Acid-Base Reactions

Acid-Base Reactions

When an acid and a base are placed together, they react to neutralize the acid and base properties, producing a salt. The H(+) cation of the acid combines with the OH(-) anion of the base to form water. The compound formed by the cation of the base and the anion of the acid is called a salt. The combination of hydrochloric acid and sodium hydroxide produces common table salt, NaCl:

The word salt is a general term which applies to the products of all such acid-base reactions.

Acids are compounds (or ions) that react with water to produce hydrogen ions (H⁺) (see Acids and Bases). Hydrogen ions account for the characteristic properties of strong acids, such as a sour taste and the ability to react with bases. Bases are compounds that yield the hydroxide ion (OH^-) in water solutions. Salts are ionic compounds that are generally formed by the reaction of an acid and a base:

                               Acid+base=Salt+water

       HCl+NaOH=NaCl+H₂O             

KHSO₄+KOH=K₂ SO₄+ H₂O

HNO₃+NH₄OH=NH₄NO₃+ H₂O

Hydrolysis

Hydrolysis is a reaction involving the breaking of a bond in a molecule using water. The reaction mainly occurs between an ion and water molecules and often changes the pH of a solution. In chemistry, there are three main types of hydrolysis: salt hydrolysis, acid hydrolysis, and base hydrolysis.

1.    1. Introduction

1.    1.1. Salt Hydrolysis

2.    1.2. Acid Hydrolysis

3.    1.3. Basic Hydrolysis

2.    2. Examples & Practice

3.    3. Solutions to Example Problems

4.    4. Use of Hydrolysis in the "Real World"

5.    5. References

6.    6. Outside Links 

7.    7. Contributors

Introduction

Salt Hydrolysis

In water, salts will dissociate completely to form ions.

Example:

NH₄Br_(s)→ NH₄⁺_(aq) + Br^-_(aq)

Here, the salt NH₄Br is put into water and dissociates into NH₄⁺and Br^-.

In the figure above,

NaCl_(s) → Na⁺_(aq) + Cl^-_(aq)

Note that water is polar, causing O to be slightly negative and H to be slightly positive.  The positively charged sodium ion is attracted to the O in water and the negatively charged chlorine ion is attracted to the H in water.

There are four possible ways of forming salts:

1. If the salt is formed from a strong base and strong acid, then the salt solution is neutral, indicating that the bonds in the salt solution will not break apart (indicating no hydrolysis occurred) and is basic.
2. If the salt is formed from a strong acid and weak base, the bonds in the salt solution will break apart and becomes acidic.
3. If the salt is formed from a strong base and weak acid, the salt solution is basic and hydrolyzes.
4. If the salt is formed from a weak base and weak acid, will hydrolyze, but the acidity or basicity depends on the equilibrium constants of K_a and K_b. If the K_a value is greater than the K_b value, the resulting solution will be acidicand vice versa.

Acid Hydrolysis

H₂O can act as an acid or a base based on the Brønsted-Lowry acid theory. If it acts as a Bronsted-Lowry acid, the water molecule would donate a proton (H⁺), also written as a hydronium ion (H₃O⁺).  If it acts as a Bronsted-Lowry base, it would accept a proton (H⁺). An acid hydrolysis reaction is very much the same as an acid dissociation reaction.

CH₃COOH +H₂O  H₃O⁺ + CH₃COO^-

In the above reaction, the proton H⁺ from CH₃COOH (acetic acid) is donated to water, producing H₃O⁺ and a CH₃COO^-. The bonds between H⁺ and CH₃COO^- are broken by the addition of water molecules. A reaction with CH₃COOH, a weak acid, is similar to an acid-dissociation reaction, and water forms aconjugate base and a hydronium ion. When a weak acid is hydrolyzed, a hydronium ion is produced.

Basic Hydrolysis

A base hydrolysis reaction will resemble the reaction for base dissociation. A common weak base that dissociates in water is ammonia:

NH₃ + H₂O  NH₄⁺ +OH^-

In the hydrolysis of ammonia, the ammonia molecule accepts a proton from the water (because water acts as a Bronsted-Lowry acid), producing a hydroxide anion (OH^-). Similar to a basic dissociation reaction, ammonia forms ammonium and a hydroxide from the addition a water molecule. 

