Acids and Bases Booklet
This package was designed to save time, in order to allow for more
exam preparation time, if any, at the end of the course.
You can purchase a booklet for $2.00, and use it as your main source
out notes. (There may be cases when additional notes and problems are
needed)
You can also sign the booklet out, like a textbook. However, all notes
and problems would need to be done in you regular notes binder. It is
your choice!
Acids and bases introduction:
1) Review of naming and writing chemical formulas for acids
• Complete review sheet (Important review)
2) Operation definitions: One based on the behaviour and
properties of something.
Give operational definition of an acid (AWC pg. 431)
Give operational definition of a base ( AWC pg. 431-432)
3) AWC page 432-433
Define the following terms and be able to understand them
for the beginning of the unit.
A) self ionization of water
B) Solvated ion
C) Hydronium ion
D) Neutral solution
E) Acidic solution
F) Basic solution
G) Ion-product constant for water
Type 4)
Acids
HCl ---Æ H+ (aq)
+
(aq) sign and contain hydrogen.
Cl-(aq)
General formula of an acid is HX (aq)
Where x is a monatomic or polyatomic ion
Anion ending example
-ide
Cl -ite
SO3 2-ate
NO3 -
naming rule
(1) hydro
ic
(2) ____ous
(3) _____ ic
example
HCl (aq)
H2SO3 (aq)
HNO3 (aq)
hydrochloric acid
sulfurous acid
nitric acid
As you can see, the negative ion determines the name given to the acid. It
tells you the rule to use.
Examples:
HClO(aq)-
hypochlorous acid (Rule 2)
ClO- is the hypochlorite ion
HCN(aq)-
hydrocyanic acid (Rule 1)
CN- is the cyanide ion – “ide”
H3PO4(aq)-
phosphoric acid
(Rule 3)
PO4-3 is the phosphate ion
*For acids, we must know how to (A) write chemical formulas and also (B)
name acids.
Acid Naming review:
1. acetic acid
2. benzoic acid
3. boric acid
4. carbonic acid
5. chloric acid
6. chlorous acid
7. perchloric acid
8. nitric acid
9. nitrous acid
10. chromic acid
11. HClO(aq)
12. HClO3(aq)
13. HCN(aq)
14. H2SO3(aq)
15. H2SiO3(aq)
16. H3BO3(aq)
17. HCl(aq)
18. H2CO3(aq)
19. HF(aq)
20. CH3COOH(aq)
Formulas and Naming Acids
If time allows and the need is present, I may request that this review
sheet, one we did in chem..11, be completed to provide more practice
with acid naming and formulas.
Name the pure substance as a hydrogen compound and as an acid when it is
mixed with water
HBr
HBr (aq)
HClO3
HClO3(aq)
H2S
H2S(aq)
H2CrO4
H2CrO4(aq)
H2SO3
H2SO3(aq)
H3PO4
H3PO4(aq)
H2C2O4
H2C2O4(aq)
H2CO3
H2CO3(aq)
HCl
HCl(aq)
HNO2
HNO2(aq)
H2Se
H2Se (aq)
HClO2
HClO2(aq)
H2S2O3
H2S2O3(aq)
HClO
HClO(aq)
Write the acid formula
Sulfuric acid
Hrdriodic acid
Sulfurous acid
Hydrosulfuric acid
Nitrous scid
Hydrobromic acid
Oxalic acid
Ethanoic acid
Phosphorous acid
Phosphoric acid
PROPERTIES OF ACIDS
Acids release a hydrogen ion into water (aqueous) solution.
Acids neutralize bases in a neutralization reaction. An acid and a base combine to make a salt
and water. A salt is any ionic compound that could be made with the anion of an acid and the
cation of a base. The hydrogen ion of the acid and the hydroxide ion of the base unite to form
water.
Acids corrode active metals. Even gold, the least active metal, is attacked by an acid, a mixture
of acids called 'aqua regia,' or 'royal liquid.' When an acid reacts with a metal, it produces a
compound with the cation of the metal and the anion of the acid and hydrogen gas.
Acids turn blue litmus to red. Litmus is one of a large number of organic compounds that change
colors when a solution changes acidity at a particular point. Litmus is the oldest known pH
indicator. It is red in acid and blue in base. The phrase, 'litmus test,' indicates that litmus has
been around a long time in the English language. Litmus does not change color exactly at the
neutral point between acid and base, but very close to it. Litmus is often impregnated onto paper
to make 'litmus paper.'
Acids taste sour. TASTING LAB ACIDS IS NOT PERMITTED BY ANY SCHOOL. The word
'sauer' in German means acid and is pronounced almost exactly the same way as 'sour' in
English. Stomach acid is hydrochloric acid. Although tasting stomach acid is not pleasant, it has
the sour taste of acid. Acetic acid is the acid ingredient in vinegar. Citrus fruits such as lemons,
grapefruit, oranges, and limes have citric acid in the juice. Sour milk, sour cream, yogurt, and
cottage cheese have lactic acid from the fermentation of the sugar lactose.
PROPERTIES OF BASES
Bases release a hydroxide ion into water solution. (Or, in the Lowry- Brønsted model, cause a
hydroxide ion to be released into water solution by accepting a hydrogen ion in water.)
Bases neutralize acids in a neutralization reaction. The word reaction is:
Acid plus base makes water plus a salt. Symbolically, where 'Y' is the anion of acid 'HY,' and 'X'
is the cation of base 'XOH,' and 'XY' is the salt in the product, the reaction is: HY + XOH
HOH + XY
Bases denature protein. This accounts for the "slippery" feeling on hands when exposed to base.
Strong bases that dissolve in water well, such as sodium or potassium lye are very dangerous
because a great amount of the structural material of human beings is made of protein. Serious
damage to flesh can be avoided by careful used of strong bases.
Bases turn red litmus to blue. This is not to say that litmus is the only acid- base indicator, but
that it is likely the oldest one.
