How to Balance Redox Reactions

To balance redox reactions, assign oxidation numbers to the reactants and products to determine how many moles of each species are needed to conserve mass and charge. First, separate the equation into two half-reactions, the oxidation portion and the reduction portion. This is called the half-reaction method of balancing redox reactions or the ion-electron method. Each half-reaction is balanced separately and then the equations are added together to give a balanced overall reaction. We want the net charge and number of ions to be equal on both sides of the final balanced equation.

For this example, let's consider a redox reaction between KMnO₄and HI in an acidic solution:

MnO₄^- + I^- → I₂ + Mn²⁺

 Separate the two half reactions:

I^- → I₂

MnO₄^- → Mn²⁺

 To balance the atoms of each half-reaction, first balance all of the atoms except H and O. For an acidic solution, next add H₂O to balance the O atoms and H⁺ to balance the H atoms. In a basic solution, we would use OH^- and H₂O to balance the O and H.

Balance the iodine atoms:

2 I^- → I₂

The Mn in the permanganate reaction is already balanced, so let's balance the oxygen:

MnO₄^- → Mn²⁺ + 4 H₂O

Add H⁺ to balance the 4 waters molecules:

MnO₄^- + 8 H⁺ → Mn²⁺ + 4 H₂O

The two half-reactions are now balanced for atoms:

MnO₄^- + 8 H⁺ → Mn²⁺ + 4 H₂O

 Next, balance the charges in each half-reaction so that the reduction half-reaction consumes the same number of electrons as the oxidation half-reaction supplies. This is accomplished by adding electrons to the reactions:

2 I^- → I₂ + 2e^-

5 e^- + 8 H⁺ + MnO₄^- → Mn²⁺ + 4 H₂O

Now multiple the oxidations numbers so that the two half-reactions will have the same number of electrons and can cancel each other out:

5(2I^- → I₂ +2e^-)

2(5e^- + 8H⁺ + MnO₄^- → Mn²⁺ + 4H₂O)

 Now add the two half-reactions:

10 I^- → 5 I₂ + 10 e^-

16 H⁺ + 2 MnO₄^- + 10 e^- → 2 Mn²⁺ + 8 H₂O

This yields the following final equation:

10 I^- + 10 e^- + 16 H⁺ + 2 MnO₄^- → 5 I₂ + 2 Mn²⁺ + 10 e^- + 8 H₂O

Get the overall equation by canceling out the electrons and H₂O, H⁺, and OH^- that may appear on both sides of the equation:

10 I^- + 16 H⁺ + 2 MnO₄^- → 5 I₂ + 2 Mn²⁺ + 8 H₂O

 Check your numbers to make certain that the mass and charge are balanced. In this example, the atoms are now stoichiometrically balanced with a +4 net charge on each side of the reaction.

Acids and Bases xD

Water

We typically talk about acid-base reactions in aqueous-phase environments -- that is, in the presence of water. The most fundamental acid-base reaction is the dissociationof water:

H2O  H+ + OH-

In this reaction, water breaks apart to form a hydrogen ion (H⁺) and a hydroxide ion (OH^-). In pure water, we can define a special equilibrium constant (K_w) as follows:

KW = [H+][OH-] = 1.00x10-14

Where K_w is the equilibrium constant for water at 25° C (unitless)
[H⁺] is the molar concentration of hydrogen
[OH^-] is the molar concentration of hydroxide
An equilibrium constant less than one (1) suggests that the reaction prefers to stay on the side of the reactants -- in this case, water likes to stay as water. Because water hardly ionizes, it is a very poor conductor of electricity.

pH

What is of interest in this reading, however, is the acid-base nature of a substance like water. Water actually behaves both like an acid and a base. The acidity or basicity of a substance is defined most typically by the pH value, defined as below:



pH = -log[H⁺]
At equilibrium, the concentration of H⁺ is 1.00 × 10^-7, so we can calculate the pH of water at equilibrium as:



pH = -log[H⁺]= -log[1.00 × 10^-7] = 7.00
Solutions with a pH of seven (7) are said to be neutral, while those with pH values below seven (7) are defined as acidic and those above pH of seven (7) as being basic.
pOH gives us another way to measure the acidity of a solution. It is just the opposite of pH. A high pOH means the solution is acidic while a low pOH means the solution is basic.



