aqueous chemistry : acids, bases, and ph

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Aqueous Chemistry: Acids, Bases, and pH

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Aqueous Chemistry : Acids, Bases, and pH. Dissociation of Water. H 3 O + = hydronium ion. OH – = hydroxide ion. Dissociation of Water. Dissociation constant:. Brønsted-Lowry Definition. An Acid can donate protons to another molecule. Brønsted-Lowry Definition. - PowerPoint PPT Presentation

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Page 1: Aqueous Chemistry : Acids, Bases, and pH

Aqueous Chemistry:Acids, Bases, and pH

Page 2: Aqueous Chemistry : Acids, Bases, and pH

Dissociation of WaterH2O H2O+ H3O+ OH–+

H3O+ = hydronium ion OH– = hydroxide ion

Page 3: Aqueous Chemistry : Acids, Bases, and pH

Dissociation of Water

[H3O+][OH–][H2O]

K=

[H3O+][OH–]Kw= = 10–14

[H3O+] = [OH–] = 10–7 M

[H3O+][OH–]Kw=

Dissociation constant:

H2O H2O+ H3O+ OH–+

pH = –log [H3O+]pH = –log [H3O+] = 7

pOH = 14 – pH= 14 – 7= 7

Page 4: Aqueous Chemistry : Acids, Bases, and pH

Brønsted-Lowry Definition

An Acid can donate protons to another molecule

HClH2O + H3O+ Cl–+

Page 5: Aqueous Chemistry : Acids, Bases, and pH

Brønsted-Lowry Definition

A Base can accept protons from another molecule

H3O+ OH–+ H2O H2O+Na+ + Na++

Page 6: Aqueous Chemistry : Acids, Bases, and pH

Strong acids dissociate completely

H+ Cl– H+ Cl–H2O + [H3O+][Cl–][HCl]

K= = 102 M

Example: Hydrochloric acid

pH = –log [H3O+] ≈ –log [HCl]

[H3O+] = [OH–] = 10–7 moles/liter

[H3O+] = 0.1 moles/liter + 10–7 moles/liter ≈ 0.1 moles/liter

In 1 liter of pure water:

Add 0.1 moles of HCl

pH = –log [0.1 moles/liter]= 1 pOH = 14–1 = 13

Page 7: Aqueous Chemistry : Acids, Bases, and pH

Strong acids dissociate completelyExample: Hydrochloric acid

pH = –log [H3O+] ≈ –log [HCl]

[H3O+] = [OH–] = 10–7 moles/liter

In 1 liter of pure water:

Add 10-8 moles of HCl to 1 liter of water

pH = –log [10–8 moles/liter] = 8 NO!!!

[H3O+] = 10–7 M + 10–8 M

pH = –log [1.1 x 10–7 moles/liter] = 6.96

= 1.1 x 10–7 M

Page 8: Aqueous Chemistry : Acids, Bases, and pH

Weak acids partially dissociate

Ac– +H+ Ac–H+ H2O

[H3O+][Ac–][HAc]

K= = 1.75 x 10-5 M

= 4.76

Example: Acetic acid

What is the pH of a 0.1 M acetic acid solution?

pK = –log K

[Ac–] + [HAc] = 0.1 M [HAc] = 0.1 M – [Ac–][H3O+][Ac–]

0.1 M – [Ac–]K=

[H3O+][Ac–]0.1 M

K=

Simplifications:

K=(0.1 M) [H3O+]2 = [H3O+]21.75 x 10-6 M2

[H3O+] = 1.32 x 10–3 M pH = –log [1.32 x 10–3 M] = 2.9

Page 9: Aqueous Chemistry : Acids, Bases, and pH

Bases bind to protonsExample: Hydroxide

H+ H2O +OH– + H2OOH–H+

Add 0.1 moles sodium hydroxide to 1 liter of water

pOH = –log [OH] ≈ –log [NaOH]

pOH = –log [0.1 moles/liter]= 1

pH = 14–1 = 13

Page 10: Aqueous Chemistry : Acids, Bases, and pH

Comparing pKa Values

pKa = 4.5 pKa = 10.5

Overall K for the reaction is:

H

A

B

H

B

A

:+ : +

K =Ka for HAKa for HB

Therefore:

pK =(pKa )for HA– (pKa )for HB

Page 11: Aqueous Chemistry : Acids, Bases, and pH

Summary• Water dissociates into hydronium ions and

hydroxide ions.

• The negative log of the hydronium ion concentration is pH, a measure of acidity.

• The negative log of the hydroxide ion concentration is pOH.

• pOH + pH = 14 for all aqueous solutions.

Page 12: Aqueous Chemistry : Acids, Bases, and pH

Summary• Adding an acid to a solution always lowers

the pH.• Strong acids release nearly all of their

protons to form hydronium ions.• Weak acids release relatively few protons to

water, although addition of a base can remove more protons.

• Adding a base to a solution always raises the pH.