Aqueous solutions can be classified as acidic, basic, or neutral. This classification scheme is based on the quantities of 2 ions, hydronium ion, H301+ and hydroxide ion, OH1-. Where do these ions come from in solutions of pure water? Water molecules in motion will randomly collide with one another. When this happens occasionally a hydrogen nucleus from one molecule will be transferred from one molecule to the other. This can be illustrated.

Notice the nucleus of one hydrogen atom, a proton, was transferred, but the electron pair was left behind. This produces the H301+ ion (hydronium) and the OH1- ion (hydroxide)

the equilibrium expression is

H2O + H20 H3O1+(aq) + OH1-(aq)

Hydronium ion (H3O1+)

Hydroxide ion (OH1-)

H20 H1+(aq) + OH1-(aq)

Which is usually shortened to:

water ionizing

The equilibrium constant for

H20 H1+(aq) + OH1-(aq)

is

Ke = [H1+][OH1-]

[H2O]

[H2O] in mol/L ismass of 1 L of water is 1000 gn = (1000g)/(18g/mol) =55.6 molL-1

Ke = [H1+][OH1-]

[H2O]

[H2O][H2O]

To simplify this equation

Here is a new constant

[H2O]Ke = Kwthe ion product constant of waterKw = [H1+][OH1-] = 1.0 x 10-14 @ 25oCat higher temperatures Kw increases

Kw = [H1+][OH1-] = 1.0 x 10-14 @ 25oCif [H1+]=[OH1-] then

[H1+]=[OH1-]= 1.0 x 10-7 which is the situation for neutral water. A scale which simplifies this very small number is the pH (potency of H1+) scale. It is based on powers of ten.

Simply take the exponent from 10-7 and then multiply it by -1. This produces the number 7.

In mathematical termspOH = -log[OH1-]so if in an aqueous solution the [OH1-] = 2.4 x 10-4, the pOH is3.62If the pOH = 4.56 then the [OH1-] =2.8 x 10-5

This is calculated using the equation [OH1-] = 10-pOH , similarly [H1+]=10-pH.

In mathematical termspH = -log[H1+]so if in an aqueous solution the [H1+] = 2.4 x 10-8, the pH is7.62Remember the whole number portion of a pH doesn’t count as a significant digit (SD), just like in the number 2.4 x 10-8 the exponent -8 doesn’t count as a SD.

If the [H1+][OH1-] = 1.0 x 10-14

then taking the log of each side pH + pOH = 14so if the pH of a solution is 2.3 the pOH is11.7This means for these 4 values[H1+], [OH1-], pH, pOH if one of them is given the other 3 can be calculated. Here is an example

If the pH of a solution is 1.45 find the [H1+], [OH1-], and the pOH.Remember here are the equations needed.

1. pH = -log[H1+]2. pOH = -log[OH1-]3. [OH1-] = 10-pOH

4. [H1+] = 10-pH.5. [H1+][OH1-] = 1.0 x 10-14

6. pH + pOH = 14

If pH = 1.456. pOH = 14 - pHpOH = 12.553. [OH1-] = 10-12.55

[OH1-] = 2.8 x 10-13M4. [H1+] = 10-1.45

[H1+] = 0.035 M

Take note the equations can be applied in varying orders and there are choices as to exactly which equations are used. These decisions are yours to make.

If the [OH1-] of a solution is 9.4 x 10-9 M find the [H1+], pH, and the pOH.

1. pH = -log[H1+]2. pOH = -log[OH1-]3. [OH1-] = 10-pOH

4. [H1+] = 10-pH.5. [H1+][OH1-] = 1.0 x 10-14

6. pH + pOH = 14

If [OH1-] = 9.4 x 10-9

2. pOH = -log 9.4 x 10-9

pOH = 8.036. pH = 14 - 8.03 = 5.974. [H1+] = 10-5.97 =

1.1 x 10-6 M

If the [H1+] of a solution is 6.2 x 10-2 M find the [OH1-], pH, and the pOH.

