Acids and Bases 2007-20081 Acids and Bases. Acids and Bases 2007-20082 Acids Svante Arrhenius, a...

Acids and Bases 2007-2008 1

Acids and Bases


Acids

• Svante Arrhenius, a Swedish chemist, defines an acid as a substance that yields hydrogen ions (H+) when dissolved in water.

• Formulas for acids contain one or

more hydrogen atoms as well as an anion.


In some cases two different names seem to be assigned to the same chemical formula.

HCl(g) hydrogen chlorideHCl(l) hydrogen chlorideHCl(aq) hydrochloric acid

The name assigned to the compound depends on its physical state. In the gaseous or pure liquid state, HCl is a molecular compound called hydrogen chloride. When it is dissolved in water, the molecules break apart into H+ and Cl- ions; in this state, the substance is called hydrochloric acid.

Naming Acids


Binary acids (formed by hydrogen and one other element) are named with a “hydro-” prefix and an “-ic” ending on the anion root.

Ex- HClHBr

Hydrochloric acidHydrobromic acid


The formulas for oxoacids, (acids that contain hydrogen and an anion containing oxygen) are usually written with the H first, followed by the anion, as illustrated in the following examples:

H2CO3 HNO3 HClO2

Carbonic acid

Nitric acid Chlorous acid

If the anion ends in “-ate” then the acid ends in “-ic”, if the anion ends in “-ite”, then the acid ends in “-ous”. Remember, “ic” goes with the higher oxidation state, N has an oxidation state of ___ in HNO3 (nitric acid) and ___ in HNO2 (nitrous acid)

+5+

3


• Acids have a sour taste; for example, vinegar owes its sourness to acetic acid, and lemons and citrus fruits contain citric acid.

• Acids cause color changes in plant dyes; for example, they change the color of blue litmus red.

• Acids react with certain metals to produce hydrogen gas.

• Acids react with carbonates and bicarbonates to produce carbon dioxide gas.

• Aqueous acid solutions conduct electricity.

Acids


Brønsted Acid• Arrhenius’s definitions of acids are

limited in that they apply only to aqueous solutions. Broader definitions were proposed by the Danish chemist Johannes Brønsted. A Brønsted acid is a proton donor.

HCl(aq) H+(aq) + Cl-

Remember, the H+ ion is really just a proton (a hydrogen atom is one proton and one electron, you pull off the electron and all you are left with is…


The size of a proton is about 10-15 m, compared to the diameter of 10-10 m for an average atom or ion. Such an exceedingly small charged particle cannot exist as a separate entity in aqueous solution owing to its strong attraction for the negative region of the polar water molecule. Consequently, the proton exists in a hydrated form as H3O+, and is referred to as the hydronium ion

H+ + H2O H3O+


Since the acidic properties of the proton are unaffected by hydration, we will generally use H+(aq) to represent the hydrated proton. This notation is for convenience only, because H3O+ is closer to reality. Keep in mind that both notations represent the same species in aqueous solution.

H+(aq) = H3O+

(aq)


Monoprotic acidseach unit of acid yields one hydrogen upon ionization

Diprotic acidseach unit of an acid gives up two H+ ions

Triprotic acidsyields three H+ ions upon ionization

HCl

H2CO3

H3PO4


Diprotic acids give up their two H+ ions in separate steps:

H2SO4(aq) H+(aq) + HSO4-(aq)

HSO4-(aq) H+(aq) + SO4

-2(aq)

Triprotic acids give up their H+ ions in three separate steps.

Is HSO4- a strong or weak acid?

ExplainWeak, only partially ionizes


• In another definition formulated by Svante Arrhenius, a base can be described as a substance that yields hydroxide ions (OH-) when dissolved in water. Some examples are

NaOHKOHBa(OH)2

Bases

Sodium hydroxidePotassium hydroxideBarium hydroxide


Ammonia (NH3) is also classified as a common base. At first glance this may seem to be an exception to the definition of a base. Note that as long as a substance yields hydroxide ions when dissolved in water, it need not contain hydroxide ions in its structure to be considered a base. In fact, when ammonia dissolves in water, the following reaction occurs:

NH3 + H2O NH4+ + OH-

Thus it is properly classified as a base.


Bases• Bases have a bitter taste• Bases feel slippery; for example,

soaps, which contain bases, exhibit this property

• Bases cause color changes in plant dyes; for example, they change the color of red litmus blue

• Aqueous base solutions conduct electricity


Brønsted Base

• A Brønsted base is defined by Johannes Brønsted as being a substance capable of accepting a proton.


Lewis Acids and Bases

So far we have discussed acid-base properties in terms of the Brønsted theory.

G.N. Lewis formulated a definition for what is now called a Lewis base – a substance that can donate a pair or electrons. A Lewis acid is a substance that can accept a pair of electrons.


The significance of the Lewis concept is that it is much more general than other definitions. For example, the reaction between boron trifluoride and ammonia is a Lewis acid-base reaction.


Strength of Acids and Bases

Strong acids are strong electrolytes, which, for practical purposes, are assumed to ionize completely in water. That means that at equilibrium, solutions of strong acids will not contain any nonionized acid molecules.

