Acid and Base Equilibrium

Chapter 16 Brown LeMay

Basic Concepts

Acids – sour or tart taste, electrolytes, described by Arrhenius as sub that inc. the H+ conc.

Bases – bitter to the taste, slippery, electrolytes, describes by Arrhenius as sub that inc. the OH- conc.

Dissociation of Water Pure water exists almost entirely of water

molecules. It is essentially a non-electrolyte.

Water ionizes to a small extent – auto-ionization

The equilibrium expression is

H20(l) <-> H+(aq) + OH-

(aq)

Kw = [H+] [OH-]

Since the H+ ion does not exist alone in water Kw is often expressed

Kw = [H3O+] [OH-]

because water conc is constant it does not appear in the expression

The proton in water

Values for Kw and [H+] and [OH-]

Kw = [H3O+] [OH-] = 1.0 x 10-14

1.0 x 10-14 = [X] [X]

1.0 x 10-14 = X2

X = 1.0 X 10-7 = [H3O+] = [OH-]

The Bronsted – Lowry definition holds true for

situations not involving water

Acids donate protons Bases accept protons

HCl(g) + NH3(g) NH4+

(g) + Cl-(aq)\

donates H+ accepts H+ conj acid conj base

BL-Acid BL-Base

notice that the reaction doesn’t happen in water and that the OH- concentration has not increased

Conjugate Acids & Bases and Amphoteric Substances

HNO2(aq) + H2O(l) NO2-(aq)+H3O+

(aq)

conjugateacid base base aciddonates - accepts protons note HNO2 is considered a Arrenius acid H+ inc in

H20 also a BL because donates a Proton all Arrenius acids and bases are also BL acids and

bases – however all BL acids and bases may or may not be arrenius acids or bases water is not an Arrenius base in this example

every acid has a conjugate base formed from the removal of a proton from that acid

every base has a conjugate acid formed from the addition of a proton to that base

Amphotheric Substances

NH3(aq)+H20(l) NH4+

(aq) + OH-(aq)

Base Acid Conj Acid Conj Base

p-acc p-donar

note water is acting as an acid in this reaction and a base in the previous one that makes it a amphortic substance

Strengths of Acid, Bases

The stronger the acid is the weaker its conjugate base (weaker acid stronger conj base)

The stronger the base the weaker its conjugate acid (weaker base stronger conj acid)

Stronger acids and bases ionize to a greater extent than do weak acids and bases.

Strong Acids – dissociate completely into ions HNO3(aq) H+(aq) + NO3

-(aq) the production of H+ ions from the acid

dominates – ignore the H+ donated from the water it is insignificant

The equilibrium lies so far to the right because HNO3 doesn’t reform

The neg log of the H+ from the acid determines pH.

Strong bases also dissociate completely and the conc of OH- from the base is the only factor considered when calculating pH.

Acids and Bases

Strong Acids

HClO4 prechloric

H2SO4 sulfuric

HI hydroiodic

HCl hydochloric

HBr hydrobromic

HNO3 nitric

Strong Bases

GI hydroxides ex. NaOH,KOH

G2 Hydroxides Sr(OH)2

GI Oxides ex.

Na2O, K2O

GI,II Amides ex. KNH2,Ca(NH2)2

The pH scale

pH – is defined as the neg log (base-10) of the H+ ion concentration

pH = -log [H+] What is the pH of a neutral solution [H+] = 1.0 x 10^-7pH = -log [1.0 x 10^-7pH = 7

Strong acids and the pH scale

An acidic solution must have a [H+] conc greater than 1.0 X 10-7 ex 1.0 X 10-6

-log [1.0 X 10-6] = pH 6 What is the pH of a basic solution?

A basic solution is one in which the [OH-]

is greater than 1X10-7.

Calc. pH of strong basic solutions Calculate the pH of a sol that has a [OH-]

con. Of 1.0 X10-5

Kw = 1x10-14 = [H+] [OH-]

Kw = 1x10-14 = [H+] [1.0X10-5]

[H+] = 1x10-14 = 1.0 X 10-9

1.0X10-5

pH = -log 1.0 x 10-9 pH = 9

The “p” Scale

The negative log of a quantity is labeled p (quantity)

Ex: we could reference the quantity of OH- directly: pOH = -log[OH-] From the definition of Kw -log Kw =(-log [H+]) (-log [OH-])= -log 1x10-4

Kw = pH + pOH = 14

Calc. pH using the p scale

Ex. OH- conc = 1.0X10-5

-log 1.0X10-5 = 5 = pOH pH + pOH = 14 pH + 5 = 14 pH = 9

Weak Acids partially ionize in aqueous solution mixture of ions and un-ionized acid in sol. WA are in equilibrium (H20 is left out

because it a pure liquid)

HA(aq) + H20(l) H30+(aq) + A-

(aq)

Ka is the acid dissociation constant Ka = [H30] [A-] = [H+] [A-] [HA] [HA]

Acid Dissociation Constant

The larger the Ka value the stronger the acid is – more product is in solution

Weak Bases

Weak bases in water react to release a hydroxide (OH-) ion and their conjugate acid:

Weak Base(aq) + H2O(l) Conjugate Acid(aq) + OH-(aq)

A common weak base is ammonia NH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)

