Acid and Base Reactions - Mister Chemistry · PDF fileDefinitions of acids and bases Svant...

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Acid and Base Reactions

Transcript of Acid and Base Reactions - Mister Chemistry · PDF fileDefinitions of acids and bases Svant...

Acid and Base Reactions

Definitions of acids and bases

Svant Arrhenius (Sweden) 1859-1927

Johannes Bronsted (Denmark) 1879-1947

G. N. Lewis (U.S.) 1875-1946

dissolves in water to yield protonsH—X H+

(aq) + X–(aq)

dissolves in water to yield hydroxideions

Y—OH Y+ (aq) + HO–

(aq)

Arrhenius definitions of acids

A base

An acid

and bases

Bronsted Definition

An acid is a proton donor

A base is a proton acceptor

An acid is a proton donor

an example is HCl

a source of hydronium ion (H3O+ )

HC

Proton transfer from HCl to water.

hydronium ion

Cl ::

:HO

H

H

:: O

H

H: H

+Cl :

::

:

A base is a proton acceptor

an example is NaOH

a source of hydroxide ions ( OH- )

HC√

A base is a proton acceptor

Cl ::

:HO

H

::

: O

H

H: : Cl :

::

:

Ammonia is a Bronsted base

OH:

:H

hydroxide ion

:N

H

H

H

O H

:::

HN

H

H

H+

A Weak Base

Acid-Base Neutralization

Acid + Base Salt + Water

Neutralization

Example of Neutralization

Complete ionic equation

HCl(aq) + NaOH(aq) NaCl(aq) + H2O (l )

Na+(aq) + Cl–(aq) + H2O(l)

Na+(aq) + OH–(aq) + H+(aq) + Cl–(aq)

Cancel Spectator ions

Acid + Base Salt + Water

OH–(aq) + H+(aq) H2O(l )

Net ionic equation

Bronsted Definition

An acid is a proton donor

A base is a proton acceptor

Conjugate acid-base pair

Conjugate Acids and Bases

• Reactions between acids and bases always yield their conjugate bases and acids.

Consider the transfer of a proton from HCl to water.

Cl ::

:HO

H

H

:: O

H

H: H

+Cl :

::

:

acidbase Conjugate base

Conjugate acid

Acetic acid

CH3COH

O

+ H2O CH3CO– + H3O+

O

acid base Conjugate base

Conjugate acid

Practice Exercise

: :CN O

H

H

::

acidbase

:CNH O

H

: ::

Conjugate base

Conjugate acid

Identify the conjugate acid-base pairs for the reaction

Equilibrium constant for proton transfer

Cl :

::

HO

H

H

:: O

H

H: H

+Cl :

::

:

[H2O]

[H3O+]

[HCl]

[Cl-] [H3O+]

[HCl]

[Cl-]K =

[H+]

[HCl]

[Cl-]K =

The Acid - Base Properties of Water

Water as an Acid and a Base

acidbase

O

H

H

::

:

O

H

:H O

H

H: H

+O

H

::

:

Conjugate base

Conjugate acid

Ion Product of Water ( Kw )

The equilibrium constant for the ionization of water into H+ and OH- ions

Ion Product of Water ( Kw )

acidbase

O

H

H

::

:

O

H

:H O

H

H: H

+O

H

::

:

Conjugate base

Conjugate acid

= [ H3O+] [ HO- ]

= 1.0 x 10-14

Kw = [ H+] [ HO-]

= 1.0 x 10-14Kw = [ H+] [ HO-]

[ H+] = [ HO-]Neutral solution:

= 1 x 10-7 M

[ H+] > [ HO-]Acidic solution:

[ H+] < [ HO-]Basic solution:

Practice Exercise

Calculate the concentration of HO- ions in a hydrochloric acid solution whose hydrogen ion concentration is 1.3 M.

Kw = 1 x 10-14 = [ H3O+] [ HO-]

1 x 10-14 = ( 1.3 ) [ HO-]

7.7 x 10-15 M = [ HO-]

pH: A Measure of Acidity

The pH Scale

pH = - log10 [ H+]

The pH of a neutral solution is 7 at 25ºC

pH = - log ( 10-6 )

solutions have pH < 7

= 6

= 8

Acidic

solutions have pH > 7

pH = - log ( 10-8 )

Basic

but we must keep in mind that a difference of 1 pH unit is equivalent to a factor of 10 acidity

the pH scale is useful in that it lets us express acidity by numbers that are not exponentials

thus, a solution with a pH of 2 is not twice as acidic as one with a pH of 4 it is 102 (or 100) times as acidic

pH

pH values of common fluids

pH values of common fluids

pH values of common fluids

Practice Exercise

The pH of a certain fruit juice is 3.33 Calculate [ H+] .

pH = 3.33

log [ H+] = -3.33

[ H+] = 10 -3.33

[ H+] = 4.7 x 10 -4M

= - log10 [ H+]

Acidsand

Bases

Other “p” Scales

• The “p” in pH tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions).

• Some similar examples are– pOH: −log [OH−]– pKw: −log Kw

© 2012 Pearson Education, Inc.

Practice Exercise

The OH- concentration of a blood sample is 2.5 x 10-7 M . What is its pH?

pOH = - log10 [ OH-]

pOH = 6.60

= - log [ 2.5 x 10-7 ]

pH = 14 - pOH

pH = 7.40

pH + pOH = 14

The pH Scale

pH = log10 =

[ H3O+]

1= log10

[ H+]

1

pH = - log10 = [ H3O+]= - log10[ H+]