Like strong acids, strong bases are all strong electrolytes that ionize completely in water.


The most common strong acids are HClO4, HCl, HNO3 and H2SO4. Hydroxides of alkali metals and alkaline Earth metals are strong bases (like NaOH, KOH and Ba(OH)2).

Other strong acids and strong bases are listed on your Relative Strengths of Acids and Bases Reference Sheet.


The strength of an acid is measured by its tendency to ionize:

HX H+ + X-

The strength of the H-X bond influences the extent to which an acid undergoes ionization. The stronger the bond (the higher the bond dissociation energy in kJ/mol), the more difficult it is for the HX molecule to break up and hence the weaker the acid.


Bond Bond Dissociation

Energy (kJ/mol)

Acid Strength

H-F 568.2

H-Cl 431.9

H-Br 366.1

H-I 298.3

Bond Dissociation Energies for Hydrogen Halides and Acid

Strengths

strong

strong

strong

weak


The Strength of Oxoacids

Oxoacids contain hydrogen, oxygen, and one other element Z, which occupies a central position. To compare oxoacid strength, it is convenient to separate the oxoacids into two groups.


Oxoacids having different central atoms that are from the same group of the periodic table and that have the same oxidation number. Within this group, acid strength increases with increasing electronegativity of the central atom.

HClO3 > HBrO3

The Cl pulls more strongly on the electron pair shared with the O, making

the O-H bond more polar, therefore making it easier to ionize (and the acid

stronger)


Oxoacids having the same central atom but different numbers of attached groups. Within this group, acid strength increases as the oxidation number of the central atom increases.

HClO4 > HClO3 > HClO2 > HClOThe greater the number of O atoms pulling

on the Cl, the more that the electrons are pulled away from the O-H bond, making the O-H bond more polar, therefore making it

easier to ionize (and the acid stronger)


You are going to remember these trends because…

•HCl is a strong acid and HF (with the higher bond dissociation energy) isn’t.

•HClO4 is a strong acid and HClO3 (where the Cl has a lower oxidation number because it has fewer oxygen atoms attached to it) isn’t.


Note: H3O+ is the strongest acid that can exist in aqueous solutions. Acids stronger than H3O+ react with water to produce H3O+ and their conjugate bases.

Thus, HCl, which is a stronger acid than H3O+, reacts with water completely to form H3O+ and Cl-.

HCl(aq) + H2O(l) H3O+(aq) + Cl-

(aq)


The OH- ion is the strongest base that can exist in aqueous solution. Bases stronger than OH- react with water to produce OH- and their conjugate acids.

For example, the oxide ion, (O-2) is a stronger base than OH-, so it reacts with water completely as follows:

O-2(aq) + H2O(l) 2OH-(aq)

For this reason, the oxide ion does not exist in aqueous solutions.


Amphoteric Compounds

NH3(g) + H2O(l) OH-(aq) + NH4+(aq)

H2SO4(aq) + H2O(l) H3O+(aq) + HSO4-(aq)

As you could see from the previous two examples, water will act as either an acid or a base, depending on the strength of the acid or base with which it is reacting. Any species that can react as either an acid or a base is described as amphoteric.

Proton acceptor (base)

Proton donor (acid)

29

Conjugate baseRemains when one proton has been removed from the acid

Conjugate acidResults from the addition of a proton to a Bronsted base

BaseProton (H+) acceptor

AcidProton (H+)

donor

An extension of the Brønsted definition of acids and bases is the concept of the conjugate acid-base pairCH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+

(aq)

A conjugate acid-base pair is defined as an acid and its conjugate base (what’s left after the H+ was removed from the acid) or a base and its conjugate acid (substance formed by the addition of the H+

to the base).** Because the acid and base are always stronger than the conjugate acid and conjugate base, the direction of the reaction proceeds from acid/base conjugate acid/conjugate base.


Identify the acid, base, conjugate acid and conjugate base in the following reaction (**Reaction proceeds from stronger to weaker…)

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

Base

Proton (H+)

acceptor(stronger base)

Acid

Proton donor

(Stronger acid)

Conjugate

Acid

Accepted proton (H+)

(weaker acid)

Conjugate

Base

What’s left after H+ was donated by

acid(weaker base)


The Acid-Base Properties of Water

Water is a very weak electrolyte and therefore a poor conductor of electricity, but it does undergo ionization to a small extent:

H2O(l) H+(aq) + OH-(aq)

This reaction is sometimes called the autoionization of water.


In the study of acid-base reactions in aqueous solutions, the hydrogen ion concentration is the key, because it indicates the acidity or alkalinity of the solution. Expressing the hydrogen ion as H+, we can write the equilibrium constant for the autoionization of water as

kw = [H+][OH-]

[H2O]

Remember, pure liquids and solids are not listed in the ionization equation, therefore…


kw = [H+][OH-]

• kw is called the ion-product constant, and is the product of the molar concentrations of H+ and OH- ions at a particular temperature.


In pure water at 25 oC, the concentrations of H+ and OH- ions are equal and found to be [H+] = 1.0 x 10-

7 M and [OH-] = 1.0 x 10-7 M. Thus,

kw = [H+][OH-]

kw = (1.0 x 10-7)(1.0 x 10-7)

kw = 1.0 x 10-14


Whether we have pure water or a solution of dissolved species, the following relation ALWAYS holds at 25 oC

kw = [H+][OH-] = 1.0 x 10-14

36

Calculate the concentration of OH- ions in an HCl solution whose hydrogen ion concentration is 1.3 M.

kw = [H+][OH-]

1.0 x 10-14 = (1.3)[OH-][OH-] = 7.7 x 10-15 M

Because HCl is a strong acid…HCl H+ + Cl-

[HCl] [H+] [Cl-]

I (initial)

C (change)E (end)

1.3 0.0 0.0

-1.3 +1.3 +1.3

0.0 1.3 1.3


pH – A Measure of AcidityBecause the concentrations of H+ and OH- ions in aqueous solutions are frequently very small numbers and therefore inconvenient to work with, Soren Sorensen in 1909 proposed a more practical measure called pH. The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration (in mol/L)

pH = -log [H+]pH is a dimensionless quantity (it will not have a label)


Since pH is simply a way to express hydrogen ion concentration, acidic and basic solutions at 25 oC can be distinguished by their pH values, as follows:

Acidic solutions: [H+] > 1.0 x 10-7 M, pH < 7.00

In an acidic solution there is an excess of H+ ions; [H+] > [OH-]

Basic solutions: [H+] < 1.0 x 10-7 M, pH > 7.00

In a basic solution there is an excess of OH- ions; [OH-] > [H+]

Neutral solutions [H+] = 1.0 x 10-7 M, pH = 7.00

Whenever [H+] = [OH-], the aqueous solution is said to be

neutral.

**Note – when concentration has two significant figures, pH will have two numbers TO THE RIGHT OF THE DECIMAL!


Remember, these are molarities

Calculate the pH of a 1.0 x 10-3 M HCl solution.

Since HCl is a strong acid, it is completely ionized in solution:

HCl(aq) H+(aq) + Cl-(aq)

HCl(aq) H+(aq) Cl-(aq)

I 1.0 x 10-3 0.0 0.0

C -1.0 x 10-

3

+1.0 x 10-

3

+1.0 x 10-

3

E 0.0 1.0 x 10-3 1.0 x 10-3

Thus, [H+] = 1.0 x 10-3 M

pH = -log(1.0 x 10-3)

pH = 3.00


The concentration of H+ ions in a bottle of table wine was 3.2 x 10-4 M right after the cork was removed. Only half of the wine was consumed. The other half, after it had been standing open to the air for a month, was found to have a hydrogen ion concentration equal to 1.0 x 10-3 M. Calculate the pH of the wine on these two occasions.

When the wine was first openedpH = -log [H+]pH = -log (3.2 x 10-4) =

3.49

After the wine sat for a monthpH = -log [H+]pH = -log (1.0 x 10-3) =

3.00Some of the ethanol converted to acetic acid, a reaction that takes place in the presence of O2.

Why did the acidity increase?


Given the pH of a solution, you can figure out the [H+] concentration by using the simple formula

[H+] = 10-pH

What is the hydrogen ion concentration of an acid with a pH of 3.00?

[H+] = 10-pH

[H+] = 10-3.00

[H+] = 1.0 x 10-3


A pH meter is commonly used in the laboratory to determine the pH of a solution. Although many pH meters have scales marked with values from 1 to 14, pH values can, in fact, be less than 1 and greater than 14.


A pOH scale analogous to the pH scale can be devised using the negative logarithm of the hydroxide ion concentration of a solution. Thus we define pOH as

pOH = -log[OH-]


Now consider again the ion-product constant for water:

kw = [H+][OH-] = 1.0 x 10-14

Taking the negative logarithm of both sides we obtain

-log[H+] + -log[OH-] = -log(1.0 x 10-14)

-log[H+] + -log[OH-] = 14.00

pH pOH pH + pOH = 14.00

“Logs make adders

multiply”


In a NaOH solution [OH-] is 2.9 x 10-4 M. Calculate the pH of the solution.

pOH = -log [OH-]pOH = -log (2.9 x 10-4)pOH = 3.54

pH + pOH = 14.00pH = 14.00 – pOHpH = 14.00 – 3.54 = 10.46

First, figure out the pOH…

Then use the pOH to figure out the pH…


Calculate the pH of a 0.020 M Ba(OH)2 solution.

Ba(OH)2 is a strong base; each Ba(OH)2 unit produces two OH-ions:

Ba(OH)2(aq) Ba+2(aq) + 2OH-(aq)

Ba(OH)2(aq)

Ba+2(aq) OH-(aq)

I (M) 0.020 0.00 0.00

C (M) -0.020 +0.020 2(+0.020)

E (M) 0.00 0.020 0.040Thus, [OH-] = 0.040 M

pOH = -log 0.040 = 1.40pH = 14.00 – pOH

pH = 14.00 – 1.40 = 12.60


To determine the hydroxide ion when given the pOH, you need to use the formula

10-pOH = [OH-]What is the molarity of a NaOH solution that has a

pH of 11.30?

