Acids & Bases

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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

Brønsted-Lowry Conjugate Acid

A product that

 is formed when a base accepts a proton

 in the reverse reaction, will donate a proton

Brønsted-Lowry Conjugate

Base

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

Example #4 —show all work

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

Example #5 —show all work

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

Example #6 —show all work

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

Example #7 —show all work

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

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