Properties of Acids

Taste sour

Contain H + ion
 pH less than 7

React with bases to form a salt and water

React with some metals to produce hydrogen gas

Properties of Acids

Turn litmus paper red

Phenolphthalein is colorless in the presence of an acid

Bromothymol blue is yellow in the presence of an acid

Found in citrus fruits in the form of citric acid

Properties of Acids

Found in soured milk and in sore muscles in the form of lactic acid

Found in vitamin C in the form of ascorbic acid

Found in carbonated beverages in the form of carbonic acid…that’s also what you exhale

Properties of Bases

Taste bitter

Contain OH ion
 pH greater than 7

React with acids to form a salt and water

React with organic material

Properties of Bases

Feel slippery because they immediately begin to dissolve the outer layer of skin tissue

Turn litmus paper blue

Phenolphthalein is fuchsia in the presence of a base

Bromothymol blue is blue in the presence of a base

Properties of Bases

Found in drain cleaners usually in the form of sodium hydroxide

Found in ammonia-based cleaners like Windex

Lye (NaOH) is used to make soaps

Classifying Acids and Bases

Svante Arrhenius—Swedish guy who put forth his definitions of acids and bases in 1884 at the age of 25
 Worked with our buddy van’t Hoff
 Received 1903 Nobel Chemistry Prize for electrolytic dissociation discoveries

Arrhenius Acids
 are substances that will dissociate in water to yield hydrogen ion (H + )

Arrhenius Bases
 are substances that will dissociate in water to yield hydroxide ion
(OH )

Classifying Acids and Bases

Johannes Brønsted (Danish) and
Thomas Lowry (English) came up with a new way to classify acids and bases and their conjugates
(pairs that have features in
common but are opposites) in the
1920’s
 never received a Nobel for furthering these acid/base concepts, and
Arrhenius never accepted them!

Brønsted-Lowry Acid

A reactant that donates a proton in a chemical reaction
 The proton is actually the hydrogen ion…since a hydrogen atom has 1 proton and 1 electron and the ion with a 1+ charge indicates that it has lost an electron

Brønsted-Lowry Base

A reactant that accepts a proton in a chemical reaction


A product that
 is formed when a base accepts a proton
 in the reverse reaction, will donate a proton


A product that
 is formed when an acid donates a proton (what’s left after the donation occurs)
 in the reverse reaction, will accept a proton

Classifying Acids and Bases

Around the same time that
Brønsted and Lowry were devising their acid/base scheme, our buddy Gilbert Lewis (yep, the same guy who did the dots) came up with yet another method of classifying them…it’s a broader method than Arrhenius, Brønsted, or Lowry ever postulated

Lewis Acid

A reactant that accepts an electron pair

Lewis Base

A reactant that donates an electron pair

Example #1
HCl (aq) + H
2
O (l)  H
3
O + (aq) + Cl (aq)
Or
HCl (aq)  H + (aq) + Cl (aq)

Example #1

HCl is an acid
 It dissociates to yield H
3
O +
(hydronium ion), which is really water with an extra H + .
(Arrhenius)
 It donates a proton (H + ) to water in the first reaction written.
(Brønsted-Lowry)

Example #1

H
2
O is an base
 It accepts a proton (H + ) from HCl in the first reaction written.
(Brønsted-Lowry)

Example #1

H
3
O + is a conjugate acid
 It is produced when the water accepts a proton (H + ) from HCl in the first reaction written.
(Brønsted-Lowry)
 In the reverse reaction, it will donate a proton (H + ) to Cl in the first reaction written. (Brønsted-
Lowry)

Example #1

Cl is a conjugate base
 It is produced when the HCl donates a proton (H + ) to water in the first reaction written.
(Brønsted-Lowry)
 In the reverse reaction, it will accept a proton (H + ) from H
3
O + in the first reaction written.
(Brønsted-Lowry)

Example #2
NH
3
(g) + H
2
O (l)  NH
4
+ (aq) + OH (aq)

Example #2

NH
3
 is a base
It accepts a proton (H + ) from water.
(Brønsted-Lowry)

Example #2

H
2
O is an acid
 It donates a proton (H + ) to ammonia. (Brønsted-Lowry)

Example #2

NH
4
+
 is a conjugate acid
It is produced when the ammonia accepts a proton (H + ) from water.
(Brønsted-Lowry)
 In the reverse reaction, it will donate a proton (H + ) to OH .
(Brønsted-Lowry)

Example #2

OH is a conjugate base
 It is produced when the water donates a proton (H + ) to ammonia.
(Brønsted-Lowry)
 In the reverse reaction, it will accept a proton (H + ) from NH
4
+ .
(Brønsted-Lowry)

Water —our special friend

Did you notice that it behaved as a base in the first example and as an acid in the second example?
 A substance that can behave as either an acid or a base is called, amphoteric or amphiprotic .