Bases taste bitter. There are very few food materials that are alkaline, but those that are taste
bitter. It is even more important that care be taken in tasting bases. Again, NO SCHOOL
PERMITS TASTING OF LAB CHEMICALS. Tasting of bases is more dangerous than tasting
acids due to the property of stronger bases to denature protein.
Bases Introduction
1. Name these bases and write dissociation equations for each
NaOH –
Ca(OH)2 –
Mg(OH)2 –
NH4OH –
Al(OH)3 -
2. Give the appropriate name or formula for the following bases.
a.
b.
c.
d.
e.
Sodium Hydroxide Ba(OH)2 Fe(OH)3 Copper (II) Hydroxide Tin (IV) Hydroxide -
Salts Introduction
HCl (aq) + NaOH (aq) ===Î H20(l) + NaCl(aq)
This is a neutralization (double replacement) reaction.
A salt is formed when an acid reacts with a base in a neutralization
reaction. H+ ion and OH- ion react together to form water molecules.
A salt is the nonmetal ion of an acid and metal ion of a base. Therefore, a
salt will not contain any H+ ions or OH- ions.
Exercise A:
Complete the following equations:
Label and name the acid, base and salt.
1)
H2CO3 (aq) + NaOH
==Î
2)
HF(aq) + Ba(OH)2 ===Î
3)
HCl(aq) + Ca(OH)2 (aq) ==Î
Exercise B:
For each of the equation above, I left space for you to write the complete
ionic equation and net ionic equation.
(Note: this is a review from the solutions unit)
Special case of the equilibrium constant (Keq)
The ionization constant for water (Kw)
Water ionizes to a tiny extent (reacts with itself)
H2O(l) Í==Î H+(aq) + OH-(aq)
Write the Keq = [H+] [OH-]
[H2O]
Since we do not use pure liquids in the expression, the bottom is eliminated. This is
why. Since water ionizes only slightly, it concentration (the amount of ionized H2O
molecules dispersed among its ions) is essentially constant.
1 molecule in a million ionizes.
Therefore;
H+(aq) x OH- (aq) = Keq x H2O(l)
[H+ ] x [OH- ] = Kw
[H+ ] x [OH- ] = 1.00 x 10-14
(ionization constant of water)
[H+ ] x [OH- ] = 1.00 x 10-14
What happens if we increase H+.
If the product must equal 1.00 x 10-14, OH – must decrease.
What happens if we increase OH-.
H+ must decrease. (A review of Le Chatilier)
If the above expression, the [H+ ]and [OH- ]are inversely proportional and
can be used in the following expression to find the concentration of H+ or
OH-. If we know one concentration we can compute the other using Kw.
Example:
Find the OH- in a 0.125 M HCl, assuming complete ionization.
HCl ----Æ H+ + OH.125M
.125M
Using basic stochiometry
H+ = .125M =
Using
[H+ ] x [OH- ] = 1.00 x 10-14
OH - = 1.00 x 10-14
H+
= 1.00 x 10-14 = 8.00 x 10 -14M
.125M
Note: Do not forget to keep a section of your binder for the class
problems will be do on the board to review the previous lesson
Activity
Each of the following compounds was dissolved in water.
Complete the table, assuming 100% ionization. (straight arrow)
Compound
Name
Type
Ionic Equation
[ ] of each ion, if
compound [ ] in
water is .050 mol/l
HClO4 (aq)
MgCl2 (aq)
Ba(OH)2 (aq)
NaOH (aq)
HBr (aq)
H3PO4 (aq)
Assume each compound is 100% ionized
a. If 0.050 mole of HCl is added to 10.0 ml of water, what is the [H+]
and [OH-]?
b. If 0.050 mole of Ba(OH)2 is added to 50.0 ml of water, what is the
[H+] and [OH-]?
c. If .40 g of NaOH is added to 100.0 ml of water, what is the [H+] and
[OH-]?
d. If 0.050 mole of Sr(OH)2 is added to 150.0 ml of water, what is the
[H+] and [OH-]?
Understand pH, acids and bases
pH means potential of hydrogen. It is the measure used to determine if a solution is
either acid or alkaline.
The pH scale goes from 0 to 14 with neutral being 7.
The scale is acid from 0 to 7 and alkaline from 7 to 14.
An acid is a molecule or an ion* that can contribute a hydronium ion (H+) to a solution.
An acid has the power to neutralize alkalis.
An alkali (which is often called a "base") is a molecule or an ion that combines with
hydronium ions to remove them from a solution.
Here are a few pHs:
Stomach's pH = 2
Small intestine's pH = 6
Large intestine's pH around 8
Our blood's pH is between 7.32 and 7.42.
The Ph Scale
Ph
0
1
3
5
7
9
11
13
14
+
-
[H ] [OH ] KW Property
100
10-14
10-14
Acid
10-1
10-13
10-14
Acid
10-3
10-11
10-14
Acid
10-5
10-9
10-14
Acid
10-7
10-7
10-14
Neutral
10-9
10-5
10-14
Base
10-11
10-3
10-14
Base
10-13
10-1
10-14
Base
10-14
100
10-14
Base
If the H+ = 1.00 x 10 –7, pH = 7, the solution is neutal
If the H+ = > 1.00 x 10 –7 , pH < 7, the solution is acidic
If the H+, = < 1.00 x 10 –7, pH > 7, the solution is basic in nature
What do we know about the nature of the pH scale? In terms of magnitude
of the values.
The pH Concept
The meaning of pH can be seen in the following relationship:
H+ = 10-pH
Taking the logrithim:
pH = log 10 H+
-
pH = - log 10 [H+]
____________________________________________________________
Sample Problem
What is the pH of 0.0066 M H2SO4 (aq) solution assuming complete
ionization?
H2SO4 -----Æ 2 H+ (aq) + S04 (aq) (Ion concentration review)
.0066M
[H+] = .0066 x 2/1(equation ratios)
= 0.013M
PH = - log 0.013
= 1.9
_____________________________________________________________
What is the pH of a 0.0010 M NaOH solution?
