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AcidEquilibrium

and pH

Sren Srensen
http://upload.wikimedia.org/wikipedia/commons/3/36/Soeren_Peter_Lauritz_Soerensen_1868-1939_2.jpg

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Acid/Base DefinitionsArrhenius Model

Acids produce hydrogen ions in aqueoussolutionsBases produce hydroxide ions inaqueous solutions

Bronsted-Lowry ModelAcids are proton donorsBases are proton acceptors

Lewis Acid ModelAcids are electron pair acceptorsBases are electron pair donors

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Acid Dissociation

HA H+ + A -Acid Proton Conjugatebase

][

]][[ HA

A H K

a

Alternately, H + may be written in itshydrated form, H 3O+ (hydronium ion)

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Dissociation of Strong AcidsStrong acids are assumed to dissociatecompletely in solution.

Large K a or

smallK

a ?Reactantfavored or

productfavored?

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Dissociation Constants: Strong Acids

Acid Formula Conjugate

Base Ka Perchloric HClO 4 ClO4- Very largeHydriodic HI I - Very large

Hydrobromic HBr Br-

Very largeHydrochloric HCl Cl - Very largeNitric HNO 3 NO3- Very largeSulfuric H 2SO 4 HSO 4- Very large

Hydronium ion H3O+ H2O 1.0

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Dissociation of Weak AcidsWeak acids are assumed to dissociateonly slightly (less than 5%) in solution.

Large K a or

smallK

a ?Reactantfavored or

productfavored?

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Dissociation Constants: Weak AcidsAcid Formula ConjugateBase Ka

Iodic HIO 3 IO 3- 1.7 x 10-1

Oxalic H 2C2O4 HC2O4- 5.9 x 10 -2

Sulfurous H 2SO 3 HSO3- 1.5 x 10-2

Phosphoric H 3PO4 H2PO4- 7.5 x 10 -3

Citric H 3C6H5O7 H2C6H5O7- 7.1 x 10-4Nitrous HNO 2 NO2- 4.6 x 10 -4

Hydrofluoric HF F - 3.5 x 10 -4

Formic HCOOH HCOO- 1.8 x 10-4

Benzoic C6H5COOH C6H5COO- 6.5 x 10 -5Acetic CH 3COOH CH3COO- 1.8 x 10-5

Carbonic H 2CO3 HCO3- 4.3 x 10 -7

Hypochlorous HClO ClO- 3.0 x 10 -8

Hydrocyanic HCN CN - 4.9 x 10 -10

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Self-Ionization of Water

H2O + H 2O H3O+ + OH -

At 25 , [H 3O+] = [OH - ] = 1 x 10 -7

K w is a constant at 25 C:

Kw = [H 3O+][OH- ]

Kw = (1 x 10 -7 )(1 x 10 -7 ) = 1 x 10 -14

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Calculating pH, pOHpH = -log 10(H3O+)

pOH = -log 10(OH- )

Relationship between pH and pOHpH + pOH = 14

Finding [H3O+], [OH - ] from pH, pOH

[H3O+] = 10 -pH

[OH- ] = 10 -pOH

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pH and pOH Calculations

H+ OH -

pH pOH

[OH -] = 1 x 10 -14

[H+]

[H+] = 1 x 10 -14[OH -]

pOH = 14 - pH

pH = 14 - pOH

p O H

= - l o g

[ O H - ]

p H = - l o g [ H

+ ]

[ O H - ] =

1 0 - p

O H

[ H + ] =

1 0 - p H

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ThepH Scale

Graphic: Wikimedia Commons user Slower

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A Weak Acid Equilibrium Problem

What is the pH of a 0.50 M solution ofacetic acid, HC 2H3O2, K a = 1.8 x 10 -5 ?

Step #1: Write the dissociation equation

HC2H3O2 C2H3O2- + H +

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Step #2: ICE it!

HC2H3O2 C2H3O2- + H +ICE

0.50 0 0- x +x +x

0.50 - x xx

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Step #3: Set up the law of mass action

HC2H3O2 C2H3O2- + H +0.50 - x xxE

)50.0()50.0())((108.1

25 x x

x x x

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A Weak Acid Equilibrium ProblemWhat is the pH of a 0.50 M solution ofacetic acid, HC 2H3O2, K a = 1.8 x 10 -5 ?

Step #4: Solve for x, which is also [H +]

HC2H3O2 C2H3O2- + H +0.50 - x xxE

)50.0(108.1

25 x x [H +] = 3.0 x 10 -3 M

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A Weak Acid Equilibrium ProblemWhat is the pH of a 0.50 M solution ofacetic acid, HC 2H3O2, K a = 1.8 x 10 -5 ?

Step #5: Convert [H +] to pH

HC2H3O2 C2H3O2- + H +0.50 - x xxE

52.4)100.3log( 5 x pH

14 acids

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