1. pH = -log[H1+]2. pOH = -log[OH1-]3. [OH1-] = 10-pOH

4. [H1+] = 10-pH.5. [H1+][OH1-] = 1.0 x 10-14

6. pH + pOH = 14

If [H1+] = 6.2 x 10-2M1. pH = -log 6.2 x 10-2 pH = 1.216. pOH = 14 - 1.21 pOH = 12.79

3.[OH1-] = 10-12.79

[OH1-] = 1.6 x 10-13M

Increasing basicityIncreasing acidity

-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Neutral

Since this is a logarithmic scale pH 9 is 10x’s more basic than pH 8, pH 12 is 1000 x’s more basic than pH 9

How much more acidic is pH 1 than pH 5?10 000 x’sWhat pH is 1000x’s more acidic than pH 2? pH of -1

Conceptual Definitions of Acids and BasesArrhenius’s Concept - acids are substances which react in water and produce hydronium ions.

HCl(g) + H20 -------> H301+(aq) + Cl1-(aq)

Bases are substances which react with water and produce hydroxide ions.

NH3(g) + H20 ------> NH41+(aq)+ OH1-(aq)

This concept has its limitations however. Can’t substances be classified as acids or bases without the involvement of water?

Bronstead’s Definition of Acids and Bases

Acids are substances which donate protons and bases are substances which accept protons. In the examples above HCl(g) is an acid because it donates protons to H2O molecules and NH3 is a base because it accepts protons from H2O molecules.

Strength of Acids and Bases is determined by the degree to which a substance produces ions in solution. A strong acid or base is a substance which completely ionizes. In other words if 100 molecules of a strong acid like HCl are placed in water all 100 of them will react with H2O producing 100 H3O1+ ions and 100 Cl1- ions. Weak acids and bases only partially ionize. Strong Acid - the reaction below goes to completion.HCl(g) + H20 --------> H301+(aq) + Cl1-(aq)

Weak Acid - the reaction occurs to a limited extent. In the example below if 100 acetic acid molecules are placed in water only a few of them will successfully react with water molecules producing hydronium ions. Most CH3COOH molecules remain intact.

CH3COOH + H20 H301+(aq)+ CH3COO1-

(aq)

Strong Acids in order of decreasing strength are HClO4, HI, HBr, H2SO4, HCl, HNO3

A table with the remaining moderate and weak acids can be found on page 803.Acid strength has to do with the ease with which an acid can lose a proton. If the binary acid strengths (HI, HBr, HCl) are compared it can be seen that HI is the strongest acid of this group because its iodide ion is the largest of the group so the force between the hydrogen ion and the iodide ion is the weakest so it loses its proton most easily.

I1-

Cl1-

Br1-

H1+

H1+

H1+

Force is strongest since the ions are closest

Force is weakest since the ions are furthest

Remember the weaker the force the stronger the acid

Strong Bases include hydroxides of group 1A and Ca2+, Ba2+, and Sr2+. A table with the remaining moderate and weak bases can be found on page 803. As with acids the weaker the bonds, the stronger the base since liberation of OH1- ions is easiest when the bonds are weakest.

Polyprotic Acids donate protons in steps. For instance carbonic acid, H2CO3 has two protons to donate and it does this in two steps:step 1H2CO3 + H20 HCO3

1- + H301+

step 2HCO3

1- + H20 CO32- + H301+

note: The arrows are constructed in this manner to show the reverse reaction has a greater tendency than the forward reaction.

Conjugate Acid - Base Pairs - When using the Bronsted concept for acids and bases it is convenient to consider all acid - base reactions as reversible equilibria. For instance when sulfurous acid, H2SO3 reacts with water the following equilibrium is established:

H2SO3 + H2O H301+ + HSO31-

conjugate pair

conjugate pair

acid base acid base

In the forward direction the H2SO3 is the proton donor so it’s the acid and the H2O is the proton acceptor so it’s the base. In the reverse direction the H301+ is the proton donor so it’s the acid and the HSO3

1- is the proton acceptor so it’s a base.

H2SO3 + H2O H301+ + HSO31-

conjugate pair

conjugate pair

acid base acid base

When looking at both forward and reverse reactions it is easy to pick out a pair of molecules which differ by a single proton (H atom without its electron). These pairs are called conjugate acid-base pairs.

acid base acid baseH2SO3 + H2O H301+ + HSO3

1-

conjugate pair

conjugate pair

Identify the acid base conjugate pairs in the equilibrium below:

PO43-

(aq) + H2O(aq) HPO42-

(aq) + OH1-(aq)

acidbase acid base

conjugate pair

conjugate pair

Show how HPO4-2 can act as both an acid and

a base.

Amphoteric (Amphiprotic) Substances can behave as both acids or bases dependent on the circumstances. Water molecules, for instance, can sometimes accept protons and behave as bases or donate protons and behave as acids.