_____________________________________________________________
Find the pH of a 0.0084 M KOH?
____________________________________________________________
POH Scale
pOH = - log 10 [OH-]
Definition of the scales:
pH + pOH = 14.00
pH = 14.00 – pOH
Sample Problem!
Find the pOH of a 0.0084 M KOH solution?
A solution has a pH of 7.52. What is the [OH-] ?
Problems:
1) The pH of a urine sample was measured to be 5.53 . What is the [H3O+], [OH-] and
pOH of the urine. Is the sample acidic, basic or neutral.
2) A sample of household ammonia is 11.9 . What is the pOH and the [OH-] of the
solution.
3) Phenol is used as a disinfectant. An aqueous solution of phenol was found to have a
pH of 4.72. Calculate the [H3O+], [OH-] and pOH of the solution.
4) A sample of baking soda was dissolved in water and the pOH of the solution was
found to be 5.81.
•
•
Is the solution acidic , basic or neutral
Calculate the pH, [H3O+] , and [OH-] of the solution.
5) The pH of an aspirin solution is 2.37. What is the [H3O+], and [OH-] of the solution.
6) If the [H3O+] of a 0.53 mol/l solution of a weak acid is 4.5 * 10 -5 mol/l what is the
pH of the solution?
Strong acids vs. Weak acids
Acids/Bases Formulas- Must Remember
Part 1
Ionization constant for water:
Kw = [H+] [OH-]
1.00 x 10-14 = [H+] [OH-]
(If you know one, you can calculate the other)
Calculating pH
pH = -log [H+]
Calculating pOH
POH = -log [OH-]
Calculating the [H+] from the pH value. (Must know [H+])
[H+] = 10 –pH
Calculating the [OH-] from the pOH value. (Must know [OH-])
[OH-] = 10 –pOH
Using the Yx button on your calculator
Relationship of pH and pOH
pH + pOH = 14
(If you know one, you can calculate the other)
Assignment
Find the [H+] , pH , and the pOH
1.
0.20 M HCl (aq)
2.
0.0030 M H2SO4 (aq)
3.
0.0050 M NaOH (aq)
Find the [H+] and [OH-]
1. Solution with a pH of 9.34
2. Solution with a pH of 2.8
Classify each as acidic or basic. Give the [H+] for the acid
solutions and give the [OH-] for the basic solutions.
a. pH = 9.0
b. pH = 11.4
c. pH = 4.0
d. pH = 1.5
Kw, pH, pOH review Assignment
(Turn sideways to View properly)
Arrhenius Acids and Bases
The Arrhenius definition of acids and bases is one of the oldest. An
Arrhenius acid is a substance that when added to water increases the
concentration of H1+ ions present. The chemical formulas of Arrhenius acids
are written with the acidic hydrogens first. An Arrhenius base is a substance
that when added to water increases the concentration of OH1- ions present.
HCl is an example of an Arrhenius acid and NaOH is an example of an
Arrhenius base.
The H1+ ion produced by an Arrhenius acid is always associated with a
water molecule to form the hydronium ion, H3O1+(aq). Arrhenius acids are
frequently referred to as proton donors, hydrogen ion donors, or hydronium
ion donors, depending on whether we are trying to emphasize the species
liberated by the acid (proton or hydrogen ion) or the species present in solution
(hydronium ion). To represent the transfer of the H1+ ion to water to form the
hydronium ion, we must include H2O in the chemical equation for acid
ionization.
Arrenhius Acids and Bases
1. Summarize the definitions of Arrenhuis Acids and Bases
2. Define and give an example of
a. monoprotic acid
b. diprotic acid
c. triprotic acid
3. Account for the monoprotic nature of ethanoic acid,
CH3COOH, despite the fact that its molecule has 4
hydrogens.
4. Give 2 ways that a solution of sodium hydroxide can be
prepared.
5. What is meant by an alkaline solution
6. Why is the [OH-] always low in solutions of Ca(OH)2 and
Mg(OH)2?
7. What is limewater (what chemical)?
8. What is the chemical name of “milk of magnesia”?
STRONG ACIDS AND STRONG BASES
The common acids that are almost one hundred percent ionized are:
HNO3 - nitric acid
HCl - hydrochloric acid
H2SO4 - sulfuric acid
HClO4 - perchloric acid
HBr - hydrobromic acid
HI - hydroiodic acid
The acids on this short list are called strong acids, because the amount of acid
quality of a solution depends upon the concentration of ionized hydrogens. You
are not likely to see much HBr or HI in the lab because they are expensive. You
are not likely to see perchloric acid because it can explode if not treated carefully.
Other acids are incompletely ionized, existing mostly as the unionized form.
Incompletely ionized acids are called weak acids, because there is a smaller
concentration of ionized hydrogens available in the solution. Do not confuse this
terminology with the concentration of acids. The differences in concentration of
the entire acid will be termed dilute or concentrated.
Strong Bases:
Likewise, there is a short list of strong bases, ones that completely ionize into
hydroxide ions and a conjugate acid. All of the bases of Group I and Group II
metals except for beryllium are strong bases. Lithium, rubidium and cesium
hydroxides are not often used in the lab because they are expensive. The bases
of Group II metals, magnesium, calcium, barium, and strontium are strong, but
all of these bases have somewhat limited solubility. Magnesium hydroxide
has a particularly small solubility. Potassium and sodium hydroxides both
have the common name of lye. Soda lye (NaOH) and potash lye (KOH) are
common names to distinguish the two compounds.
LiOH - lithium hydroxide
NaOH - sodium hydroxide
KOH - potassium hydroxide
RbOH - rubidium hydroxide
CsOH - cesium hydroxide
Mg(OH)2 - magnesium hydroxide
Ca(OH)2 - calcium hydroxide
Sr(OH)2 - strontium hydroxide
Ba(OH)2 - barium hydroxide
Memorize the strong acids and strong bases.
All other acids or bases are weak.
Strong Acids (The Basics)
A strong acid means 100% ionization/dissociation. Acid strength is a
measure of the amount of dissociation.
HNO3 ---Æ H+(aq) + NO3-(aq) (strong acid)
VS.
HIO3 Í=Î H+(aq) + IO3-(aq) (Weak acid)
Which acid do you believe will give more H+ in solution?
The strong acids will give more H+ ions in solution. Since it has a single
arrow in its equation, it shows that reactants are transformed into products.
With weak acids, the double arrow indicates an equilibrium system.
Therefore, there is an appreciable amount of both reactants and products.
There is not a complete transformation to products.
The strong acids: (Memorize) recognize when to use
----Æ or Í===Î
HClO4(aq) – perchloric acid (Stay away)
HCl (aq) – Hydrochloric acid
H2SO4 (aq) – Sulfuric acid
HNO3 (aq) – Nitric acid
HI (aq)
HBr (aq)
Halogen acids
--------H3PO4 (aq) – Phosphoric acid (Borderline 100%)
Try not to confuse the following later:
The above are the only strong acids. However, in every acid-base reaction,
there is a stronger and weaker acid. For example, when the reaction
contains 2 weak acids, one of the acids is a stronger acid and one is the
weaker acid.
To this point, we have mostly been using strong acids and bases in
our kw, pH, pOH calculations. Why?
Writing Ka expressions
WEAK ACIDS AND WEAK BASES
We can write the chemical equation for the dissociation of a weak acid, using 'A-' to
represent the conjugate base, as;
HA
Í==Î
A- + H+
And, similarly, we can write the chemical equation for the dissociation of a weak base,
using 'X+' to represent the conjugate acid, as;
XOH
Í==Î
OH- + X+
The equilibrium expression for the dissociation of a weak acid is; Note, weak acids and
bases are equilibriums, like we dealt with in our other unit. The Ka is very similar to the
Keq
In language, the equilibrium expression reads; "The dissociation constant of an acid is
equal to the concentration of hydrogen ions times the concentration of the conjugate
base of the acid divided by the concentration of un-ionized acid."
XOH
Í==Î
OH- + X+
Kb is very similar to the Ka!
Write balanced ionization equations (1 ionization) and Ka expressions for these acids in
water.
a) H3PO4
b) C6H5COOH
d) HCOOH
c) HSO4equation
Ka expression
a)
a)
b)
b)
c)
c)
d)
d)
Calculating Ka (for weak acids)
1) A student prepared a 0.10 M solution of formic acid, HCHO2 (aq). Its pH
was determined to be 2.38 at 25 oC. Calculate the Ka for this acid at this
temperature.
2) A 0.25 M solution of benzoic acid (Use formula HB) is found to have a
[H+] = 4.0 x 10-3 M.
Assuming this reaction
HB (aq) Í==Î H+(aq) + B-(aq)
Calculate the Ka
3) The pH of a 0.50 mol/L HNO2 solution is 1.38.
What is the Ka of this acid?
Ka problems: #1
Calculating the [H+] from a given Ka value
1) The ionization constant (Ka) for hypobromous acid,
HbrO, = 2.0 x 10-9.
Calculate the [H+] in a solution of 0.0010 M HBrO solution.
2) Ka for hypoiodous acid, HIO, is approximately 2.5 x 10-11.
Would would you expect the [H+] of a 0.010 M solution of
this acid to be?
Ka problems: Set #2
Calculating the [H+] from a given Ka value
1) Given that the Ka for hydrofluoric acid, HF, is
7.0 x 10-4. (Note: OK to use approximate)
Calculate the [H+] of a 0.100M HF solution.
2) Calculate the [H+] and the pH of a 0.10M HCN
Solution.
Ka for HCN = 7.0 x 10-10
3. Propanoic acid, C2H5COOH(aq), gives swiss cheese a characteristic
flavor. The PH of a 0.200 M solution of propanoic acid is 2.780.
Calculate the value of Ka for this acid
Bronsted-Lowry acids and bases
A Bronsted-Lowry (BL) acid is defined as any substance that can donate a hydrogen ion
(proton) and a Bronsted-Lowry base is any substance that can accept a hydrogen ion
(proton). Thus, according to the BL definition, acids and bases must come in what is
called conjugate pairs. For example, consider acetic acid dissolved in water:
Notice that we have written
explicitly in these reactions. The reason is that
acid/base dissociation occurs by a proton transfer reaction from an acid species to a
specific water molecule. The transfer occurs through a hydrogen bond between the acid
molecule and a solvating water molecule. (remember chem..11)
Here, CH3COOH is a BL acid because it can donate a proton, and CH3COO its
conjugate base because it can accept a proton. Note that H2O and H3O+ also form such a
conjugate pair.
rather than
Note that the
water in the above reaction.
has been used to denote the nature of
ions in
Similarly when ammonia is dissolved in water, one has
Here, NH3 is the BL base and its conjugate acid is NH4+ . Similarly, H2O acts as a BL
acid and OH- acts as a BL base.
An interesting ambiguity comes up within the BL definition, namely, that some species
can act either as a BL acid or a BL base. Such beasts are called amphoteric. An
example is the hydrogen carbonate ion, HCO . When dissolved in water, two posible
reaction can occur:
or
In the first of these, HCO3- acts as a BL acid with CO32- as its conjugate base, while in
the second it acts as a BL base with H2CO3 as its conjugate acid.
Bronsted Lowry definition of acids and bases:
PartA: Definition
Acid: A species that
Base: A species that
Examples:
HCl
+
H2O -----Æ H3O+ + Cl-
HCl + NH3 Í==Î NH4+ + Cl-
Part B) Conjugate acid base pairs.
HPO42- + H2O Í==Î H2PO4- + OHWrite down the identity of the acid and base for the forward reaction?
Write down the identity of the acid and base for the reverse reaction?
Note:
The acid on the left, after donating the H+ to the acceptor, becomes the base
on the right.
That is, H2PO4- / HPO42- is a conjugate acid and base pair.
That is, H2O / OH- is a conjugate acid / base pair.
In summary:
Activity:
Label the acid, base, con. Acid, con. Base and list the acid base pairs
Review:
HCl + H2O ---Æ H3O+ + Cl- Label acid, base, con. acid, con. base
Part C) Extent of reaction
HCl + H2O ---Æ H3O+ + ClWhere keq is very large; this means?
** If HCl + H2O ---Æ H3O+ + Cl- occurs to a great extent;
Then,
Cl - + H3O+ ---Æ HCl + H2O
Note to memorize:
HCl + H2O -----Æ
H3O+ + Cl-
Why?
Part D) Referring to the same system as in the previous sample problem:
HCl + H2O -----Æ
H3O+ + ClWhich acid will donate H+ to the greater extent?
Which base will accept H+ to the greater extent?
Note:
Example
H3O+ + HS- -----Æ H2S + H2O
Sa
sb
wa wb
Part E:
The extent of a bronsted acid-base reaction (position are equilibrium)
depends upon the relative strengths of the acids involved.
Note:
From the acid/base pairs
Write the equation so that products are favored
H2S / HS-
and
H3O+ / H2O
Wa (Hint)
____________________________________________________________
The Summary: The five Key points to remember:
Examine the following reactant molecules and ions in aqueous Example;
A.
HSO4 -
+
ClO -
_______
+
_______
HSO4 -
+
ClO -
SO42-
+
HClO
Bronsted
Acid
Bronsted
Base
Conjugate Base
Conjugate
Acid
Complete the reactions, label the acid, base, conjugate acid, conjugate base
and the acid / pairs
B.
NH4 +
+
Cl -
+
C.
HSO3 -
+
PO43-
+
D.
H2S
+
NO3 -
+
E.
H3O +
+
CN-
+
In each of the following equations, label the acids, bases, and
conjugate acid-base pairs:
a) HCl + H2O ↔ Cl- + H3O+
b) HClO4 + H2SO4 ↔ ClO4- + H3SO4+
c) HPO42- + H2SO4 ↔ H2PO4- + HSO4- d) NH3 + HNO3 ↔ NH4+ + NO3-
a)
b)
c)
d)
Bronsted / Lowry Theory:
1. A.
HIO3 (aq) + HCO3-(aq) <===Î H2CO3(aq) + IO3-(aq)
Using this acid-base theory, determine
a.
b.
c.
d.
the acid
the base
the conjugate acid
the conjugate base
B. List the acid/base pairs
C. Pretend that HIO3(aq) is the stronger acid, label the sb,wa,wb
D. Are the reactants or products favored at equilibrium
2. The HCO3- ion is capable of acting as either a Bronsted acid or base.
Show this behavior by writing the equations for its reactions with CN-(aq)
and with HNO3(aq). (Exam prep-2pts)
Acids and Bases Review
Explain what is the same and different between the Arrenhius definition of
acids and bases and the Bronsted-Lowry definition of acids and bases.
3. Autoionization is the formation of a cation and an anion from two molecules of
the same substance. One example is the autoionization of water.
H20 + H20 Í==Î H30+ + OHAnother example is the autoionization of NH3
NH3 + NH3 Í==Î NH4+ + NH2While the Arrenhius theory of acids and bases does not apply to the autoionization of
ammonia, the bronsted Lowry theory of acids and bases applies quite well.
A. Explain why the arrhenius theory of acids and bases does not apply to
the autoionization of ammonia.
B. Explain how the bronsted/lowry theory of acids and bases does apply
to the autoionization of ammonia. Include in you answer, both
conjugate acid and base pairs.
C. What are the conjugate acid-base pairs for the autoonization of
ethanol, C2H5OH?
4. A. Write the bronsted-lowry equation for PFOA, perfluorooctanoic
acid, C7F15COOH in aqueous solution. Identify the acid-base pairs.
B. Using the structure below, explain why PFOA has a low solubility in
water.
F F F F F F F O
F -C - C - C - C – C- C - C -C - OH
F F F F F F F F
C. Does the low solubility of PFOA in water give you any information
about the strength of the acid? Explain?
Calculating the Keq for general Bronsted acid – base reaction
Review: (Remember the usefulness of the Ka)
The strengths of acids are given by the magnitude of Ka. Ka indicates the extent to
which an acid donates a proton, H+ to water in the reaction.
HF + H2O ------Æ H3O+ + FHCN + H2O -----Æ H3O+ + CN-
Ka = 6.7 x 10-4
Ka = 4.8 x 10-10
What do these values tell us about the strengths of the acids?
Because Ka, HF greater than
Ka, HCN
HF is a stronger acid
Thus in a reaction involving the a –b pairs HF, F- and HCN, CNthe favored reaction is:
HF + CN- Í===Î
Sa
Sb
Wa
HCN + FWb
Remember, weaker is favored at equilibrium. Products are fovored. Be sure to study
the 5 key points to remember about the Bronsted/Lowry definition of acids and bases
To find the Keq for this reaction: That is to find the Keq equilibrium constant for any
acid / base reaction. Remember this formula for the exam
Keq =
Ka of the acid on the left side of equation
Ka of the acid on the right side of the equation
Example from above
HF + CN- Í===Î
Keq = Ka HF
Ka HCN
HCN + F-
=
This keq can obtained by using Ka’s, given to you or found in tables on acids
Summary on usefulness of the Ka constant:
1) tells the relative strength of an acid
2) allows the prediction of direction of an acid – base reaction
(weaker acid and base are favored at equil.)
3) allows the determination of the Keq of an acid – base reaction.
Ka and Keq Calculation
1) For this theoretical acid – base reaction, calculate the keq
HX + CN - Í====Î HCN + XKa HX = 5.0 x 10 -5
Ka HCN = 4.8 x 10 -10
HF (Ka = 6.6 x 10-4) and HCN (Ka = 6.2 x 10- 10)are two weak
acids that appear in this equilibrium.
HCN(aq) + F- (aq) Í==Î HF (aq) + CN-(aq)
A)Use the information provided to explain which direction of the
reaction above is favored.
B) Use the Ka expressions, equation, and Ka values provided,
calculate the numerical value of the
Chem. 12 Outcome 4a
Trace the development of acid-base theories from the original to the
Arrhenius definition to the concept of the Bronsted Lowry theory, finishing
with the Lewis theory
____________________________________________________________
A) We started 1st with the operational definition of acids and bases. This was the
original concept of acids and bases, only based upon what you can see; the operational
definition.
B) Expanded to the Arrenhius, which expands on the previous operational definition.
This is a conceptional definition.
An acid provides H+ ions in solution.
A base provides OH- ions in solution.
A very generalized concept based solely on the fact if the proper ions are present in
solution.
EX. HCl ===Î H+ + ClArrenhius Acid
+
NaOH ===Î Na + OH Arrenhius Base
C) Bronsted Lowry- We spent extensive time on this concept.
Proton donor and proton acceptor: There are always 2 on the left and 2 on the right in an
acid – base reaction.
EX. NH3 (g) + H2O (l) <==Î NH4+ (aq) + OH-(aq)
H+
H+
Acceptor
Donor
D) Lewis Acids –
A Lewis acid accepts an electron pair forming a covalent bond.
Lewis acids and bases
The most general definition of acids and bases, which encompasses the
Arrhenius and Bronsted-Lowry definitions is due to our old friend, Lewis
and his dot structures. A Lewis acid is defined to be any species that accepts
lone pair electrons. A Lewis base is any species that donates lone pair
electrons. Thus,
is a Lewis acid, since it can accept a lone pair, while
and NH are Lewis bases, both of which donate a lone pair:
Interestingly, however, is that species which have no hydrogen to donate (a
la the Bronsted-Lowry scheme) can still be acids according to the lewis
scheme. As an example, consider the molecule BF . If we determine Lewis
structure of BF , we find that B is octet deficient and can accept a lone pair.
Thus it can act as a Lewis acid. Thus, when reacting with ammonia, the
reaction would look like:
In fact octet deficient molecules are often strong Lewis acids because they
can achieve an octet configuration by accepting a lone pair from a Lewis
base. Compounds involving elements in periods lower then the second
period can act as Lewis acids as well by expanding their valence shells.
Thus, SnCl acts as a Lewis acid according to the reaction:
Amphiprotic Substances:
Also called Amphoteric substances:
This term refers to the fact that a substance can behave as both an acid and a
base.
Ex. H2O
HF + H2O Í==Î
base
H3O+ + F-
CH3NH2 + H2O Í===Î
methylamine
Acid
CH3NH3+ + OH-
Some metallic hydroxides can neutralize both acids and bases, and in doing
so are acting as both an acid and a base.
Examples:
Al(OH)3 (s) + 3HCl ==Î AlCl3 (aq) + 3 H20(l)
Base
Al(OH)3 (s) + NaOH(aq) Í==Î NaAlO2 (aq) + 2 H2O(l)
Review: The above examples show the property of an operation
definition of an acid and base. Because there are so many
theories, the substances can be Amphiprotic.
Neutralization reaction
- reaction of an acid and a base in aqueous solution giving a salt and
water. They are always double replacement reactions.
- Note: we briefly covered this topic when we introduced the concept
of salts.
Salt – an ionic compound composed of a positive ion other than H+ and a
negative ion other than OH-.
Ex. NaCl, NH4Br, NaHCO3, CaSO4, are all salts by definition.
How do we get a neutral Solution for an above reaction type?
Note:
A neutral solution results from a neutralization reaction if the acid and base
are combined in the mole ratios as specified by the balanced equation and
provided that there are no other ions present that would reaction with the
water itself to from H+ or OH-. This would throw off the balance of H+ =
OH- equality of a neutral solution.
Ex. Ca(OH)2 (aq) + 2HCl(aq) -------Æ CaCl2 (aq) + 2 H20(l)
Sample
1. HNO3 (aq) + LiOH (aq)
Practice problems:
Solutions unit review with a twist. Writing the 3 types of equations
1) molecular equation
2) complete ionic equation
3) the net equation
A) HNO2(aq) + KOH (aq)
B) HCl (aq) + NaOH (aq)
C) CH3COOH (aq) + KOH (aq)
D)
HClO4 (aq) + Mg(OH)2 (aq)
Neutralization Stoichiometry
Summary:
H2SO4 (aq) + 2NaOH (aq) ----Æ 2 H2O (l) + Na2SO4 (aq)
1.5 mol
x
3.0 mol of NaOH needed to neutralize 3.0 mol of H2SO4
We use the balanced chemical equation to find the chemical amount of the
other species in the equation.
In this neutralization situation, there are chemical equivalent amounts of
reactants. Note that sometimes the amounts are the same and sometimes
the values differ. This is determined by the balanced chemical equation.
This is also called stoichiometric equivalence!
Neutralization problems:
1. How many moles of HNO3 is needed to neutralize 0.20 moles of NaOH?
2. How many moles of sulfuric acid, H2SO4, are required to neutralize 0.87
moles of sodium hydroxide?
This is basic chem. 11 stochiometry!
Other Neutralization problems: (Exam Prep)
Putting it all together
1) Calculate the [H3O+] (aq) ion concentration of the resulting
solution if 20.0 ml of 0.200 mol/L HCl is mixed with 30.0
ml of 0.350 mol/L NaOH.
2) Calculate the pH of a solution made by mixing 20.0 ml of
0.300 mol/L HCl (aq) and50.0 ml of 0.115 mol/L
NaOH (aq). The mixture is diluted to a final volume of 1.00
L.
3.Calculate the pH of a solution made by mixing 60.0 ml of .500
mol/L H2SO4 (aq) and 50.0 ml with 50.0 ml of .350 mol/ L LiOH.
Exam Prep:
A) How many more times more acidic than normal rain (5.6) if
the pH of the acid rain is 4.2.
B) Environmentalists are more concerned with sulfuric acid and
nitric acid than they are with nitrous acid. Using 0.100 M
solutions of the three acids, evaluate the validity of the statement.
C) One method of dealing with acid rain is to neutralize it with
calcium carbonate, CaCO3. What mass of CaCO3 would be
required to neutralize 1000L of acid rain at pH = 4.2.
CO3-2 (aq) + H3O+ Í==Î HCO3- (aq) + H2O(l)
Acid-Base indicators:
An indicator is a solution that changes color at a specific pH. There are
many individual indicators and a standard solution of indicators called
the universal indicator.
Background Notes:
An acid base indicator is a complex organic substance that is a weak acid.
It is represented by the acid-base pair Hln, ln-.
The molecular species Hln and the anion ln- have different colors.
The relative concentrations of Hln and ln- determines the color.
To illustrate: consider that Hln is red and ln- is blue; also suppose Ka for the
indicator is 1.0 x 10 –8.
Hln Í===Î H+ + InRed
Blue
Suppose a few drops of this indicator is added to an acidic solution where
the [H+] = 0.10 = 1.0x 10-1.
If H+ ions are added to the system, the system will shift to relieve the stress
by producing more HIn, by making H+ react with In- thus changing the color
of the solution to red
Hln Í===Î H+ + InRed
Blue
If OH- is added, the equilibrium shifts to from more ln- and eventually [Hln] and
[ln-] approach equality, giving a transition color which usually spans about 2 pH
units.
(The OH- reacts with the H+ producing H2O. This reduces the [H+], causing it to
shift to the product side of the equation.)
If more OH- is added, [ln-] increases until the color of ln- dominates. When this
product is favored, the color is blue, indicating a base.
The above is commonly referred to as “The Common Ion Effect”. Which is an
application of Le Chatilier’s principle.
Please refer to you chemistry data booklet, Page 9 to view the
table of indicators which is used for this course
Example for bromothymol blue
- If the pH < 3 the indicator is yellow
- If the pH > 4.6 the indicator is blue
- If the pH 3 < pH < 4.6 varies from yellow through green to blue
-
Ex. The transition color is green
-
Example Phenol red
If pH < 6.6 the indicator is yellow
If pH > 8.0 the indicator is red
6.6 < pH < 8.0 varies from yellow through orange to red
Ex. The transition color is orange
Note:
Please keep this in mind when we start to discuss titrations, an application
of neutralization stoichiometry.
Titration pH at endpoint vs. Transition color
Key: Choose an indicator that has a transition color which has the same pH
as the endpoint of the titration. You want the indicator to change color
when there is stoichiometric equivalence!
For example:
If a titration has an endpoint at a pH of 6.1, which indicator would you use?
Chlorophenol red – Orange from 5.2-6.8
Titrations:
A method used to find the unknown concentration of an acid or a base. A
solution of known concentration [ ] is reacted with a solution of unknown
concentration [ ].
This method is an application of a neutralization stoichiometry problem, which
will include our solution work from that unit. It is a very important exam topic.
Example Titration:
Diagram
1) Equation (All stochiometry problems need an equation)
2) Neutralization:
Note: We mentioned this earlier that an indicator is used tell us when we
have reached a titration endpoint, this is when we see the indicators
transition color.
Is enough info given to find out neutralization mole amounts for the reactant in the
reaction?
3)
Is enough info given to find out how much of the other reactant was
neutralized?
4) Find the [ ] of the unknown.
TITRATION
The word "titration" rhymes with "tight nation," and refers to a commonly used
method of (usually) finding the concentration of an unknown liquid by
comparing it with a known liquid. An acid-base titration is good to consider when
learning the method, but there are more uses for the technique. The measure of
oxalate ion using potassium permanganate in a warm acid environment is a good
example of a redox titration. The Mohr titration is a determination of chloride
concentration using known silver nitrate solution and sodium dichromate an
indicator.
A measured amount of the unknown material in a flask with indicator is usually
combined with the known material from a buret (rhymes with "sure bet"). The
buret is marked with the volume of liquid by a scale with zero on top and
(usually) fifty milliliters on the bottom. The buret has some type of valve at the
bottom that can dispense the contained liquid.
It is not necessary to start the titration with the known liquid level in the buret at
the zero mark, but the level must be within the portion of the buret that is marked.
The buret on the left shows about 1.8 ml of the yellow liquid in it. Most
laboratory burets can be read to an accuracy of one hundredth of a milliliter. (The
drawing on the left is a bit crude. Most burets show the ten divisions of a
milliliter and you can interpolate between the marks.) One reads the buret by
getting at eye level to the bottom of the meniscus (curve in the liquid) and
comparing the bottom of the meniscus to the marks on the glass. A reading of the
buret is taken before and at the end of the titration. The amount of knowconcentration liquid used is the difference of the beginning and ending buret
reading.
The endpoint of the titration is usually shown by some type of indicator. A pH
indicator is a material, usually an orgainc dye, that is one color above a
characteristic pH and another color below that pH. There are many materials that
can serve as pH indicators, each with its own ph range at which it changes color.
Some have more than one color change at distinct pH's. Litmus and
phenolphthalein are common pH indicators. Litmus is red in acid (below pH 4.7)
and blue in base (above pH 8.1). Phenolphthalein (The second 'ph' is silent and
the 'a' and both 'e's are long, if that is any help.) is clear in acid (below pH 8.4)
and pink- purple in base (above pH 9.9). These ranges may seem large, but near
the equivalence point, the point at which the materials are equal, there is a large
change in pH. The equivalence point may not occur at pH 7, neutral pH, so the
appropriate pH indicator must be chosen for the type of acid and base being
titrated.
The volume of the material of unknown concentration is known by how much is
put into the reaction vessel. The concentration of the standard is known, and its
volume is known from the measurement of liquid used in the titration.