10-pH = [H+]10-11.30 = 5.0 x 10-11 = [H+]

[H+][OH-] = kw

(5.0 x 10-11)[OH-] = 1.0 x 10-14

[OH-] = 2.0 x 10-4

Because NaOH is a strong base, the [OH-] at the end is equal to the initial concentration of

NaOH.The molarity of the NaOH = 2.0 x 10-4 M


Most acids are weak acids, which ionize only to a limited extent in water. At equilibrium, aqueous solutions of weak acids contain a mixture of nonionized acid molecules, H3O+ ions, and the conjugate base.

The limited ionization of weak acids is related to the equilibrium constant for ionization, which is represented as Ka.

Weak Acids and Acid Ionization Constants


Consider a weak monoprotic acid, HA. Its ionization in water is represented by

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

or simply

HA(aq) H+(aq) + A-(aq)


Write the equilibrium expression for the ionization of HA.

Ka= [H+][A-]

[HA]

Ka, the acid ionization constant, is the equilibrium constant for the ionization of an acid.


At a given temperature, the strength of the acid HA is measured quantitatively by the magnitude of Ka. The larger Ka, the stronger the acid – that is, the greater the concentration of H+ ions at equilibrium due to its ionization. Keep in mind, however, that only weak acids have Ka values associated with them.


You have a reference sheet that lists a number of weak acids and their Ka values at 25 oC. Although all of the acids on that sheet are weak, within the group there is great variation in their strengths.

For example, Ka for HF (6.8 x 10-4) is

about 1.5 million times greater than that for HCN (6.2 x 10-10).


Generally, we can calculate the hydrogen ion concentration or pH of an acid solution at equilibrium, given the initial concentration of the acid and its Ka value.

Alternatively, if we know the pH of a weak acid solution and its initial concentration, we can determine its Ka.


Suppose we are asked to calculate the pH of a 0.50 M HF solution at 25 oC. The ionization of HF is given by

HF(aq) H+(aq) + F-(aq)

From your reference sheet we can write

Ka =[H+][F-]

[HF]= 6.8 x 10-4


The first step is to identify all the species present in solution that may affect its pH. Because weak acids ionize to a small extent, at equilibrium the major species present are nonionized HF and some H+ and F- ions.

Another major species is H2O, but its very small Kw (1.0 x 10-14) means that water is not a significant contributor to the H+ ion concentration. Therefore, unless otherwise stated, we will always ignore the H+ or OH- ions produced by the autoionization of water.


HF(aq) H+(aq) + F-(aq)

We can summarize the changes in the concentrations of HF, H+, and F- in the table below:

HF(aq) H+(aq) F-(aq)

I (M) 0.50 0.00 0.00

C (M) -x +x +x

E (M) 0.50 – x

x x


The equilibrium concentrations of HF, H+ and F-, expressed in terms of the unknown x, are substituted into the ionization constant expression to give

Ka =

(x)(x)

0.50 - x= 6.8 x 10-4

Rearranging this expression, we write

x2 + 6.8 x 10-4x – 3.4 x 10-4 = 0


This is a quadratic equation which can be solved using the quadratic formula. Or, we can try using a shortcut to solve for x.

Because HF is a weak acid and weak acids ionize only to a slight extent, we reason that x must be small compared to 0.50. Therefore we can make the approximation

0.50 – x ≈ 0.50


Now the ionization constant expression becomes

x2

0.50 - x≈

x2

0.50= 6.8 x 10-

4

Rearranging, we get

x2 = (0.50)(6.8 x 10-4) = 3.4 x 10-4

x = √3.4 x 10-4 = 0.018 M


Thus we have solved for x without having to use the quadratic equation. At equilibrium, we have:

[HF] =[H+] =[F-] =

(0.50 – 0.018) M = 0.48 M0.018 M0.018 M

And the pH of the solution is

pH = -log(0.018) = 1.74

This is determined by going back to the ICE

chart

61

How good is this approximation? The approximation is valid if the following expression is equal to or less than 5%

0.018 M

0.50 M

X 100 = 3.6%

Molarity of H+ at equilibrium

Initial concentration of weak acid

If this is greater than 5%, you must use the quadratic formula


The Quadratic Equation

-b± √b2 – 4ac 2a

x =

The values from the equation shown below, (from slide 43), can now be substituted in to the quadratic equation.x2 + 6.8 x 10-4x – 3.4 x 10-4 = 0

a = 1; b = 6.8 x 10-4; c = -3.4 x 10-4

63

-b± √b2 – 4ac 2a

x =

-6.8 x 10-4 ± √(6.8 x 10-4)2 – 4(1)(-3.4 x 10-4)

2(1)x =

-6.8 x 10-4 ± .0014 2(1)

x =

x = .018 M or -.018 M

The second solution is physically impossible because the concentration of ions produced as a result of ionization cannot be

negative.pH = -log(0.018) = 1.74

Note – this is the same

value as we

estimated earlier!


Percent Ionization

We have seen that the magnitude of Ka indicates the strength of an acid. Another measure of the strength of an acid is its percent ionization, which is defined as

Percent ionization =

H+ concentration at equilibrium

Initial concentration of acid

X 100


The stronger the acid, the greater the percent ionization.