HBr(g) + H2O H301+(aq) + Br1-(aq)

base

NH3(g) + H2O 0H1-(aq) + NH41+(aq)

acid

Any substance which alters the pH of an aqueous solution is regarded as an acid or a base. Acids and Bases are classified into 2 categories, strong and weak.Strong Acids and Bases are completely ionized. This means in solution no molecules are present. All the molecules break apart (dissociate), to form ions. Weak acids and bases only partially dissociate and ionize and so are involved in equilibria.

H1+ Cl1-

H1+

Cl1-

H1+

Cl1-

H1+

Cl1-H1+

Cl1-H1+

Cl1-H1+

Cl1-H1+

Cl1-H1+

Cl1-H1+

Cl1-H1+

Cl1-

H1+F1-

H1+F1-

H1+F1-

H1+F1-

H1+F1-

H1+F1-

H1+F1-

H1+F1-

H1+F1-

H1+F1-

H1+F1-

H1+F1-

H1+F1-

Is this a Strong or Weak Acid?

Is this a Strong or Weak Acid?

Find the pH and pOH of a 0.28 molL-1 solution of HClO4.Since this is a strong acid the ionization is complete. This can be represented by …

HClO4 H1+(aq) + ClO41-(aq)

(molL-1)initial 0.28 0.0 0.0

Before any HClO4 has dissolved

-0.28 0.28 0.28shift

final 0.00 0.28 0.28

HClO4 H1+(aq) + ClO41-(aq)

(molL-1)initial 0.28 0.0 0.0

-0.28 0.28 0.28shift

final 0.00 0.28 0.28

Remember pH = -log[H1+] sopH = - log 0.28pH = 0.55pH + pOH = 14 sopOH = 14 - 0.55pOH = 13.45

If the pOH of a solution of HCl is found to be 12.9 what is the concentration of this solution.Since this is a strong acid the ionization is complete. This can be represented by …

HCl H1+(aq) + Cl1-(aq)

(molL-1)initial 0.08 0.0 0.0

-0.079 0.079 0.079shift

final 0 0.079 0.079In this case the pOH can be used to determine the final [H1+], pH = 14 - 12.9, [H1+] = 10-1.1

Find the pH and pOH of a 0.12 molL-1 solution of Sr(OH)2.Since this is a strong base the ionization is complete. This can be represented by …

Sr(OH)2Sr2+(aq) + 2 OH1-(aq)

(molL-1)initial 0.12 0.0 0.0

-0.12 0.12 0.24shift

final 0.00 0.12 0.24Remember pOH = -log[OH1-] sopOH = - log 0.24 = 0.62pH = 14 - 0.62 = 13.38

Find the pH and pOH of a 1.68 molL-1 solution of HCN (Ka = 6.2 x 10-10.Since this is a weak acid the ionization is incomplete. This can be represented by …

HCN H1+(aq) + CN1-(aq)

(molL-1)initial 1.68 0.0 0.0

Before any HCN has dissociated

- x x xshift

final 1.68 - x x x

To determine the value of x the Ke for this reaction must be known. It is found in tables called Ka (equilibrium constants for acids). For HCN Ka = 6.2 x 10-10.

HCN H1+(aq) + CN1-(aq)

Ka = [H1+][CN1-]

[HCN] =

[x]2

[1.68-x]

Since the amount of acid which ionizes is very small this x is negligible

6.2 x 10-10 = x2

1.68

x = 3.2 x 10-5, pH = 4.49, pOH = 9.51

What is the percentage ionization of the 1.68 mol/L HCN solution?

If an unknown 0.34 mol/L monoprotic acid is 1.3 x 10-5% ionized what is its Ka?

If the pH of a solution of HOCl is found to be 4.9 what is the concentration of this solution. Ka = 3.0 x 10-8

Since this is a weak acid the ionization is incomplete. This can be represented by …

HOCl H1+(aq) + OCl1-(aq)

(molL-1)initial x 0.0 0.0

1.26 x 10-5 1.26 x 10-5shift

final x - 1.26 x 10-5 1.26 x 10-5 1.26 x 10-5

In this case the pH can be used to determine the final [H1+], [H1+] = 10-4.9 = 1.26 x 10-5.