Examples & Practice

1) H₂CO₃ + H₂O  H₃O⁺ +HCO₃^-
    a. Identify which of these is the conjugate base and which is the weak acid.
    b. Does the weak acid hydrolyze?
2)
    a. Write out the chemical equation for the hydrolysis HF.
    b. Is water acting as a Bronsted-Lowry acid or Bronsted-Lowry base?
3)
    a. Write out the equation for the dissociation of the salt NH₄Br.
    b.Write out the hydrolysis of the cation that is produced from the dissociation of the ammonium bromide.
    c. From what kinds of acids and bases is ammonium bromide (NH₄Br) made from? Strong acid/strong base? Strong acid/weak base?     Strong base/weak acid? Weak base/weak acid?
    d. State whether salt hydrolyzes.
    e. State whether solution is acidic or basic.
4). CH₃COO^- +H₂O  CH₃COOH + OH^-
    What is the pH of 0.30 M of sodium acetate?
    (Hint: First find Kb value)
    Given: K_a of CH₃COOH= 1.8 x 10^-5
5)
    a. Does sodium acetate (from previous problem) hydrolyze?
    b. Is solution acidic or basic?

Solutions to Example Problems

1)
    a. The conjugate base is the HCO₃^-. The weak acid is the H₂CO₃.
    b. Yes it hydrolyzes.
₂₎
    a.HF + H₂O  H₃O⁺ + F^-
    b.Water is acting as a Bronsted-Lowry base because it is accepting a proton (H⁺) from the HF.
3)
    a. NH₄Br → NH₄⁺+ Br^-
    b. Br^- does not hydrolyze; it is an ion.
        NH₄⁺ +H₂O  H₃O⁺ +NH₃^-  <--- Hydrolysis of NH₄+
    c. HBr is a strong acid. Ammonia is a weak base. So NH₄Br  is made of a strong acid and weak base.
    d. Yes it hydrolyzes.
    e. Acidic
4)
    CH₃COO^- +H₂O  CH₃COOH + OH^-

	CH3COO-	CH3COOH	OH-
I	0.30M
C	-x	x	x
E	0.30 - x	x	x

K_b=       K_w          
      K_a of CH₃COOH
K_b= 1.0 x10^-14
      1.8 x 10^-5
K_b= 5.6 x 10^-10
K_b= [CH₃COOH][OH^-]
          [CH₃COO^-]
(Use the given K_b and the concentrations from the ICE table)
5.6 x 10^{-10 =     x2}    
                    0.30-x     (Assume x << 0.30)
x²= 1.68 x 10^-10
x=1.30 x 10^-5 = [OH^-]
pOH= -log (1.30 x 10^-5) = 4.89
pH = 14 - pOH
pH = 14.00 - 4.89 = 9.11
5)
    a. Yes it hydrolyzes
    b. Basic solution

Use of Hydrolysis in the "Real World"

In nature, living organisms are only able to live by processing fuel to make energy. The energy that is converted from food, is stored into ATP molecules (Adenosine Triphosphate). Life requires many processes in order to sustain itself such as cellular respiration, respiration, muscle contraction, distribution of hormones, transmittance of neuro-transmitters in the brain, etc. All of these important processes require an input of energy. To distribute this energy, the energy from the ATP molecules must be released. To release the energy stored in the bonds of ATP molecules, hydrolysis must occur to break a phosphate group off of an ATP molecule, thus releasing energy from the bonds. ATP now becomes ADP (Adenosine Diphosphate) from losing a phosphate group through hydrolysis.

Redox Reactions

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Redox reactions, or oxidation-reduction reactions, have a number of similarities to acid-base reactions. Fundamentally, redox reactions are a family of reactions that are concerned with the transfer of electrons between species. Like acid-base reactions, redox reactions are a matched set -- you don't have an oxidation reaction without a reduction reaction happening at the same time. Oxidation refers to the loss ofelectrons, while reduction refers to the gain of electrons. Each reaction by itself is called a "half-reaction", simply because we need two (2) half-reactions to form a whole reaction. In notating redox reactions, chemists typically write out the electrons explicitly:

Cu (s) ----> Cu²⁺ + 2 e^-

This half-reaction says that we have solid copper (with no charge) being oxidized (losing electrons) to form a copper ion with a plus 2 charge. Notice that, like the stoichiometry notation, we have a "balance" between both sides of the reaction. We have one (1) copper atom on both sides, and the charges balance as well. The symbol "e^-" represents a free electron with a negative charge that can now go out and reduce some other species, such as in the half-reaction:

2 Ag⁺ (aq) + 2 e^- ------> 2 Ag (s)

Here, two silver ions (silver with a positive charge) are being reduced through the addition of two (2) electrons to form solid silver. The abbreviations "aq" and "s" mean aqueous and solid, respectively. We can now combine the two (2) half-reactions to form a redox equation:

We can also discuss the individual components of these reactions as follows. If a chemical causes another substance to be oxidized, we call it the oxidizing agent. In the equation above, Ag⁺ is the oxidizing agent, because it causes Cu(s) to lose electrons. Oxidants get reduced in the process by a reducing agent. Cu(s) is, naturally, the reducing agent in this case, as it causes Ag⁺ to gain electrons.

As a summary, here are the steps to follow to balance a redox equation in acidic medium (add the starred step in a basic medium):

1.     Divide the equation into an oxidation half-reaction and a reduction half-reaction

2.     Balance these

o    Balance the elements other than H and O

o    Balance the O by adding H₂O

o    Balance the H by adding H⁺

o    Balance the charge by adding e^-

3.     Multiply each half-reaction by an integer such that the number of e^- lost in one equals the number gained in the other

4.     Combine the half-reactions and cancel

5.     **Add OH^- to each side until all H⁺ is gone and then cancel again**

In considering redox reactions, you must have some sense of the oxidation number (ON) of the compound. The oxidation number is defined as the effective charge on an atom in a compound, calculated according to a prescribed set of rules. An increase in oxidation number corresponds to oxidation, and a decrease to reduction. The oxidation number of a compound has some analogy to the pH and pK measurements found in acids and bases -- the oxidation number suggests the strength or tendency of the compound to be oxidized or reduced, to serve as an oxidizing agent or reducing agent. The rules are shown below. Go through them in the order given until you have an oxidation number assigned.

1.     For atoms in their elemental form, the oxidation number is 0

2.     For ions, the oxidation number is equal to their charge

3.     For single hydrogen, the number is usually +1 but in some cases it is -1

4.     For oxygen, the number is usually -2

5.     The sum of the oxidation number (ONs) of all the atoms in the molecule or ion is equal to its total charge.

As a side note, the term "oxidation", with its obvious root from the word "oxygen", assumes that oxygen has an oxidation number of -2. Using this as a benchmark, oxidation numbers were assigned to all other elements. For example, if we look at H₂O, and assign the value of -2 to the oxygen atom, the hydrogens must each have an oxidation number of +1 by default, since water is a neutral molecule. As an example, what is the oxidation number of sulfur in sulfur dioxide (SO₂)? Given that each oxygen atom has a -2 charge, and knowing that the molecule is neutral, the oxidation number for sulfur must be +4. What about for a sulfate ion (SO₄ with a total charge of -2)? Again, the charge of all the oxygen atoms is 4 x -2 = -8. Sulfur must then have an oxidation number of +6, since +6 + (-8) = -2, the total charge on the ion. Since the sulfur in sulfate has a higher oxidation number than in sulfur dioxide, it is said to be more highly oxidized.

Working with redox reactions is fundamentally a bookkeeping issue. You need to be able to account for all of the electrons as they transfer from one species to another. There are a number of rules and tricks for balancing redox reactions, but basically they all boil down to dealing with each of the two half-reactions individually. Consider for example the reaction of aluminum metal to form alumina (Al₂O₃). The unbalanced reaction is as follows:

Looking at each half reaction separately:

This reaction shows aluminum metal being oxidized to form an aluminum ion with a +3 charge. The half-reaction below shows oxygen being reduced to form two (2) oxygen ions, each with a charge of -2.

If we combine those two (2) half-reactions, we must make the number of electrons equal on both sides. The number 12 is a common multiple of three (3) and four (4), so we multiply the aluminum reaction by four (4) and the oxygen reaction by three (3) to get 12 electrons on both sides. Now, simply combine the reactions. Notice that we have 12 electrons on both sides, which cancel out. The final step is to combine the aluminum and oxygen ions on the right side using a cross multiply technique:

Taking care of the number of atoms, you should end up with: