Name: Class Notes #____
Acids and Bases Notes
I. Strength of Acids and Bases
A. Bases: Strong ______: metal hydroxides of Group I and II metals (except Be) that are soluble in water and dissociate (separates into ions) completely in dilute aqueous solutions
Weak Bases: a molecular substance that ionizes only slightly in water to produce an alkaline (basic) solution (ex. NH3)
B. Acids: Strong ______: an acid that ionizes (separates into ions) completely or very nearly completely in aqueous solutions (will not have an ionization constant (Ka) value).
Weak Acids: an acid that ionizes only slightly in dilute aqueous solutions (will have an ionization constant (Ka) value).
1. ______or hydrohalic acids – HCl, HBr, and HI “hydro____ic acid” are strong acids. Other binary acids are weak acids (HF and H2S). Although the H-F bond is very polar, the bond is so strong (due to the small F atom) that the acid does not completely ionize.
2. ______– contain a polyatomic ion
a. strong acids (contain 2 or more oxygen per hydrogen)
HNO3 – nitric from nitrate
H2SO4 - sulfuric from sulfate
HClO4 - perchloric from perchlorate
b. weak acids (acids with l less oxygen than the “ic” ending)
HNO2 – nitrous from nitrite
H3PO3 - phosphorous from phosphite
H2SO3 - sulfurous from sulfite
HClO2 - chlorous from chlorite
c. weaker acids (acids with “hypo ous” have less oxygen than the “ous” ending)
HNO - hyponitrous
H3PO2 - hypophosphorus
HClO - hypochorous
O
׀
d. Organic acids – have carboxyl group -C-OH - usually weak acids
HC2H3O2 - acetic acid
C7H5COOH - benzoic acid
II. Characteristics of Acids and Bases (page 453-458)
There are multiple definitions for acids and bases and I have summarized them all here. Keep in mind that there are some exceptions and this is just a guide to follow when trying to determine if a substance is an acid or a base.
ACDIS BASES
-usually begin with ______-usually contain ______(if they don’t they produce OH¯ in water)
-______taste -______taste
-litmus paper turns ______-litmus paper turns ______
-pH paper ______-pH paper ______
-Phenolphthalein – ______- Phenolphthalein – ______
-feels like water -feels slippery
-react with metals to produce hydrogen gas
-______-______
-pH < ______-pH > ______
-Arrhenius Acid: donates (or produces) hydronium -Arrhenius Base: donates (or produces) hydroxide
ions (H3O+) in water or hydrogen ions (H+) in water ions (OH-) in water
-Bronsted-Lowry Acid: donates a proton (H+) in water, -Bronsted-Lowry Base: accepts a proton in water,
H3O+ has an extra H+, if it donated it to another molecule OH- needs an extra H+ if it accepts one from
it would be H2O (page 467) another molecule it would be H2O (page 468)
HNO3 + H2O ® H+ + NO3- KOH + H2O ® K+ + OH-
HNO3 + H2O ® H3O+ + NO3- NH3 + H2O ® NH4+ + OH-
HCl + H2O ® H+ + Cl-
HCl + H2O ® H3O+ + Cl-
-Lewis Acid: not all acids contain H, any atom, ion, -Lewis Base: not all bases contain OH, any atom,
or molecule that accepts an electron pair to form a ion, or molecule that donates an electron pair to form
covalent bond is an acid a covalent bond is a base
Conjugates
HF + H2O Û F- + H3O+
Acid Base Conjugate Base Conjugate Acid
Here HF donated a proton (H+) to the water and the water accepted the proton (H+).
HF is referred to as the ______and water is referred to as the ______.
The fluorine ion, F- is referred to as the ______of HF. F- can accept a proton (H+) to be stable.
The hydronium ion, H3O+ is referred to as the ______of water. H3O+ can donate a proton (H+) to be stable.
Example: 1. Determine the acid, base, conjugate acid, and conjugate base in each of the following equations:
HCl + H2O Þ Cl- + H3O+
H2SO4 + H2O Þ HSO4- + H3O+
NH3 + H2O Þ OH- + NH4+
2. What is the conjugate base of the following substances?
a. H2O ______b. NH4+______c. HNO2______d. HC2H3O2______
3. What is the conjugate acid of the following substances?
a. HCO3-______b. H2O______c. HPO42-______d. NH3______
III. Water
Water is an amphoteric substance (it can act as an acid or as a base).
Autoionization of water:
H2O + H2O ® H3O+ + OH-
Ionization constant for water (Kw) Kw = [H30+][OH-] or Kw = [H+][OH-]
Because in water at 25°C [H+] = l.0 X l0-7 M and [OH-] = l.0 X l0-7 M the Kw = 1.0 X l0-14 mol2/L2
No matter what an aqueous solution contains, at 25°C [H+] [OH-] = l.0 x l0-14 mol2/L2
Kw varies with temperature.