pOH = -log[OH^-]pH + pOH = 14.00 at 25°C

Definitions of acids and bases

• Arrhenius
  acid: generates [H⁺] in solution
  base: generates [OH^-] in solution
  normal Arrhenius equation: acid + base  salt + water
  example: HCl(aq) + NaOH(aq)  NaCl(aq) + H₂O(l)

• Brønsted-Lowery:
  acid: anything that donates a [H⁺] (proton donor)
  base: anything that accepts a [H⁺] (proton acceptor)
  normal Brønsted-Lowery equation: acid + base  acid + base
  example: HNO₂(aq) + H₂O(aq)  NO₂^-(aq)+ H₃O⁺(aq)
  Each acid has a conjugate base and each base has a conjugate acid. These conjugate pairs only differ by a proton. In this example: NO₂^- is the conjugate base of the acid HNO₂ and H₃O⁺ is the conjugate acid of the base H₂O.

• Lewis:
  acid: accepts an electron pair
  base: donates an electron pair
  The advantage of this theory is that many more reactions can be considered acid-base reactions because they do not have to occur in solution.

Salts

A salt is formed when an acid and a base are mixed and the acid releases H⁺ ions while the base releases OH^- ions. This process is called hydrolysis. The pH of the salt depends on the strengths of the original acids and bases:

Acid	Base	Salt pH
strong	strong	pH = 7
weak	strong	pH > 7
strong	weak	pH < 7
weak	weak	depends on which is stronger

These salts are acidic or basic due to their acidic or basic ions. When weak acids or weak bases react with water, they make strong conjugate bases or conjugate acids, respectively, which determines the pH of the salt.

Acid-Base Character

For a molecule with a H-X bond to be an acid, the hydrogen must have a positive oxidation number so it can ionize to form a positive +1 ion. For instance, in sodium hydride (NaH) the hydrogen has a -1 charge so it is not an acid but it is actually a base. Molecules like CH₄ with nonpolar bonds also cannot be acids because the H does not ionize. Molecules with strong bonds (large electronegativity differences), are less likely to be strong acids because they do not ionize very well. For a molecule with an X-O-H bond (also called an oxoacid) to be an acid, the hydrogen must again ionize to form H⁺. To be a base, the O-H must break off to form the hydroxide ion (OH^-). Both of these happen when dealing with oxoacids.Strong Acids: These acids completely ionize in solution so they are always represented in chemical equations in their ionized form. There are only seven (7) strong acids:



HCl, HBr, HI, H₂SO₄, HNO₃, HClO₃, HClO₄
To calculate a pH value, it is easiest to follow the standard "Start, Change, Equilibrium" process.
Example Problem: Determine the pH of a 0.25 M solution of HBr.

Answer:
	HBr (aq)	→	H+(aq)	+	Br-(aq)
Start:	.25 M		0 M		0 M
Change:	-.25		+.25		+.25
Equilibrium:	0		.25		.25
	pH = -log[H+] = -log(.25) = 0.60

Weak Acids: These are the most common type of acids. They follow the equation:



HA(aq)  H⁺(aq) + A^-(aq)
The equilibrium constant for the dissociation of an acid is known as K_a. The larger the value of K_a, the stronger the acid.

Ka =	[H+][A-] [HA]

Example Problem: Determine the pH of 0.30 M acetic acid (HC₂H₃O₂) with the K_a of 1.8x10^-5.

Answer:

Write an equilibrium equation for the acid:

HC2H3O2  H+ + C2H3O2-

Write the equilibrium expression and the Ka value:

Ka = [H+][C2H3O2-] [HC2H3O2] = 1.8x10-5

"Start, Change, Equilibrium":

HC2H3O2 H+ + C2H3O2-     Start: 0.30 M 0 M 0 M     Change: -x +x +x     Equilibrium: 0.30 - x x x

Substitute the variables (disregard the "-x" because it is so small compared to the 0.30) and solve for [H+]:

Ka = 1.8x10-5 = (x)(x) (.30 - x) = x2 .30

x = [H+] = 2.3x10-3

pH = -log[H+] = 2.64

Strong Bases: Like strong acids, these bases completely ionize in solution and are always represented in their ionized form in chemical equations. There are only eight (8) strong bases:



LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)₂, Sr(OH)₂, Ba(OH)₂
Example Problem: Determine the pH of a 0.010 M solution of Ba(OH)₂.

