ASAM BASA

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 dissociation of 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 <>
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 the molarity 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.

CHEMICAL NOMENCLATURE

Oxidation and Reduction

When forming compounds, it is important to know something about the way atoms will react with each other. One of the most important manners in which atoms and/or molecules react with each other is the oxidation/reduction reaction. Oxidation/Reduction reactions are the processes of losing and gaining electrons respectively. Just remember, "LEO the lion says GER:" Lose Electrons Oxidation, Gain Electrons Reduction. Oxidation numbers are assigned to atoms and compounds as a way to tell scientists where the electrons are in a reaction. It is often referred to as the "charge" on the atom or compound. The oxidation number is assigned according to a standard set of rules. They are as follows:

1. An atom of a pure element has an oxidation number of zero.
2. For single atoms in an ion, their oxidation number is equal to their charge.
3. Fluorine is always -1 in compounds.
4. Cl, Br, and I are always -1 in compounds except when they are combined with O or F.
5. H is normally +1 and O is normally -2.
6. The oxidation number of a compound is equal to the sum of the oxidation numbers for each atom in the compound.

Forming Ionic Compounds

Knowing the oxidation number of a compound is very important when discussing ionic compounds. Ionic compounds are combinations of positive and negative ions. They are generally formed when nonmetals and metals bond. To determine which substance is formed, we must use the charges of the ions involved. To make a neutral molecule, the positive charge of the cation (positively-charged ion) must equal the negative charge of the anion (negatively-charged ion). In order to create a neutral charged molecule, you must combine the atoms in certain proportions. Scientists use subscripts to identify how many of each atom makes up the molecule. For example, when combining magnesium and nitrogen we know that the magnesium ion has a "+2" charge and the nitrogen ion has a "-3" charge. To cancel these charges, we must have three magnesium atoms for every two nitrogen atoms:

3Mg²⁺ + 2N^3- --> Mg₃N₂

Knowledge of the charges of ions is crucial to knowing the formulas of the compounds formed.

• alkalis (1st column elements) form "+1" ions such as Na⁺ and Li⁺
• alkaline earth metals (2nd column elements) form "²⁺" ions such as Mg²⁺ and Ba²⁺
• halogens (7th column elements) form "-1" ions such as Cl^- and I^-

Other common ions are listed in the table below:

Positive ions (cations)	Negative ions (anions)
1+	1-
ammonium (NH4+)	acetate (C2H3O2-)
copper(I) (Cu+)	azide (N3-)
hydrogen (H+)	chlorate (ClO3-)
silver (Ag+)	cyanide (CN-)
	dihydrogen phosphate (H2PO4-)
2+	hydride (H-)
cadmium (Cd2+)	bicarbonate (HCO3-)
cobalt(II) (Co2+)	hydroxide (OH-)
copper(II) (Cu2+)	nitrate (NO3-)
iron (Fe2+)	nitrite (NO2-)
lead (Pb2+)	perchlorate (ClO4-)
manganese(II) (Mn2+)	permanganate (MnO4-)
mercury(I) (Hg22+)	thiocyanate(SCN-)
mercury(II) (Hg2+)
nickel (Ni2+)	2-
tin (Sn2+)	carbonate (CO32-)
zinc (Zn2+)	chromate (CrO42-)
	dichromate (Cr2O72-)
3+	hydrogen phosphate (HPO42-)
aluminum (Al3+)	oxide (O2-)
chromium(III) (Cr3+)	peroxide (O22-)
iron(III) (Fe3+)	sulfate (SO42-)
	sulfide (S2-)
	sulfite (SO32-)
	3-
	nitride (N3-)
	phosphate (PO43-)
	phosphide (P3-)

Naming Ionic Compounds

The outline below provides the rules for naming ionic compounds:

Positive Ions

1. Monatomic cations (a single atom with a positive charge) take the name of the element plus the word "ion"
   
   Examples:
   
   • Na⁺ = sodium ion
   • Zn⁺² = zinc ion
   
   
2. If an element can form more than one (1) positive ion, the charge is indicated by the Roman numeral in parentheses followed by the word "ion"
   
   Examples:
   • Fe²⁺ = iron(II) ion
   • Fe³⁺ = iron (III) ion

Negative Ions

1. Monatomic anions (a single atom with a negative charge) change their ending to "-ide"
   
   Examples:
   
   • O^2- = oxide ion
   • Cl^- = chloride ion
   
   
2. Oxoanions (negatively charged polyatomic ions which contain O) end in "-ate". However, if there is more than one oxyanion for a specific element then the endings are:
   
   

   Two less oxygen than the most common starts with "hypo-" and ends with "-ite"	One less oxygen than the most common ends with "-ite"	THE MOST COMMON OXOANION ENDS WITH "-ATE"	One more oxygen than the most common starts with "per-" and ends with "-ate"
   ClO- = hypochlorite	ClO2- = chlorite NO2- = nitrite SO32- = sulfite	Most common oxyanions with four oxygens SO42- = sulfate PO43- = phosphate CrO42- = chromate Most common oxyanions with three oxygens NO3- = nitrate ClO3- = chlorate CO32- = carbonate	ClO4- = perchlorate

