.61, 17.63, 17.64,

17.65, 19.79, 19.80, 19.81, 19.83

Closing Time