Since H2O is a pure liquid it is not expressed in the equilibrium Kb expression

Kb = [NH4+][OH-] (base dissociation

[NH3] constant) Kb always refers to the equilibrium in

which a base reacts with H2O to form the conjugate acid and OH-

Lewis Acids and Bases Review - An Arrhenius acid reacts in water

to release a proton - base reacts in water to release a hydroxide ion

In the Bronstead-Lowry description of acids and bases: acid reacts to donate a proton - a base accepts a proton

G.N.Lewis defination - Lewis acid is defined as an electron-

pair acceptor Lewis base is defined as an electron-

pair donor

In the example with ammonia, the ammonia is acting as a Lewis base (donates a pair of electrons), and the proton is a Lewis acid (accepts a pair of electrons)

Lewis is consistent with the description by Arrhenius, and with the definition by Bronstead-Lowry. However, the Lewis description, a base is not restricted in donating its electrons to a proton, it can donate them to any molecule that can accept them.

Calculating the pH of a Weak Acid

What is the pH of an aq sol that is 0.0030M pyruvic acid HC2H3P3? Ka = 1.4x10-4 at 25oC

HC2H3P3 H+ + C2H3P3-

I 0.0030 0 0 C -X X X E 0.0030-X +X +X

Ka = [H+] [C2H3P3-] pluggin in the values

[HC2H3P3] from the table

1.4x10-4 = X2

(0.0030-X)

1.4x10-4 (0.0030) = X2

4.2x10-7-1.4x10-4x = X2

X2 + 1.4x10-4x-4.2x10-7 = 0 a quadratic

ignore the neg sol x = 5.82 x10-4

pH = -log 5.82 x10-4 pH = 3.24

Learning Check What is the pH at 25oC of a solution made

by dissolving a 5.00 grain tablet of aspirin (acetylsalicylic acid) in 0.500 liters of water? The tablet contains 0.325g of the acid HC9H7O4. Ka = 3.3x10-4 mm = 180.2g/l

H+ = 9.4x10-4 pH = 3.03

Buffers

A solution that resists changes in pH when a limited amount of an acid or base is added to it.

Buffers contain either a weak acid and it’s conj. base or a weak base and it’s conj. acid.

Examples Ex. Weak acid and conj. base equal molar amounts Strong Acid added H+ + A- HA conj. base weak acid the conj base interacts with the H+ ions from the strong acid changing them to a weak acid

Strong base added OH + HA HOH + A-

weak acid conj base the weak acid interacts with the OH- ion

from the base to form water and the conj. base

If the concentration of A- and HA are large and the amount of H+ or OH- is small the

solution will be buffered or the change in pH will be minimized.

Buffering capacity – the amount of acid or base a buffer can react with before a significant change in pH occurs

Ratio of acid to conj base – unless the ratio is close to 1 ( between 1:10 and 10:1) will be too low to be useful.

Calculating the pH of a buffer

Note: a solution of 0.10 M acedic acid and its conj base 0.20 M acetate from sodium acetate is a buffer solution pH = 5.07

Ex. Calc. the pH of a buffer by mixing 60.0 ml of 0.100 M NH3 with 40.0ml of 0.100 M NH4Cl.

1St cal the conc. of each species

M = moles/liters

mol of NH3 0.10M = X/0.060 l = 0.0060mol

mol of NH4 0.10M = X/0.040 l = 0.0040mol

[NH3] = 0.0060 mol/ 0.100 l = 0.060 M

[NH4] = 0.0040 mol/ 0.100 l = 0.040 M

NH3 + H2O NH4+ + OH-

I 0.060M O.040 0

C -X +X +X

E 0.060-x 0.040+x X

Kb = [NH4+] [OH-] ( 0.040+X)X

1.8X10-5 [NH3] (0.060-X)

Ignore X 1.8X10-5 = 0.040X X = 2.7X10-5

0.060

-log (2.7x10-5) = 4.6 pOH pH = 9.4

Or using Henderson - Hasselbalch equation

pOH = pKb + log [conj acid] = 4.74 + log ( [0.04]

[B] [0.06])

= 4.6 pOH pH= 9.4

What is the pH of a buffer prepared by adding 30.0ml of .15M HC2H3O2 to 70ml of .2M NaC2H3O2?[HC2H3O2] = .15M = x/.03 = .0045/.01 = .045

[C2H3O2] = .20M =x/.07 = .0140/.01 = .140

ka = 1.7x10-5HHeq pH = pKa + log [conj base] [acid] pH = 4.77 + log(.140 = 5.3 .045)

Adding an acid or a base to a buffer

Calc the ph of 75ml of the buffer solution of(0.1M HC2H3O2 and 0.2M NaC2H3O2)

to which 9.5 ml of 0.10M HCl has been added. Compare the change to that of adding HCl to pure water.

H+ + C2H3O2 -

HC2H3O2

H+ = 0.10M = n/.0095l = 0.00095 moles

C2H3O2 - = 0.2M = n/.075 = 0.0150 moles

HC2H3O2 = 0.10M = n/.075 = 0.0075 moles

Neutralization Reaction C2H3O2

= C2H3O2 moles – H+ moles

0.0150n - .00095n = 0.014 HC2H3O2

= Orginial Conc. + Conc Contributed by reaction

0.075 moles + 0.00095 = 0.0085mol

[C2H3O2] = 0.014mol/0.085l = 0.16M

[HC2H3O2] = 0.0085/0.085 = 0.10M

pH = 4.76 + log (.16/.10) = 4.96