The extent to which a weak acid ionizes depends on the initial concentration of the acid. The more dilute the solution, the greater the percent ionization.


Diprotic and Polyprotic Acids

The treatment of diprotic and polyprotic acids is more involved than that of monoprotic acids because these substances yield more than one hydrogen atom per molecule.

These acids ionize in a stepwise manner, that is, they lose one proton at a time.

An ionization constant expression should be written for each ionization step.


Oxalic acid (H2C2O4) is a poisonous substance used chiefly as a bleaching and cleansing agent (for example, to remove bathtub rings). Calculate the concentrations of all the species present at equilibrium in a 0.10 M solution.


H2C2O4 H+ + HC2O4-

H2C2O4(aq)

H+(aq) HC2O4-(aq)

I (M) 0.10 0.00 0.00

C (M) -x +x +x

E (M) 0.10 – x x x


Ka = [H+][HC2O4

-]

[H2C2O4]= 5.6 x 10-

2

Ka = (x)(x)

0.10 - x= 5.6 x 10-

2

Let me save you some work, you need to use the quadratic formula for this one

x2 + 5.6 x 10-2x - 5.6 x 10-3 = 0

x = 0.052


When the equilibrium for the first stage of ionization is reached, the concentrations are:

[H+] =

[HC2O4-]

=

[H2C2O4] =

0.052 M

0.052 M

(0.10 - 0.052) M = 0.048 M


Next we consider the second stage of ionization.

At this stage, the major species will be HC2O4

-, (this serves as the acid in the second stage), H+, and C2O4

-2 (the conjugate base).


HC2O4- H+ + C2O4

-2

HC2O4-(aq) H+(aq) C2O4-2(aq)

I (M) 0.052 0.052 0.00

C (M) -y +y +y

E (M) 0.052 – y 0.052 + y y


Ka = [H+][C2O4

-2]

[HC2O4-]

= 5.4 x 10-

5

Ka = (0.052 + y)(y)

0.052 - y= 5.4 x 10-

5

Let me save you some work, you DON’T need to use the quadratic formula for this one, 0.052 + y and 0.052 – y ≈

0.052

Ka = (0.052)(y)

0.052 = 5.4 x 10-

5

y = 5.4 x 10-5


Testing the approximation - checking the 5% rule

5.4 x 10-5 M

0.052 MX 100 =.10%

The approximation is valid.


At equilibrium

[H2C2O4] =

[HC2O4-] =

[H+] =

[C2O4-2] =

0.048 M

(0.052 – 5.4 x 10-5) M = 0.052 M

(0.052 + 5.4 x 10-5) M = 0.052 M

5.4 x 10-5 M


This example shows that for diprotic acids, if Ka1 » Ka2, then we can assume that the concentration of H+ ions is the product of only the first stage of ionization.


Weak bases, like weak acids, are weak electrolytes.

Ammonia is a weak base that ionizes only to a limited extent in water:

NH3(aq) + H2O(l) NH4+ + OH-

(aq)

Weak Bases and Base Ionization Constants


The equilibrium constant is given by

Kb =[NH4

+][OH-]

[NH3]

Where Kb is called the base ionization constant.


Follow the same procedures you used with weak acids when solving problems involving weak bases.

The main difference is that you will calculate [OH-] first, rather than [H+].


The Relationship Between the Ionization Constants of Acids and Their

Conjugate Bases

For any conjugate acid-base pair it is always true that

KaKb = Kw


Calculate the Kb of the conjugate base of acetic acid

CH3COOH H+ + CH3COO-

Kb =Kw

Ka

Kb =1.0 x 10-14

1.8 x 10-5

Kb = 5.6 x 10-10

Conjugate base


Salts (which are one of the products of an acid-base neutralization reaction) are strong electrolytes that completely dissociate into ions in water. The term salt hydrolysis describes the reaction of an anion or a cation of a salt, or both, with water. Salt hydrolysis usually affects the pH of a solution.

Acid – Base Properties of Salts


Salts that Produce Neutral Solutions

NaNO3 Na+ + NO3-

The cation of this salt came from

a strong base

(NaOH)

The anion of this salt came from a

strong acid (HNO3)

H2O



NaNO3 Na+ + NO3-H2O

The Na+ ion and the OH- from the water

would not stay together (NaOH

would be a strong base and would

dissociate)

The NO3- ion and

the H+ from the water would not stay together,

(remember, HNO3 is a strong acid)



NaNO3 Na+ + NO3-H2O

Consequently, NaNO3 and other salts formed from a strong acid and a strong base do not affect the pH

of a solution.


Salts that Produce Basic Solutions

NaCH3COO Na+ + CH3COO-H2O


a strong base

(NaOH)


weak acid (CH3COOH)



NaCH3COO Na+ + CH3COO-H2O

The Na+ ion and the OH- from the water

would not stay together (NaOH

would be a strong base and would

dissociate)

The CH3COO- ion and the H+ from the water WOULD stay

together, (remember,

CH3COOH is a weak acid, meaning it

only ionizes slightly)



CH3COO- OH- + CH3COOHH2O

Because the CH3COO- ions would bond with the H+ ions, the OH- ions (which are left behind when the H+

ions come off of the water molecules) affect the pH of the solution. In other

words, solutions produced by salts made from strong bases and weak

acids will be basic in nature.


