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Notes on Equilibria, Buffers - General Chemistry II | CHEM 1200, Study notes of Chemistry

Material Type: Notes; Professor: Thomas; Class: General Chemistry II; Subject: Chemistry; University: Auburn University-Montgomery; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 08/16/2009

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Equilibria
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Equilibria
Chapter 17
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EquilibriaEquilibriaEquilibriaEquilibria

C

hapter 17

Buffers Buffers

Buffers Buffers

Solutions that resist pHchange when small amountsof acid and base added Two types weak acid + its saltweak base + its salt

Common ion effect

Buffers Buffers

Buffers Buffers

HA

(aq)

+ H

2

O

(l)

H

3

O

(aq)

+A

(aq)

Add acid

left shift

Buffers Buffers

Buffers Buffers

HA

(aq)

+ H

2

O

(l)

H

3

O

(aq)

+A

(aq)

HHHHC

C

C C

2222

HHHH

3 3

3 3

O O

O O

2222

H

H

H

H

2222

O O

O O

º º

º º

H H

H H

3333

OOOO

  • C

  • C

  • C

  • C

2222

H H

H H

3333

OOOO

2222

Buffer pH depends on ratio ofconjugate acid-base pair

ADDINGADDINGADDINGADDING ACID/BASE

ACID/BASEACID/BASEACID/BASE TO

TOTOTO Buffers

Buffers

Buffers Buffers

BuffersBuffers andBuffersBuffers

and bloodandand

blood

blood blood

Oxygentransportedby hemoglobinCO

2

transportedin plasma andred blood cellsHCO

3

is buffer forcontrolling bloodpH

SolubilitySolubilitySolubilitySolubility product

productproductproduct

K

sp

equilbrium constant for lowsolubility ionic compounds

AgCl

(s)

Ag

(aq)

+ Cl

(aq)

K

sp

=[Ag

][Cl

]

SolubilitySolubilitySolubilitySolubility product

productproductproduct

AgCl

(s)

Ag

(aq)

+ Cl

(aq)

At equilibrium the system is asaturated solution of Ag

& Cl

Low

K

sp

means low solubility

SolubilitySolubilitySolubilitySolubility product

productproductproduct

Solubility

AgCl: 0.00188 g/L

Molar Solubility Ag

2

SO

4

: 0.015 mol/L

Given solubility or K

sp

, find other

Problem 1Problem 1Problem 1Problem 1

Find

K

sp

for Ag

2

CrO

4

Given solubility is 7.8 x 10

mol/L

Ag

2

CrO

4

2 Ag

CrO

4

2-

1 mol

2 mol +

1 mol

7.8 x 10

mol

²²²²

2 x 7.8 x 10

mol + 7.8 x 10

mol

K

sp

= [Ag

]

2

[CrO

4

2-

]

K

sp

= [2 x 7.8 x 10

]

2

[7.8 x 10

]

= 1.9 x 10

Problem 2Problem 2Problem 2Problem 2

Convert from mol/L to g/L

2.1 x 10

x 78 = 1.6 x 10

g/L