Unit 12 Acids and Bases

Unit 14 Acids and Bases

When we discussed the types of bonding, we learned that ionic solutes were electrolytes. Electrolytes are solutes that dissociate into positive and negative ions in aqueous solution. For example, salt, NaCl is an electrolyte. When it is dissolved in water, it dissociates into Na^+1 and Cl^-1 ions.

NaCl (s) <--> Na⁺¹(aq) + Cl^-1(aq)

Covalently bonded materials like dextrose do not dissociate and are NOT electrolytes.

C6H12O6(s) <--> C6H12O6(aq)

Three kinds of electrolytes are acids, bases and salts.

Acid / Base Theories

Arrhenius (1887)

Acids are electrolytes that release H⁺¹ ions when they are added to water. For example

HCl (g) <-----> H⁺¹ (aq) + Cl ^-1 (aq)

H2SO4 (l) <--> 2 H⁺¹ (aq) + SO4^-2 (aq)

Bases are electrolytes that release OH^-1 ions when added to water. For example

NaOH (s) <--> Na⁺¹ (aq) + OH^-1 (aq)

Ca(OH)₂ (s) <--> Ca⁺² (aq) + 2 OH^-1 (aq)

Arrhenius's definition of acids and bases is quite specific and works well in most instances. His theory can't explain why ammonia (NH₃) is a base (Where is the OH ?). Also the H⁺¹ ion would be extremely unstable due to its small size and large charge giving it an extremely large charge density.

Bronsted - Lowry (1923)

Acids are any substance that can donate H⁺¹ ions in aqueous solution.

Bases are any substances that can accept H⁺¹ ions in aqueous solution.

HCl + H₂O <--> H₃O⁺¹ + Cl^-1

acid base conjugate conjugate
acid base

In the above equation, HCl is the B-L acid because it releases H⁺¹ ion into solution. Notice the H⁺¹ ion doesn't exist in solution, but instead joins with water. The water is the B-L base because it accepts the H⁺¹ ion.

Applying the B-L definition to the reverse reaction, H₃O⁺¹ acts as the acid and Cl^-1 acts as the base. To avoid confusion between the acids and bases of the forward and reverse reaction, we designate the conjugate acid as the acid in the reverse reaction and the conjugate base as the base in the reverse reaction.

Another way of defining conjugate acids and bases is

Conjugate Base - what is left over of the acid after it releases its H⁺¹ ion.

Conjugate Acid - the base in the forward reaction becomes the conjugate acid after it has accepted the H⁺¹ ion

We can now explain how ammonia (NH₃) is basic, without the OH^-1 ion.

NH₃ + H₂O <--> NH₄⁺¹ + OH^-1

base acid conjugate conjugate
acid base
Ammonia is a B-L base because it accepts a H⁺¹ from the B-L acid H₂O. The conjugate acid is the NH₄⁺¹ ion because it acts as the acid in the reverse reaction. The OH^-1 ion acts as the conjugate base.

Notice that in the reaction with HCl, water was acting as a B-L base but in the reaction with NH₃, water was the B-L acid. Water is still considered to be neutral, but can act as a weak acid or base.

Strength of Acids and Bases

In everyday life when we refer to a strong cup of coffee we are talking about the concentration of the coffee. In acid - base chemistry, strength does NOT refer to the concentration of the acid or base. Acid - Base strength refers to the amount of dissociation that occurs. A strong acid or base dissociates nearly 100 %. A weak acid or base dissociates significantly less than 100 %.

HCl and NaOH are strong because when they are added to water, there is essentially none of the un-ionized form left in the solution. Acetic acid, HC₂H₃O₂ and ammonia, NH₃ are considered weak because they are only about 5 % ionized. The un-ionized form is the predominate species in solution.

Strong Acid - ionized species are predominant

HCl <--> H⁺¹ + Cl^-1

Weak Acid - un-ionized species is predominant

HC₂H₃O₂ <--> H⁺¹+ C2H3O2^-1

We observed acid base strength when we tested the conductivity of a 0.1 M HCl solution and a 0.1 M HC₂H₃O₂ solution. Because the HCl is a strong acid and nearly completely ionized, the solution was an excellent conductor. The HC₂H₃O₂ solution barely conducted because of the small amount of ionization.

Because the acids in a reaction become conjugate bases when a H⁺¹ is lost, we can compare their strengths. Strong acids yield weak conjugate bases and vice versa. Also strong bases yield weaker conjugate acids and vice versa.