Answer:
	Ba(OH)2(aq)	→	Ba2+(aq)	+	2OH-(aq)
Start:	.010 M		0 M		0 M
Change:	-.010		+.010		+.010
Equilibrium:	0		.010		.010
pOH = -log[OH-] = -log(.010) = 1.70
pH = 14.00 - 1.70 = 12.30

Weak Bases: These follow the equation:



Weak Base + H₂O  conjugate acid + OH^-example: NH₃ + H₂O  NH₄⁺ + OH^-

K_b is the base-dissociation constant:

Kb =	[conjugate acid][OH-] [weak base][H2O]

example: Kb =

[NH4+][OH-] [NH3[H2O]

K_a x K_b = K_w = 1.00x10^-14

To calculate the pH of a weak base, we must follow a very similar "Start, Change, Equilibrium" process as we did with the weak acid, however we must add a few steps.
Example Problem: Determine the pH of 0.15 M ammonia (NH₃) with a K_b=1.8x10^-5.

Answer:

Write the equilibrium equation for the base:

NH3 + H2O  NH4+ + OH-

Write the equilibrium expression and the Kb value:

Kb = [NH4+][OH2-] [NH3][H2O] = 1.8x10-5

"Start, Change, Equilibrium":

NH3 + H2O NH4+ + OH-     Start: 0.15 M -- 0 M 0 M     Change: -x -- +x +x     Equilibrium: 0.15 - x -- x x

Substitute the variables (disregard the "-x" because it is so small compared to the 0.15) and solve for [OH-]:

Kb = 1.8x10-5 = (x)(x) (.15 - x) = x2 .15

x = [OH-] = 1.6x10-3 M

pOH = -log[OH-] = 2.80

pH = 14.00 - 2.80 - 11.20

When dealing with weak acids and weak bases, you also might have to deal with the "common ion effect". This is when you add a salt to a weak acid or base that contains one of the ions present in the acid or base. To be able to use the same process to solve for pH when this occurs, all you need to change are your "start" numbers. Add themolarity of the ion, which comes from the salt, and then solve the K_a or K_b equation as you did earlier.


Example Problem: Find the pH of a solution formed by dissolving 0.100 mol of HC₂H₃O₂ with a K_a of 1.8x10^-8 and 0.200 mol of NaC₂H₃O₂ in a total volume of 1.00 L.

Answer:
	HC2H3O2(aq)	→	H+(aq)	+	C2H3O2-(aq)
Start:	.10 M		0 M		.20 M
Change:	-x		+x		+x
Equilibrium:	.10 - x		x		.20 + x
Ka = 1.8x10-8 = (x)(.20 + x) (.10 - x) = (x)(.20) (.10)
x = [H+] = -9.0x10-9
pH = -log(9.0x10-9) = 8.05

Acid-Base Titrations

An acid-base titration is when you add a base to an acid until the equivalence point is reached which is where the number of moles of acid equals the number of moles of base. For the titration of a strong base and a strong acid, this equivalence point is reached when the pH of the solution is seven (7) as seen on the following titration curve:
For the titration of a strong base with a weak acid, the equivalence point is reached when the pH is greater than seven (7). The half equivalence point is when half of the total amount of base needed to neutralize the acid has been added. It is at this point where the pH = pK_a of the weak acid.
In an acid-base titration, the base will react with the weak acid and form a solution that contains the weak acid and its conjugate base until the acid is completely gone. To solve these types of problems, we will use the K_a value of the weak acid and the molarities in a similar way as we have before. Before demonstrating this way, let us first examine a short cut, called the Henderson-Hasselbalch Equation. This can only be used when you have some acid and some conjugate base in your solution. If you only have acid, then you must do a pure K_a problem and if you only have base (like when the titration is complete) then you must do a K_b problem.

pH = pKa + log

[base] [acid]

Where:
pH is the log of the molar concentration of the hydrogen
pK_a is the equilibrium dissociation constant for an acid
[base] is the molar concentration of a basic solution
[acid] is the molar concentration of an acidic solution
Example Problem: 25.0 mL of 0.400 M KOH is added to 100. mL of 0.150 M benzoic acid, HC₇H₅O₂ (K_a=6.3x10^-5). Determine the pH of the solution.

Answer:

Determine where in the titration we are:

0.400 M x 0.025 L = 0.0100 mol KOH added

0.150 M x 0.100 L = 0.0150 mol HC7H5O2 originally

because only 0.100 mol of base has been added, that means the thitration is not complete; this means there are two ways to solve this problem: the normal way and the way using the Henderson-Hasselbalch Equation.