Salts that Produce Acidic Solutions

NH4Cl NH4+ + Cl-

H2O


a weak base


strong acid


Salts that Produce Acidic Solutions

NH4Cl NH4+ + Cl-

H2O

Because the Cl- ions wouldn’t bond with the H+ ions, and the H+ ion that would separate from the NH4

+, it would leave excess H+ in solution. In other words, solutions produced by salts made from

strong acids and weak bases will be acidic in nature.

NH3 + H+


NO2- has an affinity

for the H+ produced by the autoionization of water, leaving an

excess of OH- behind.

The NH4+

dissociates producing H+

ions

When the cation of a salt comes from a weak base and the anion comes from a weak acid, you need to compare the Ka and Kb values to determine if the solution will be acidic or basic. For example,

NH4NO2 NH4+ + NO2

-

Ka = 5.7 x 10-10 NH4+ NH3 + H+

Kb = 1.4 x 10-11 NO2- HNO2 +

OH-

therefore an aqueous solution of NH4NO2

will be acidic.

If Kb > Ka, the solution will be basic,

if Ka > Kb, the solution will be acidic..


Oxides can be classified as acidic, basic, or amphoteric.

• All alkali metal oxides and all alkaline earth metal oxides except BeO are basic.

• Beryllium oxide and several metallic oxides in the boron family (Group 3A) and carbon family (Group 4A) are amphoteric.

Na2O + H2O 2NaOH

2Ba(OH)2BaO + H2O


• Nonmetalic oxides in which the oxidation number of the representative element is high are acidic. The nonmetalic oxides that react with water to form acids are sometimes referred to as acidic anhydrides. Representative elements in which the oxidation number is low (for example, CO and NO) show no measurable acidic properties.

CO2 + H2O SO3 + H2O

N2O5 + H2O P4O10 + H2O Cl2O7 + H2O

H2CO3

H2SO4

2HNO3

4H3PO4

2HClO4


Buffer Solutions

A buffer solution is a solution of (1) a weak acid or a weak base and (2) its salt; both components must be present. The solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.

Buffers are very important to chemical and biological systems. The pH in the human body varies greatly from one fluid to another; for example, the pH of blood is about 7.4, whereas the gastric juice in our stomach has a pH of about 1.5. These pH values, which are crucial for proper enzyme function and the balance of osmotic pressure, are maintained by buffers in most cases.


NaCH3COO CH3COO- + Na+H2O

CH3COOH CH3COO- + H+

If you add a base to this solution, the OH- will be neutralized by the acetic

acid in the buffer, therefore you will not notice a significant

difference in the pH of the solution.


NaCH3COO CH3COO- + Na+H2O

If you add an acid to this solution, the acetate ion will bond to the H+, so no appreciable change in pH will

be observed because it is such a small increase in H+



The buffering capacity, that is, the effectiveness of the buffer solution, depends on the amount of acid and conjugate base from which the buffer is made. The larger the amount, the greater the buffering capacity.


In general, a buffer system can be represented as salt-acid or conjugate base-acid. Thus the sodium acetate-acetic acid buffer system we discussed can be written as

CH3COONa/CH3COOH or simply

CH3COO-/CH3COOH.


Which of the following are buffer systems?

KF/HF

NaClO4/HClO4


(a)Calculate the pH of a buffer system containing 1.0 M CH3COOH and 1.0 M CH3COONa.

(b)What is the pH of the buffer system after the addition of 0.10 mole of gaseous HCl to 1 L of the solution. Assume that the volume of the solution does not change when the HCl is added.


CH3COOH CH3COO- H+

I 1.0 1.0 0.0

C -x + x x

E 1.0 - x 1.0 + x 0.0 + x

Assume ionization

is negligibl

e


Ka =[H+][CH3COO-]

[CH3COOH]= 1.8 x 10-5


[H+] =

Rewriting the above equation…

[CH3COOH]

[CH3COO-][H+] =

ka

1.8 x 10-5(1.0)

(1.0)


[H+] = 1.8 x 10-5 M

-log[H+] =

-log(1.8 x 10-5)

pH = 4.74

When the concentration of the acid and the conjugate base are the same, the pH

of the buffer is equal to the pKa, which is determined by taking the –log of the ka, of

the acid pH = pka = -log ka


(b)After the addition of HCl, complete ionization of the 1.0 M HCl acid occurs;

HCl H+ + Cl- 0.10 mol 0.10 mol

This 0.10 mol of H+ ions will bond with 0.10 mol of the CH3COO- ions,

DECREASING the amount of CH3COO- by .10 mol and

INCREASING the amount of CH3COOH by .10 mol


So now instead of 1.0 mol of CH3COO- ions, there will only be 0.90 mol (1.0 - .10)

And since there were originally 1.0 mol of CH3COOH, you now have 1.10 mol of

CH3COOH with the added .10 mol

[CH3COO-] = .90 M

[CH3COOH] =

1.10 M


Ka =[H+][CH3COO-]