Nomenclature

Acid Nomenclature

We can essentially break acids into two categories.

Binary acids contain only hydrogen and one other element. They are named by using the following system:

Hydro ________ ic acid
For example

HCl Hydrochloricacid

HF Hydrofluoricacid

H₂S Hydrosulfuricacid

Acids with (-ate) polyatomic ions have hydrogen and a polyatomic ion, usually containing oxygen. They are named by using the following system:
_________ic acid
For example
H₂SO₄ Sulfuric acid

HNO₃ Nitric acid

H₃PO₄ Phosphoric acid

Recall that there were a few polyatomic ions that contained the same elements in different ratios (-ite). They are named by using the following system:
_________ous acid
For example

H₂SO₃ Sulfurous acid

HNO₂ Nitrous acid

Base Nomenclature

For bases, we simply name the positive ion, then add hydroxide.

NaOH Sodium Hydroxide

Ca(OH)₂ Calcium Hydroxide

NH₄OH Ammonium Hydroxide

Neutralization

Acids and bases react with each other in a double displacement type reaction. The properties of each are destroyed and water and a salt are produced. In this case the term salt doesn't necessarily mean sodium chloride. The word salt is a general term describing an ionic substance. Sodium chloride is one of many different salts.

Some neutralization reactions:

NaOH + HCl --> NaCl + H₂O
LiOH + HF --> LiF + H₂O
2 NH₄OH + H₂SO₄ --> (NH₄)₂SO₄ + 2 H₂O

Salts are named according to the rules we developed in the beginning of the year.

A neutralization always produces water and a salt, but doesn't always result in a neutral product. While water is neutral, the salt may be neutral, slightly acidic or basic.

When a strong acid and strong base react, the salt is neutral.

When a strong acid reacts with a weak base, the salt will be somewhat acidic.

When a weak acid reacts with a strong base, the salt will be somewhat basic.

Ionization Constants For Weak Acids

Weak acids and bases do not completely dissociate but instead reach an equilibrium well before the reaction can get to completion. Let's consider the reaction of a hypothetical weak acid, HA when added to water. For simplicty, we'll use the Arrhenius theory when we write out the equation.

HA <--> H⁺¹ + A^-1

Because this equilibrium involves an acid, we give the equilibrium constant a special designation, K_a.

Ka = [H⁺¹][A^-1]
[HA]

We know that H⁺¹ doesn't exist freely in solution but is actually combined with water to form H₃O⁺¹. With that in mind, we can modify the equation for Ka in the following way.

Ka = [H₃O⁺¹][A^-1]
[HA]

For weak bases we end up with an equilibrium expression but call it K_b.

The smaller the K_a, the weaker the acid. K_a's are only useful for weak acids.

Polyprotic Acids

Some acids have more than 1 ionizable hydrogen. Two examples include H₂SO₄ and H₃PO₄. These hydrogens ionize one at a time. Generally as each hydrogen leaves the acid, it makes it more difficult to lose the next hydrogen. For H₃PO₄, the first K_a is the largest, the second is significantly smaller and the third is even smaller.

K_a1 = 7.11 x 10^-3

K_a2 = 7.99 x 10^-8

K_a3 = 4.8 x 10^-13

Ionization of Water

From our tests with the conductivity apparatus, we observed that solutions of electrolytes conducted whereas solutions of covalently bonded compounds did not. We also noticed that tap water conducted electricity but deionized water did not. If we were to use a more sensitive device, we would find that even the most pure deionized water is a slight conductor. How could this be?

While water is considered a covalently bonded molecule, about one molecule in a billion will undergo an ionization. The water molecule dissociates to form an H⁺¹ ion and an OH^-1 ion. The H⁺¹ ion being highly unstable immediately combines with a neighboring water molecule to form H₃O⁺¹, the hydronium ion.

The equilibrium constant for this reaction is

Keq = [H₃O⁺¹][OH^-1] = [H₃O⁺¹][OH^-1]
[H2O][H2O] [H2O]²

Since very little water dissociates, its concentration remains essentially constant. As we saw in the equilibrium section, substances whose concentration remain constant (like solids) in an equilibrium system are left out of the equilibrium expression. This special case of the equilibrium constant is called the ion product of water, Kw.