Neutral solution [H+] = [OH-] Acidic solution [H+] > [OH-] Basic solution [H+] < [OH-]
Kw Problems: If the molarity of a strong acidic or basic is known we can use the Kw to find the [H+] or [OH-].
Kw = [H+][OH-] or l.0 x l0-14 mol2/L2 = [H+] [OH-]
1. Determine the hydrogen ion and hydroxide ion concentrations in a solution that is 3.0X10-2M NaOH.
2. Determine the hydrogen ion and hydroxide ion concentrations in a solution that is 4.0X10-3M HNO3.
IV. pH Scale [ ] – brackets mean concentration or Molarity
1. The pH scale indicates the ______ion concentration, [H3O+], of a solution. (In other words how
many H3O+ ions are in a solution. If there are a lot we assume it is an acid, if there are very few it is a base.)
pH
acidic basic
[H3O+] 1 0.1 0.01 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14
a lot of H3O+ not a lot of H3O+
2. pOH Scale
The pOH scale indicates the ______ion concentration, [OH-], of a solution. (In other words how many OH- ions
are in a solution. If there are a lot we assume it is a base, if there are very few it is an acid.)
pOH
basic acidic
[OH-] 1 0.1 0.01 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14
a lot of OH- not a lot of OH-
Example: 1. Lemon juice (citric acid) pH = 2.0, pOH = ______2. Pure water pH = 7.0, pOH = ______
3. Milk of magnesia pH = 10.0, pOH = ______
3. Calculations Involving pH, pOH, [H3O+], & [OH-] of Strong Acids and Strong Bases
1st determine which ion will be produced, either OH- or H3O+ (Acids produce H3O+and bases produce OH-.) 2nd
use formula to determine pOH or pH. 3rd check if answer is reasonable
pH = -log [H3O+] pOH = -log [OH-] pOH + pH = 14
pH = -log [H+]
Example Problems:
1. What is the pH of a 0.001M NaOH solution?
1st step: Hydroxide will be produced and the [OH-] = 0.001M
2nd step: pOH = -log [0.001]
pOH = 3.0
pH = 14.0-3.0 = 11.0
(Significant figures and pH: When conuting significant figures, only count what comes APTER the decimal in a pH or pOH value. For example the number of sig figs in a pH of 12.01 is 2, the number of sig figs in a pH of 12.0 is 1. This is only for pH and pOH values!!!)
2. What is the pH of a 3.4X10-5M HCl solution?
3. What is the pH of a solution if the pOH = 5?
4. What is the pH of a 10 liter KOH solution if 5.611 grams of KOH were used to prepare the solution?
5. What is the pOH of a 1.1X10-5M HNO3 solution?
6. If the pH of a KOH solution is 10.75, what is the molar concentration of the solution? What is the pOH? What is the [H+]?
The pH of a strong acid cannot be greater than 7. If the acid concentration [H3O+] is less than 1.0X10-7, the water becomes the important source of [H3O+] or [H+] and the pH is 7.00. Just remember to check if you answer is reasonable!
7. What is the pH of a 2.5X10-10M HCl solution?
8. What is the pH of a 1.0X10-11M HNO3 solution?
4. Ka and Kb and Calculations Involving pH of Weak Acids and Weak Bases
A. Ka and Weak acids - Very little of a weak acid ionizes. Most of it stays in its molecular form:
HC2H3O2 + H20 = H30+ + C2H3O2-
Ionization Constants for Weak Acids (Ka)
For the reaction: HA + H20 à H30+ + A- , then:
[H30+][A-] or [H+][A-]
Ka = ------Ka = ------
[HA] [HA]
NOTE: H20 is not part of the equilibrium equation.
Example: HC2H3O2 + H20 = H30+ + C2H3O2-
Ka = ------
Weak acid strength is compared by the Ka values of the acids. The smaller the Ka, the weaker the acid. (see handout of Ka and Kb values) Strong acids do not have Ka values because the [HA] is so small that the Ka for the acid cannot be measured. Remember that strong acids completely ionize in water so there would be almost no unionized [HA].
B. Kb and Weak Bases - ionize only slightly in water to produce an alkaline (basic) solution (ex. NH3):
NH3 + H20 = NH4+ + OH-
Ionization Constants for Weak Bases (Kb)
For the reaction: B + H20 à BH+ + OH- , then:
[BH+][OH-]
Kb = ------
[B]
Example: NH3 + H20 = NH4+ + OH-
Kb = ------
C. Calculations Involving pH of Weak Acids - RICE
Calculating the pH of weak acids involves setting up an ______. Always start by writing the equation, setting up the acid equilibrium expression (Ka), defining initial concentrations, changes, and final concentrations in terms of x, substituting values and variables into the Ka expression and solving for x. The RICE method will be used.