[CH3COOH]= 1.8 x 10-5

[H+] =

Rewriting the above equation…Ka[CH3COOH]

[CH3COO-][H+] =

(1.8 x 10-5)(1.10)

(0.90)


[H+] = 2.2 x 10-5 M

pH = -log(2.2 x 10-5)

pH = 4.66

This is a very slight change in pH, before the HCl was added, the [H+] = 1.8 x 10-5, after the addition of HCl,

the [H+] = 2.2 x 10-5, this is an increase by a factor of 1.2 with a pH

change from 4.74 to 4.66

2.2 x 10-5

1.8 x 10-5=

1.2


Before the addition of HCl: After the addition of HCl:

0.10 M

1.0 x 10-7 M= 1.0 x 106

This would be a millionfold increase as the pH changed from 7.00 to

1.00!

[H+] = 1.0 x 10-7

[H+] = 0.10 M


A "very convenient" equation for dealing with buffer solutions is the

Henderson-Hasselback equation.

pH = pKa + log[A-][HA]


log 1 = 0

Previous question - Calculate the pH of a buffer system containing 1.0 M CH3COOH

and 1.0 M CH3COONa.

pH = pka + log[A-][HA]

pH = pka + log[1.0][1.0]

pH = pka

pH = -log(ka) pH = -log(1.8 x 10-

5)pH = 4.74


What is the pH of the previous buffer system after the addition of 0.10 mole of gaseous HCl to 1 L of the solution. Assume that the volume of the solution does not change when the HCl is added.

pH = pka + log[A-][HA]

pH = pka + log[.90]

[1.10]

pH = -log(1.8 x 10-5) + -.041

pH = 4.66

Remember, the .10 M H+ from the HCl would shift

the equilibrium to

the left taking .1M CB,

decreasing the amount of CB to .90, and increasing the amount of A

to 1.10


Acid – Base Titrations

Quantitative studies of acid-base neutralization reactions are most conveniently carried out using a technique known as titration. In titration, a solution of accurately known concentration, called a standard solution, is added gradually to another solution of unknown concentration, until the chemical reaction between the two solutions is complete.


If we know the volumes of the standard solution, we can calculate the concentration of the unknown solution.

Sodium hydroxide is one of the bases commonly used in the laboratory. However, it is difficult to obtain solid sodium hydroxide in a pure form because it is hygroscopic, (it has a tendency to absorb water from air), and its solution reacts with carbon dioxide. For these reasons, a solution of sodium hydroxide must be standardized before it can be used in accurate analytical work.


We can standardize the sodium hydroxide solution by titrating it against an acid solution of accurately known concentration. The acid often chosen for this task is a monoprotic acid called potassium hydrogen phthalate (abbreviated as KHP), for which the molecular formula is KHC8H4O4.


The procedure for the titration of KHP and NaOH is as follows:

1. Add a known amount of KHP to an Erlenmeyer flask. Add some distilled water to make up a solution.

2. Next, carefully add NaOH solution from a buret until the equivalence point is reached. The equivalence point is the point at which the acid has completely reacted with or been neutralized by the base.


The equivalence point is usually signaled by a sharp change in the color of an indicator. In an acid-base titration, indicators are substances that have distinctly different colors in acidic and basic solutions. One common indicator is phenolphthalein.

Phenolphthalein indicates the presence of a/n _________. Phenolphthalein is ________ in acidic solutions, _______ in neutral solutions and ________ in basic solutions.

basecolorless

pinkcolorless


At the equivalence point, all the KHP present has been neutralized by the added NaOH and the solution is still colorless. However, if we add just one more drop of NaOH solution from the buret, the solution will immediately turn pink because the solution is now basic.

119

Apparatus for acid-base titration.

A NaOH solution is added from the buret to a KHP solution in an

Erlenmeyer flask.

A faint pink color appears when the

equivalence point is reached. If your

solution turns fuchsia, you have gone past

the equivalence point


The neutralization reaction between NaOH and KHP is one of the simplest types of acid-base neutralization known. A neutralization reaction is a reaction between an acid and a base.

Aqueous strong acid-strong base reactions produce water and a salt, (an ionic compound made up of a cation other than H+ and an anion other than OH- or O-2)


The reaction between KHP and sodium hydroxide is:

KHC8H4O4(aq) + NaOH(aq) KNaC8H4O4(aq) + H2O(l)

acid base salt water


You can use the following format to solve acid-base neutralization problems

MM mol

mol

L L

Acid Base


MM mol

mol

L L

Acid Base

What is the molarity of the acid if 16.1 mL of 0.610 M NaOH was required to neutralize 20.0 mL of H2SO4?

.0161.610.0200


MM mol

mol

L L

Acid Base

.0161.610.0200

X = .00982

.610 M = .610 mol =

1 L

x

.0161 L

.00982


MM mol

mol

L L

Acid Base

.0161.610.0200 .00982

H2SO4 + 2NaOH Na2SO4 + 2H2O

.00982 mol NaOH

x1 mol H2SO4

2 mol NaOH

.00491

= .00491 mol H2SO4


MM mol

mol

L L

Acid Base

.0161.610.0200 .00982

.00491

M = .00491 mol =

.0200 L

.0246 M

.0246


The reaction between HCl, a strong acid, and NaOH, a strong base, can be represented by

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

The pH profile of the titration of this neutralization reaction is known as

a titration curve.


Consider the addition of 0.10 M NaOH solution (from a buret) to an Erlenmeyer flask containing 25 mL of 0.10 M HCl.

Before the addition of NaOH, the pH of the acid is given by –log(0.10), or 1.00

When the NaOH is added, the pH of the

solution increases slowly at first

Near the equivalence point, the pH begins to rise steeply, and at the

equivalence point the curve rises almost vertically

Beyond the equivalence point, the pH again increases slowly with the addition of NaOH


It is possible to calculate the pH of a solution at every stage of titration.

What is the pH of the solution after the addition of 10.0 mL of 0.10 M NaOH to 25.0 mL of 0.10 M HCl?


0.10 mol NaOH = 1 L

x .0100 L

x = 0.00100 mol NaOH

0.10 mol HCl = 1 L

x .0250 L

x = 0.00250 mol HCl

0.00250 mol HCl - 0.00100 mol NaOH 0.00150 mol HCl Excess

acid


To determine the pH, you need to calculate [H+]

.0350 L

.00150 mol HCl =.043 M HCl

pH = -log (.043)

pH = 1.37

Total volume of the

original HCl and NaOH


At the equivalence point of a titration between a weak acid and a strong base, the pH will be greater than 7.

133

Because…

CH3COOH + NaOH CH3COONa + H2OCH3COONa CH3COO- + Na+

This acetate ion has an affinity for the H+ ion in the

water, (thus leaving the OH- behind, making

the solution basic.)

at neutralization


The flat portion of the titration curve before the equivalence point is called the buffer region. In this part of the pH scale, the acid and conjugate base are both present in significant concentrations and the solution resists changes in pH. As base is added to a solution in this buffer region, acetic acid reacts with it to form acetate ion, without a large change in pH.

The pKa of a weak acid can be determined

experimentally


In the middle of the buffer region lies the half-equivalence point. Here the volume of base added is half that required to reach the equivalence point and half the acetic acid has been converted to the conjugate base, acetate ion. This means that the concentrations of acetic acid and acetate ion are equal. If we examine the equilibrium expression at the half-equivalence point, we find something interesting…

At the half-equivalence point, [CH3COOH] = [CH3COO-], so

Ka = [H3O+] 1/2 way


At the half-equivalence point, [CH3COOH] = [CH3COO-], so

Ka = [H3O+]

Taking the negative log of both sides yields pKa = pH

Ka = 10-pH

This gives us an experimental way to determine the Ka of a

weak acid, and using a Ka table, the identity of an unknown weak acid.


A slightly different curve results when you titrate a strong acid vs a weak base. At the equivalence point of a titration between a strong acid and a weak base, the pH will be less than 7.


Because…

HCl + NH3 NH4Cl

NH4Cl NH3 + H+

The pH of less than 7 is due to the presence of H+ ions formed

by the hydrolysis of

NH4+

at neutralization

139

Acid-Base IndicatorsAn indicator is usually a weak organic acid or

base that has distinctly different colors in its nonionized (molecular) form and ionized form. The end point of a titration occurs when the indicator changes color.

However, not all indicators change color at the same pH, so the choice of indicator for a particular titration depends on the nature of the acid and base used in the titration (that is, whether they are strong or weak).

By choosing the proper indicator for a titration, we can use the end point to determine the equivalence point.

140

Let us consider a weak monoprotic acid that we will call HIn. To be an effective indicator, HIn and its conjugate base, ___, must have distinctly different colors.

In-

HIn H+ + In-

One color

A different

color


If the indicator is in a sufficiently acidic environment, the equilibrium, according to Le Chatelier’s principle, shifts to the __________ and the predominant color will be that of ______,

leftHIn

HIn H+ + In-


In a basic environment, the equilibrium shifts to the right because…

HIn H+ + In-

The H+ and the OH- will form water, thus removing the H+ from the

system and therefore shifting the equilibrium to the right. The

predominant color will then be that of In-.


The end point of an indicator does not occur at a specific pH; rather, there is a range of pH within which the end point will occur. In practice, we choose an indicator whose end point lies on the steep part of the titration curve. Because the equivalence point also lies on the steep part of the curve, this choice ensures that the pH at the equivalence point will fall within the range over which the indicator changes color.


Some Common Acid-Base Indicators

Indicator In acid In base pH range

Thymol blue Red Yellow 1.2 - 2.8

Bromophenol blue

Yellow Bluish purple 3.0 – 4.6

Methyl orange Orange Yellow 3.1 – 4.4

Methyl red Red Yellow 4.2 – 6.3

Chlorophenol blue

Yellow Red 4.8 – 6.4

Bromothymol blue

Yellow Blue 6.0 – 7.6

Cresol red Yellow Red 7.2 – 8.8

Phenolphthalein Colorless Reddish pink 8.3 – 10.0

The pH range is defined as the range over which the indicator changes from the acid color to the

base color.


This is THE END of your CHEM II

NOTES!!!

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