Kw = [H₃O⁺¹][OH^-1]

It has been determined that at 25 C, the Kw for water and dilute solutions is 1 x 10^-14

Kw = [H₃O⁺¹][OH^-1] = 1 x 10^-14

An acidic solution has more [H₃O⁺¹] than [OH^-1].
A basic solution has more [OH^-1] than [H₃O⁺¹].
In a neutral solution [H₃O⁺¹] = [OH^-1]

In each case, the product of [H₃O⁺¹] and [OH^-1] is 1 x 10^-14

We can solve for either concentrations by rearranging the Kw equation.

[H₃O⁺¹] = Kw / [OH^-1]

[OH^-1] = Kw / [H₃O⁺¹]

The pH Scale

Expressing the acidity of a solution in terms of H₃O⁺¹ concentration can be cumbersome with such small numbers in exponential notation so the pH scale has been created to simplify things. As you probably remember from other science classes, the pH scale runs from 0 to 14 with pH's less than 7 acidic, more than 7 basic and 7 being neutral.

We define pH as the negative base 10 logarithm of the hydronium concentration.

pH = - log [H₃O⁺¹]

If the concentration of hydronium ion was 0.02 M, the pH would be - log [0.02] = 2

If the pH of a solution is 9, we would solve for the hydronium ion concentration by:

- log [H₃O⁺¹] = 9

[H₃O⁺¹] = 1 x 10^-9 M

pH Calculations

Strong Acids and Bases

If we know the concentration of the strong acid or base, we know the concentration of the [H₃O⁺¹] or [OH^-1] because strong acids and bases completely dissociate.

From the concentration of [H₃O⁺¹] we can immediately solve for pH. pH = - log [H₃O⁺¹]

From the concentration of [H₃O⁺¹, we would need to use the Kw expression to solve for the [H₃O⁺¹] and then determine the pH. 1) [H₃O⁺¹] = Kw / [OH^-1]

2) pH = - log [H₃O⁺¹]

Weak Acids

We can determine the pH of a solution of a weak acid if we know its concentration. The calculations involving a weak acid are more complex because we cannot make the same assumptions about the [H₃O⁺¹] being equal to the concentration of the acid. Remember that a weak acid undergoes incomplete dissociation, which means that the concentration of [H₃O⁺¹] will be quite different from the acid.

If we know the Ka and the concentration of the weak acid we can easily determine the pH. Let's look at a sample problem.

HA is a weak acid with a Ka of 1 x 10^-5. Calculate the pH of a 0.1 M HA solution.

1) Write out the dissociation of the acid

HA + H₂O <--> H₃O⁺¹ + A^-1

2) Write out the Ka expression

Ka = [H₃O⁺¹][A^-1]
[HA]

3) Let x = [H₃O⁺¹] = [A^-1] Remember, they should be equal due to the 1:1 ratio in the dissociation equation.
Let [HA] = 0.1 - x The amount of HA dissociated should be equal to the amount of H₃O⁺¹ formed.

4) Substitute the numbers and variables into the Ka expression and solve for x. ***

1 x 10^-5 = x² __
(0.1 - x)

*** If the Ka is small like it is here, we can assume that very little of the acid dissociates and therefore 0.1 - x ~ 0.1. The equation simplifies to:

1 x 10^-5 = x² __
0.1

x² = 1 x 10^-6

x = 1 x 10^-3 = [H₃O⁺¹]

pH = - log (1 x 10^-3) = 3

Buffers

It is important in just about every living organism to maintain a constant pH. The internal fluids such as blood and lymph in humans is kept at a pH of about 7.4. Many foods that we consume have an acidic pH. Most enzymes only work optimally in a narrow pH range, so the organism must have a means to keep the pH in this range. Buffers can absorb moderate amounts of base or acid without a change in pH.

A buffer consists of a weak acid and its salt or a weak base and its salt. Buffers work optimally when the concentration of the weak acid (or weak base) is equal to its salt.

Lets consider a buffering system with the hypothetical weak acid HA and its salt NaA. Since HA is a weak acid, it is hardly dissociated, so the predominant species in solution from the acid is HA. The salt NaA is soluble and completely dissociates, which means the predominant species from the salt is A-. We can now see what happens if small quantities of an acid or base are added.

Buffer + added acid:

A- + HCl --> HA + Cl^-1

Notice the HCl cannot make the solution more acidic because the A^-1 ion "absorbs" the H^+1.

Buffer + added base:

HA + OH^-1 --> A^-1 + HOH

The weak acid from the buffer converts the OH^-1 ion from the base into water.