Normal way:

HC7H5O2 + OH- C7H5O2- + H2O     before reaction: 0.015 mol 0.0100 mol 0 mol --     change: -0.0100 -0.0100 +0.0100 --     after reaction: 0.0050 0 0.0100 --

Ka = [H+][C7H5O2-] [HC7H5O2] = 6.3x10-5 = (x)(0.0100) 0.0050

x = [H+] = 3.2x10-5 M

pH = -log(3.2x10-5) = 4.49

Henderson-Hasselbalch Way:

[HC7H5O2] = 0.0050 mol 0.125 L = 0.040 M

[C7H5O2-] = 0.0100 mol 0.125 L = 0.0800 M

pH = pKa + log [base] [acid]

pH = -log(6.3x10-5) + log 0.0800 0.0400 = 4.20 + 0.30 = 4.50

This equation is used frequently when trying to find the pH of buffer solutions. A buffer solution is one that resists changes in pH upon the addition of small amounts of an acid or a base. They are made up of a conjugate acid-base pair such as HC₂H₃O₂/C₂H₃O₂^- or NH₄⁺/NH₃. They work because the acidic species neutralize the OH^- ions while the basic species neutralize the H⁺ ions. The buffer capacity is the amount of acid or base the buffer can neutralize before the pH begins to change to a significant degree. This depends on the amount of acid or base in the buffer. High buffering capacities come from solutions with high concentrations of the acid and the base and where these concentrations are similar in value.


Practice weak acid problem:
C₆H₅COONa is a salt of a weak acid C₆H₅COOH. A 0.10 M solution of C₆H₅COONa has a pH of 8.60.

1. Calculate [OH^-] of C₆H₅COONa
2. Calculate K_b for: C₆H₅COO^- + H₂O <---> C₆H₅COOH + OH^-
3. Calculate K_a for C₆H₅COOH

See the weak acid solution.


Practice titration problem:
20.00 mL of 0.160 M HC₂H₃O₂ (K_a=1.8x10^-5) is titrated with 0.200 M NaOH.

1. What is the pH of the solution before the titration begins?
2. What is the pH after 8.00 mL of NaOH has been added?
3. What is the pH at the equivalence point?
4. What is the pH after 20.00 mL of NaOH has been added?

See the titration solution.

Lets start things off easy :)

Chemistry

      -  is the science of matter and the changes it undergoes. The science of matter is also addressed by physics, but while physics takes a more general and fundamental approach, chemistry is more specialized, being concerned with the composition, behavior, structure, and properties of matter, as well as the changes it undergoes during chemical reactions. It is a physical science which studies various substances, atoms, molecules, crystals and other aggregates of matter whether in isolation or combination, and which incorporates the concepts of energy and entropy in relation to the spontaneity of chemical processes.

Matter

      - is a general term for the substance of which all physical objects consist. Typically, matter includes atoms and other particles which have mass. A common way of defining matter is as anything that has mass and occupies volume. In practice however there is no single correct scientific meaning of "matter," as different fields use the term in different and sometimes incompatible ways.

Phases of Matter

      - phases of matter can also be described as the states of matter. there are many phases or states of matter such as:

a. Solid

      - It is characterized by structural rigidity and resistance to changes of shape or volume. Unlike a liquid, a solid object does not flow to take on the shape of its container, nor does it expand to fill the entire volume available to it like a gas does. The atoms in a solid are tightly bound to each other, either in a regular geometric lattice (crystalline solids, which include metals and ordinary water ice) or irregularly (an amorphous solid such as common window glass).

b. Liquid

      - It is able to flow and take the shape of a container. Some liquids resist compression, while others can be compressed. Unlike a gas, a liquid does not disperse to fill every space of a container, and maintains a fairly constant density. A distinctive property of the liquid state is surface tension, leading to wetting phenomena.

c. Gas

      - It is made up of individual atoms, elemental molecules made from one type of atom (e.g. oxygen), or compound molecules made from a variety of atoms (e.g. carbon dioxide). A gas mixture would contain a variety of pure gases much like the air. What distinguishes a gas from liquids and solids is the vast separation of the individual gas particles. This separation usually makes a colorless gas invisible to the human observer. The interaction of gas particles in the presence of electric and gravitational fields are considered negligible as indicated by the constant velocity vectors in the image.