-1.26 x 10-5

Ka = [H1+][OCl1-]

[HOCl] =

[1.26 x 10-5]2

[x - 1.26 x 10-5 ] 3.0 x 10-8 = 1.58 x 10-10

x

HOCl H1+(aq) + OCl1-(aq)

(molL-1)initial x 0.0 0.0

1.26 x 10-5 1.26 x 10-5shift

final x - 1.26 x 10-5 1.26 x 10-5 1.26 x 10-5

-1.26 x 10-5

This amount is negligible

x = 5 x 10-3 molL-1

Strong acids totally ionize because the force of attraction between the two ions is relatively weak and water molecules pulling on them can separate them.

H1+

NO31 NO3

1H1+

This means NO31- has a weak pull on H1+ ions

Conversely weak acids have negative ions with relatively strong attractions for H1+ ions

NO21-H1+

These ions cannot be easily ripped apartWater molecules can’t do it

If a salt like NaNO2 is added to water it is easily pulled apart since sodium ions are quite large compared to H1+ ions making the attraction between Na1+ and NO2

1- fairly weak.

Na1+ NO21- NO2

1-Na1+

NO21

H1+

H1+

O2-NO2

1

H1+

H1+

O2-NO2

1

H1+

H1+

O2-

NO21 H1+

H1+

O2-NO21- H1+

H1+

O2-

All negative ions except those found in strong acids, Cl1-, Br1-, I1-, NO3

1-, ClO4

1-, HSO41-, are capable of

creating OH1- ions, this means they act like bases because they raise pH

The equilibrium for NO21- where it alters pH is

NO21-

(aq) + H20 HNO2 (aq)

+ 0H1-(aq)

Show how K2CO3 alters pH

CO32-

(aq) + H20 HCO31-

(aq) +0H1-

(aq)

Show how NaF alters pH

F1-(aq) + H20 HF(aq)

+0H1-(aq)

Show how NaCl alters pH it doesn’t, Cl1- has too weak a pull on H1+

remember, Cl1- is part of the strong acid HCl

What is the pH of a 0.24 molL-1 solution of potassium sulfite Kb of sulfite ion is 1.5 x 10-7?When potassium sulfite is placed in water it dissociates into ions. This can be shown by:

K2SO3(s) 2K1+(aq) + SO32-(aq)

Since SO32- is not part of a strong acid it is

capable of pulling apart (hydrolyzing) water molecules. This can be shown by:

SO32-(aq) + H20 HSO3

1-(aq) + 0H1-molL-1

initial 0.24 0.00 0.00shift -x +x +x

@E 0.24-x +x +x

SO32-(aq) + H20 HSO3

1-(aq) + 0H1-molL-1

initial 0.24 0.00 0.00shift -x +x +x

@E 0.24-x +x +x

The Kb of SO32- is 1.5 x 10-7. So to find x

1.5 x 10-7 =x2

0.24 -x X is so small compared to 0.24 it can be regarded as negligible

x = 1.9 x 10-4 = [OH1-]

pOH = -log (1.9 x 10-4)pOH = 3.72 pH = 14 - 3.72 = 10.28

Positive ions like NH41+, can donate

H1+ ions to water molecules.This increases the amount of hydronium (H301+) creating an acidic affect. This can be illustrated by the following:

H2O

NH41+

H3O1+

NH3

NH41+ + H2O NH3 + H3O1+

Multivalent metal ions with a large charge density can also donate H1+ . They do this by attracting water molecules.Once the water molecules are attached to these highly charged ions the electron pairs shared by the hydrogen and oxygen atoms are pulled toward the oxygen end of the water molecule making the H1+ ions vulnerable to removal by other colliding water molecules.This can be illustrated by the following:

H

HO

Al3+

H

HO

Al3+

Al3+

H

HO

H

HO

H

HO

H

HO

H

HO

H

HO

O

H

HH

[Al (H2O)6]3+ + H2O

[Al(H2O)5OH]2+ +H3O1+

Notice the electron pairs moving closer to the Al3+ Only small

trivalent ions and Be2+ act in this way

H

HH O

If the pH of a solution of NH4Cl is found to be 5.23 what is the concentration of this solution Ka NH4

1+ = 5.7 x 10-10 ?Since the pH is below 7 this substance is acting like an acid. It is not among the strong acids so it must be a weak acid. Since it has no H1+ to donate it must be acting like an acid by hydrolyzing water. Positive ions can behave this way. When NH4Cl is dissolved it dissociates into ions. This is shown by

NH4Cl(s) NH41+(aq) + Cl1-(aq)

The NH41+ acts like a weak acid so it must ….

(molL-1)initial x 0.0 0.0

- 5.9 x 10-6 5.9 x 10-6 5.9 x 10-6shift

final x - 5.9 x 10-6 5.9 x 10-6 5.9 x 10-6

In this case the pH can be used to determine the final [H1+], pH = 5.23, [H1+] = 10-5.23 = 5.9 x 10-6

NH41+ + H2O NH3 + H3O1+

Ka = [H1+][NH3]

[NH41+]

=[5.9 x 10-6]2

[x - 5.9 x 10-6] 5.7 x 10-10 = 3.47 x 10-11

x

This amount is negligible

x = 0.061 molL-1

There are experimental situations where real life situations are simulated. Living organisms have fluids which are maintained within very narrow pH ranges so these simulated experiments require solutions in which pH changes are minimal despite the introduction of acids and bases.Mixtures which resist changes in pH when acids or bases are added are called buffers. Buffers are equilibrium mixtures which shift when acids or bases are added.

Buffers contain 1. a weak acid and a soluble salt

containing the conjugate base of the weak acid

OR

2. A weak base and a soluble salt containing the conjugate acid of the weak base

If the weak acid is HF then a soluble salt containing its conjugate base partner isNaF.Remember that conjugate acid - base partnerships are different by a single H, bases gain H’s, acids lose H’s.If the weak acid is HC2H3O2 then a soluble salt containing its conjugate base partner isKC2H3O2 .If the weak base is NH3, then a soluble salt containing its conjugate acid partner is NH4Cl.

Let’s now look at how buffers work. If the buffer is HF and NaF the equilibrium mixture is

HF H1+ + F1-

If the buffer is HC2H3O2 and NaC2H3O2 the equilibrium mixture is

HC2H3O2 H1+ + C2H3O21-

If the buffer is H2CO3 and NaHCO3 the equilibrium mixture is

H2CO3 H1+ + HCO31-

H2CO3 H1+ + HCO31-

Why do buffers require a salt? If you can recall the definition of a weak acid they only ionize a small amount, so there is only a small amount of HCO3

1- present at equilibrium. If the equilibrium is going to resist a change in pH when acids or bases are added there must be sufficient quantities of substances on both sides of this equilibrium. When acids are added (a source of H1+, the equilibrium shifts

H2CO3 H1+ + HCO31-

to get rid of it. If a base is added, the OH1- produced reacts with the H1+ forming water. This means the equilibrium shiftsto replace the H1+ removed by the addition of the OH1-.A buffer’s capacity to resist large changes in pH is determined by the quantity of weak acid and conjugate base present. When either of these quantities is used up the solution can no longer act as a buffer.

Buffer Problems1. 1.0 L of a buffer is prepared by mixing acetic

acid and sodium acetate. If both of these substances are 1.0 molL-1 at equilibrium find the pH if the Ka for acetic acid is 1.8 x 10-5?Equilibrium is

HC2H3O2 H1+ + C2H3O21-

@E 1.0 1.0 x

Ke =[H1+] [C2H3O2

1-][HC2H3O2]

1.8 x 10-5 =[x] [1.0]

[1.0]pH = -log 1.8 x 10-5

pH = 4.74

1. If 1.0 mL of 2.0 M HCl is now added to this buffer what is the new pH?(assume no volume change of

1.0L.)

HC2H3O2 H1+ + C2H3O21-

@E 1.0 1.0 1.8 x 10-5

upset 2.0 M x 0.0010L=0.0020

shift +0.0020 -0.0020 -0.0020

@E 1.002 0.998x

Assume all the HCl is consumed by the equilibrium shifting right to left

1. If 5.0 mL of 2.0 M NaOH is now added to this buffer what is the new pH?(assume no volume change.)

HC2H3O2 H1+ + C2H3O21-

@E 1.0 1.0 1.8 x 10-5

upset 2.0 M x 0.0050 L= - 0.010

shift -0.010 +0.0100 +0.010

@E 0.99 1.010x

Assume all the OH1- is consumed by the H1+ equilibrium shifts left to right to replace the H1+

HC2H3O2 H1+ + C2H3O21-

@E 0.99 1.01x

Ke =[H1+] [C2H3O2

1-][HC2H3O2]

1.8 x 10-5 =[x] [1.01]

[0.99]

x =[1.8 x 10-5] [1.01]

[0.99]

pH = -log x = 4.75