Buffering in the Blood

Blood is buffered by a combination weak acid H2CO3 and its salt NaHCO3. The weak acid is generated by CO2 dissolving in the blood and reacting with the water.
CO2 + H2O <--> H2CO3

The H2CO3 then undergoes dissociation

H2CO3 <--> H⁺¹ + HCO3^-1

HCO3^-1 can further dissociate

HCO3^-1 <--> H⁺¹ + CO3^-2

The HCO3^-1 ion can act as a base and neutralize added acid.

HCO3^-1 + H⁺¹ <--> H2CO3

Since it has an ionizable H, it can act as an acid and "assist" H2CO3 to neutralize any added base.

HCO3^-1 + OH^-1 --> CO3^-2 +HOH

Lets look at a specific example of buffering.

A solution consists of 0.05 M CH₃COOH and 0.05 M NaCH₃COO. The Ka of acetic acid is 1.76 x 10^-5. What is the pH?

We know the salt will dissociate completely so

[CH₃COO^-1] = [NaCH₃COO] = 0.05 M

CH₃COOH is a weak acid and has a Ka = [H+][CH₃COO^-1] =

[CH₃COOH]

[H₃O⁺¹][CH₃COO^-1]
[CH₃COOH]

Let x = [H₃O⁺¹]weak acid = [CH₃COO^-1] weak acid

Let 0.05 - x = [CH₃COOH] ~ 0.05 (because x <<< 0.05)

The total [CH₃COO^-1] = [CH₃COO^-1]salt + [CH₃COO^-1]weak acid

= 0.05 + x ~ 0.05

Revisiting the Ka, we see that the concentration of the weak acid [CH₃COOH] equals the ion [CH₃COO^-1]. This reduces the Ka expression to

Ka = [H₃O⁺¹] -->

pH = -log [H₃O⁺¹] = -log Ka = - log 1.76 x 10^-5 = 4.75

Titration

Because acids and bases react with each other in specific mole to mole ratios, an acid of known concentration can be used to determine the concentration of a base of unknown concentration and vice versa. In order to accomplish this, the endpoint of the titration must be known. This can be accomplished by using an indicator like phenolphthalein or a pH meter. If we visualize a graph of pH vs. added base we would see the following:

Near the end of the titration, the H₃O⁺¹ ions are being used up and are replaced by an excess of OH^-1 ions. This causes a dramatic change in pH at the endpoint. An indicator changes color in the area of the graph where there is a very large change in the pH. Phenolphthalein changes from colorless to red at a pH of above 8, which is within the area with a large pH change. Phenolphthalein works well as an indicator in the first two titrations. It is not a good indicator for the third titration because the graph begins to level out before it reaches the pH where the indicator changes.

Titration Calculations

When the endpoint is reached in a titration the moles of hydronium ions is equal to the mole of hydroxide ions

moles H₃O⁺¹ = moles OH^-1

use stoich approach BCA table????

In this case, the volumes were converted to liters because the concentrations are in molar units which are in moles per liter.

Although the equation Ma*Va = Mb*Vb works well in the above problem, it can cause problems if the acid or base have more than one ionizable H or OH.

A better approach to titration would be to think in terms of H₃O⁺¹ ion and OH^-1 ion. When doing titration calculations, we need to remember two things.

At the endpoint the moles H₃O⁺¹ = moles OH^-1.

The product of Molarity and volume in liters is moles.

It takes 50 ml of a 0.1 M H3PO4 solution to neutralize a 100 ml of a solution of Ca(OH)2 that has an unknown concentration. What is the concentration of the Ca(OH)2 solution?

Step 1 Find the number of moles of H3O^+1 (the solution with known concentration)

0.015 moles H₃O⁺¹

Step 2 Set the number of moles OH^-1 ion equal to the moles H₃O⁺¹

moles H₃O⁺¹ = moles OH^-1

Step 3 Use the moles OH^-1 ion and volume of the base to determine the concentration of the OH^-1 ion.

[OH^-1] = moles OH^-1 / volume base = 0.015 moles OH^-1 | 1000 ml =
100 ml | 1 l

= 0.15 moles OH^-1 / liter

Step 4 Determine the concentration of the base.

0.15 moles OH^-1 | 1 mol Ca(OH)₂ = 0.075 moles Ca(OH)₂ / liter
liter | 2 moles OH^-1

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