There are two types of weak acids: (l) those that ionize less than 5.00 % and (2) those that ionize more than 5.00 %. To find the percent of ionization:
[concentration of ions]
% ionization = ------X 100
[concentration of acid]
When the Ka of an acid is about l0-5, this usually means they are ionizing less than 5.00%, unless it is a very dilute solution.
Example: Calculate the pH of a 3.5 M solution of acetic acid.
HC2H302 is a weak acid and in water:
Reaction: HC2H302 + H20 = H30+ + C2H302-
Initial 3.5 0 0
Change -x +x +x where x = mol/L or concentration
Equilibrium 3.5 – x x x of acid that ionizes
[H30+][C2H302-] (x) (x)
Ka = ------so l.8 x l0-5 = ------
[HC2H302] (3.5 – x) neglect x
Because Ka is so small (l.8 x l0-5), x is going to be very small compared to the concentration of HC2H3O2, so we can approximate [HC2H3O2 – x] as just [HC2H3O2].
Solve for x: l.8 x l0-5 = x2/3.5 and x = 7.94 x l0-3 M where x = [C2H302-] = [H30+]. Before going on, make sure it was okay to neglect x.
X 7.94 x l0-3
% ionization = ------= ------x l00 = 0.23 %
[HC2H302] 3.5
This is obviously less than 5.0 % so it was okay to ignore the x. Now to find the pH.
The pH is due to the [H30+] so pH = -log [7.94 x l0-3] and pH = 2.10.
Practice: Find the pH of a l.50 M solution of benzoic acid.
Example of when you CANNOT neglect x: Calculate the pH of a l.00 x l0-4 M solution of acetic acid. (Ka of acetic acid = l.8 x l0-5)
[H30+][C2H3O2-]
HC2H3O2 + H20 = H30+ + C2H302- Ka = ------= l.8 x l0-5
[HC2H302]
Reaction: HC2H3O2 + H20 = H30+ + C2H3O2-
Initial l.00 x l0-4 0 0
Change -x +x +x
Equilibrium l.00 x l0-4 –x x x
(x) (x)
l.8 x l0-5 = ------x = 4.2 x l0-5
l.00 x l0-4 –x
4.2 x l0-5
% ionization = ------x l00 = 42% Ionization is > 5.00% so x cannot be neglected and the
l.00 x l0-4 quadratic formula must be used.
Quadratic Formula: ax2 + bx + c = 0
x = -b ± √ (b2 – 4ac)
2a
x2 + l.8 x l0-5 x – l.8 x l0-9 = 0
x = -l.8 x l0-5 ± √ (l.8 x l0-5)2 – 4(l)(-l.8 x l0-9) x = 3.5 x l0-5 and –5.2 x l0-5
2(l)
Since a concentration cannot be negative, then x = 3.5 x l0-5 M. x = [H30+] = 3.5 x l0-5 and pH = -log 3.5 x l0-5 = 4.46
Practice (when x cannot be neglected): Calculate the pH of a 2.35 x l0-4 M solution of acetic acid. (Ka of acetic acid = l.8 x l0-5)
D. Calculations Involving pH of Weak Bases - RICE
Determination of the pH of a weak base is very similar to the determination of the pH of a weak acid. Follow the same steps - RICE. Remember, however, that x is the [OH-].
Practice: Calculate the [OH-] and the pH for a 15.0 M NH3 solution. The Kb for NH3 is l.8 x l0-5
2. What is the pH of a 3.6 M methylamine solution?
a. What is the % ionization?
V. Acid Rain
Many industrial processes produce gases such as NO, NO2, CO2, SO2, and SO3. These compounds can dissolve in atmospheric water to produce acidic solutions that fall to the ground in the form of rain or snow. Marble found in many buildings and statues is composed of calcium carbonate, when acid snow or rain falls on these structures a great deal of damage is caused. (page 475)
VI. Neutralization Reactions
______– the reaction of an acid with a base to produce water and a salt. (This occurs when H3O+ and OH- ions are supplied in equal numbers by the reactants.)
______- an ionic compound composed of a cation from a base and an anion from an acid.
Example: HCl + NaOH à HOH + NaCl
Complete ionic equation: H+ + Cl- + Na+ + OH- à HOH(l) + Na+ + Cl-
Net ionic equation: H+ + OH- à HOH
Practice: Write net ionic equations for the following.
1. H2SO4 + KOH ® K2SO4 + H2O
2. HCl + NaOH ® NaCl + H2O
3. phosphoric acid and ammonium hydroxide
VII. Titrations
______– A neutralization reaction of an acid by a base or vice versa; it is usually used to find the concentration, molarity, of an unknown acid or base. The concentration of the other is known.