Example #3
H
H—N—H + H +  H—N—H
H H

Example #3

NH
3
 is a base
It donates a pair of electrons to H + .
(Lewis)

Example #3

H + is an acid
 It accepts a pair of electrons from
NH
3
. (Lewis)

Autoionization of Water

Every acid and base will dissociate in water…even water (since it’s amphoteric)!
2H
2
O (l)  H
3
O + (aq) + OH (aq) or
H
2
O (l)  H + (aq) + OH (aq)

Autoionization of Water

Usually, the 2 nd reaction is the one we will use since H
3
O + is just water with an extra H + .

Write the K expression for the 2 nd reaction, keeping in mind that we only include gaseous and aqueous phases.

Autoionization of Water

K = [H + ][OH ]
 note that water is not included because it is a liquid
 This expression is known as the K w
, or equilibrium constant for water, expression
 K w
= [H + ][OH ]

Autoionization of Water

The value of K w
25 ° C.
is 1 x 10 -14 M 2 at

This is a small K value.
 If the temperature changes, so does the value of K w

Autoionization of Water

Make an equilibrium chart for the dissociation, or autoionization, of water.

Autoionization of Water
Initial
Change
Eq.
[H
2
O] [H + ] [OH ]
-0 0

Autoionization of Water
Initial
Change
Eq.
[H
2
O] [H + ] [OH ]
-0 0
-+x +x

Autoionization of Water
Initial
Change
Eq.
[H
2
O] [H + ] [OH ]
-0 0
-+x +x
-x x

Autoionization of Water

Determine the equlibrium concentrations of both the hydrogen ion and the hydroxide ion by plugging into the K w expression.
1 x 10 -14 M 2 = [x][x]

Autoionization of Water
1 x 10 -14 M 2 = x 2
1 x 10 -7 M = x
[H + ] = 1 x 10 -7 M
[OH ] = 1 x 10 -7 M

Autoionization of Water

Because the H + and OH concentrations are equal, the solution is neutral.

If [H + ] > [OH ], then the solution is an acid.

If [H + ] < [OH ], then the solution is a base.

Autoionization of Water

Since every acid or base dissociation we will entertain occurs in water, then the K w expression is applicable to any of these dissociations.

Autoionization of Water

Thus, if you know the [H + ] concentration of a solution, you can determine the [OH ] concentration.

And, if you know the [OH ] concentration of a solution, you can determine the [H + ] concentration.

pH

Represents the “power of hydrogen”

Calculated by taking the opposite of the logarithm of the hydrogen ion concentration… pH = -log [H + ]

pH

Calculate the pH of water at
25 ° C knowing that the [H + ] is 1 x
10 -7 M.
pH = -log [1 x 10 -7 ] pH = 7

pH

If you already know the pH of a solution, then you can find the
[H + ] using:
[H + ] = 10 -pH
So,
[H + ] = 10 -7
[H + ] = 1 x 10 -7 M

pOH

Represents the “power of hydroxide”

Calculated by taking the opposite of the logarithm of the hydroxide ion concentration… pOH = -log [OH ]

pOH

Calculate the pOH of water at
25 ° C knowing that the [OH ] is
1 x 10 -7 M.
pOH = -log [1 x 10 -7 ] pOH = 7

Ooh, ah!

So, the sum of pH and pOH for all aqueous solutions will be 14.
pH + pOH = 14


Find the pH, pOH, and [OH ] and state whether the solution is acidic, neutral or basic if the hydrogen ion concentration is
3.48 x 10 -4 M.
pH = 3.46
pOH = 10.54
[OH ] = 2.87 x 10 -11 M acidic


Find the pOH, [H + ], and [OH ] and state whether the solution is acidic, neutral or basic if the pH is
9.84.
pOH = 4.16
[H + ] = 1.45 x 10 -10 M
[OH ] = 6.91 x 10 -5 M basic


Find the pH, [H + ], and [OH ] and state whether the solution is acidic, neutral or basic if the pOH is 12.7.
pH = 1.30
[H + ] = 5.01 x 10 -2 M
[OH ] = 1.99 x 10 -13 M acidic


Find the pH, pOH, and [H + ], and state whether the solution is acidic, neutral or basic if the
[OH ] is 5.26 x 10 -2 M.
pH = 12.7
pOH = 1.28
[H + ] = 1.90 x 10 -13 M basic