   
   
3. Polyatomic anions (a negatively charged ion containing more than one type of element) often add a hydrogen atom; in this case, the anion's name either adds "hydrogen-" or "bi-" to the beginning
   
   Example:
   CO₃^2- becomes HCO₃^-
   "Carbonate" becomes either "Hydrogen Carbonate" or "Bicarbonate"
   
   
4. When combining cations and anions into an ionic compound, you always put the cation name first and then the anion name (the molecular formulas are also written in this order as well.)
   
   Examples:
   
   • Na⁺ + Cl^- --> NaCl
     sodium + chloride --> sodium chloride
     
   • Cu²⁺ + SO₄^2- -->CuSO₄
     copper(II) + sulfate --> copper(II) sulfate
     
   • Al³⁺ + 3NO₃^- --> Al(NO₃)₃
     aluminum + nitrate --> aluminum nitrate

Arrangement of Atoms

In naming ions, it is important to consider "isomers." Isomers are compounds with the same molecular formula, but different arrangements of atoms. Thus, it is important to include some signal within the name of the ion that identifies which arrangement you are talking about. There are three main types of classification, geometric, optical and structural isomers.

1. Geometric isomers refers to which side of the ion atoms lie. The prefixes used to distinguish geometric isomers are cis meaning substituents lie on the same side of the ion and trans meaning they lie on opposite sides. Below is a diagram to help you remember.
2. Optical isomers differ in the arrangement of four groups around a chiral carbon. These two isomers are differentiated as L and D.
3. Structural isomers differentiate between the placement of two chlorine atoms around a hexagonal carbon ring. These three isomers are identified as o, m, and p. Once again we have given you a few clues to help your memory.

A pop-up nomenclature calculator is available for help when naming compounds and for practice problems.

Naming Binary Molecular Compounds

Molecular compounds are formed from the covalent bonding between non-metallic elements. The nomenclature for these compounds is described in the following set of rules.

1. The more positive atom is written first (the atom which is the furthest to the left and to the bottom of the periodic table)
2. The more negative second atom has an "-ide" ending.
3. Each prefix indicates the number of each atom present in the compound.
   

   Number of Atoms	Prefix	Number of Atoms	Prefix
   1	mono	6	hexa
   2	di	7	hepta
   3	tri	8	octa
   4	tetra	9	nona
   5	penta	10	deca

   
   Examples:
   CO₂ = carbon dioxide
   P₄S₁₀ = tetraphosphorus decasulfide

Naming Inorganic Acids

1. Binary acids (H plus a nonmetal element) are acids that dissociate into hydrogen atoms and anions in water. Acids that only release one hydrogen atom are known as monoprotic. Those acids that release more than one hydrogen atom are called polyproticacids. When naming these binary acids, you merely add "hydro-" (denoting the presence of a hydrogen atom) to the beginning and "-ic acid" to the end of the anion name.
   
   Examples:
   HCl = hydrochloric acid
   HBr = hydrobromic acid
   
2. Ternary acids (also called oxoacids, are formed by hydrogen plus another element plus oxygen) are based on the name of the anion. In this case, the -ate, and -ite suffixes for the anion are replaced with -ic and -ous respectively. The new anion name is then followed by the word "acid." The chart below depicts the changes in nomenclature.
   
   

   Anion name	Acid name
   hypo___ite	hypo___ous acid
   ___ite	___ous acid
   ___ate	___ic acid
   per___ate	per___ic acid

   
   Example:
   ClO₄^- to HClO₄ => perchlorate to perchloric acid
   ClO^- to HClO => hypochlorite to hypochlorous acid

Naming Compounds

A detailed treatise on naming organic compounds is beyond the scope of these materials, but some basics are presented. The wise chemistry student should consider memorizing the prefixes of the first ten organic compounds:

Number of Carbons	Prefix
1	meth-
2	eth-
3	prop-
4	but-
5	pent-
6	hex-
7	hept-
8	oct-
9	non-
10	dec-

There are four basic types of organic hydrocarbons, those chemicals with only carbon and hydrogen:

1. Single bonds (alkane): suffix is "ane", formula C_nH_2n+2
2. Double bonds (alkene): suffix is "ene", formula C_nH_2n
3. Triple bonds (alkyne): suffix is "yne", formula C_nH_2n-2
4. Cyclic compounds: use prefix "cyclo"

   So, for example, an organic compound with the formula "C₆H₁₄" would be recognized as an alkane with six carbons, so its name is "hexane".

   
   Examples:
   N₂O₄ = dinitrogen tetraoxide
   S₂F₁₀ = disulfur decafluoride