Titrations:
Part A) Titration is a method of finding the amount of a reactant that reacts
with a known amount of another substance.
A method of quantitative chemical analysis.
Part B) In a titration, one reactant is added to the other reactant until
chemical equivalent quantities are present, given by the mole ratio in the
balanced equation. Note, this does not mean that the mole values are
always equal. They are only equal when the ratio in the balanced equation
is 1:1.
Part C) The point at which chemically equilivant quantities of reactants are
present is called the end point or equivalence point.
This is called the theoretical endpoint, according to the chemical equation.
Part D) In a titration a reactant is admitted to the reaction container in
controlled quantities from a slender graduated tube called a buret. Allows
you to add a fine stream, to tiny drops, or drop by drop.
Part E) In an acid base titration there is a rapid pH change when equivalent
quantities of reactants are present. Note: At or near the end point, a tiny
addition of acid or base produces an enormous change in pH. This is the
key in letting us know when enough reactant has been added.
Part F) Chemically equivalent quantities of a particular acid and base
coincide with a particular pH.
We can have a strong acid / strong base (pH 7) titration, weak acid / strong
base (pH below 7) titration, and strong acid / weak base (pH above 7)
titration. You will need to remember the 3 titration curves and their
associated indicators which we will cover later in this section.
Part F) An indicator is used to signal when the end point is reached. An
indicator is a compound that changes color over that range of the rapidly
changing pH at the end point. Remember, the closer you are to a neutral
pH, the easier it is to overshoot the endpoint.
Part G) A solution of known concentration is called a standard solution.
The procedure by which its concentration is known is called
standardization.
Titration practice problems:
1) If 10.1ml of standard 0.10 mol/L of NaOH (aq) was titrated to an
equivalence point with 25.2 ml of HCl (aq), find the [ ] of the acid.
2) 20.0 ml of NaOH solution are found to be chemically equivalent to 15.0
ml of .010 M HCl solution.
Find the molarity of the solution.
Experimental vs. Theoretical Endpoint
Know the difference between the 2 terms:
Theoretical endpoint – The pH when there are chemical equivalent quantities of
acid and base are present according to the balanced chemical equation.
Titration experimental endpoint: The pH when the transition color of the
indicator in observed. This can mean disaster is you did not choose the correct indicator
to suit you titration.
Example:
NaOH and HCl would be a strong acid/base titration with a theoretical endpoint at
pH=7. Pick an indicator (Methyl Red)
However, if you used Methyl orange, you would see an experimental endpoint at pH=3.
Note: Make sure the indicator matches the pH endpoint.
Choosing a proper Indicator- to show us the experimental endpoint
Strong acid
Strong Base
Indicator:
Weak Acid
Weak base
Indicator
Strong Acid
Weak Base
Weak acid
Strong Base
The Hydrolysis of Salts
A salt solution may be basic, acidic or neutral based on three different
situations. When dealing with a solution, people will often forget about the
H2O present is the solution.
Situation 1:
A salt solution is basic if it contains an anion that is sufficiently strong base
to remove a proton (H+) from water, giving OHExample:
Though the forward reaction occurs only slightly, it is enough to make
OH- > H+.
A negative ion (anion) which is the conjugate base of a weak acid is a
strong enough base to remove a proton from water. Ex. NO2-, SO32-,
CN-.
Anions of strong acids (HCl, HNO3, HClO4, HBr, HI) do not hydrolyse.
Situation 2:
A salt solution is acidic if it contains a cation that donates a proton to water,
giving H3O+.
Ex. Hydrated cations such as
Cation with a high charge to size ratio (High charge density)
Situation 3:
A salt solution is neutral if none of its ions hydrolyse, or if the anion and
cation hydrolysis occur to an equal extent, keeping it neutral.
Note: The principle from before that a Bronsted acid will donate a proton
to any base whose conjugate acid is weaker than the donar acid.
Hydrolysis Generalizations:
A) Salt containing an anion that does hydrolyse and a cation that does
not, gives a solution that is basic.
B) Salt containing an anion that does not hydrolyse and a cation that
does, gives a solution that is acidic.
C) Salt containing neither an anion nor cation that hydrolyses, gives a
solution that is neutral
D) Salt containing an anion and cation which both hydolyse gives a
solution which is an acid or a base depending on which hydrolyses to
the greater extent. The solution is neutral if both ions hydrolyse to
the same extent.
Activity:
A “salt solution” may be acidic, basic or neutral
Compound
HCl(aq)
NaOH(aq)
KNO3(aq)
Indicator Test
Why?
In some solutions we must remember the presence of water
NaCO3 (aq)
CuSO4 (aq)
1) 20.0 ml of NaOH is needed to reach the endpoint with 15.0 ml of 0.10M
H2SO4. Find the concentration of the NaOH solution
2)
It requires 75.0 ml 0.0500 mol/L NaOH (aq) to neutralize 200.0 ml of gastric
juice. We can assume that HCl (aq) is the only acid in gastric juice.
a) Calculate the concentration of HCl in gastric juice
b) Calculate the pH of gastric juice
c) Calculate the number of grams of HCl(aq) per 200 ml of gastric juice.
Case study – Haloacetic acids
A. Chloroacetic acid, a haloacetic acid, ionizes as shown in the equation
below
CH2ClCOOH (aq) + H2O (l) Í=Î H3O+ (aq) + CH2ClCOO_ (aq)
If 0..310 mol/L CH2ClCOOH (aq) has a pH of 2.230, what is the numerical
value for Ka for this acid
The chemical treatment of swimming pools uses the disinfectant
hypochlorous acid, HOCl, which helps to maintain low levels of bacteria
and algae in the water. Sodium hypochlorite, NaOCl, is often used for this
purpose. The NaOCl dissolves to form OCl- ions, and the following
equilibrium is established in the pool.
OCl-(aq) + H2O(l) Í==Î HOCl (aq) + OH-(aq)
a. Write the Kb expression
b. The initial concentration of OCl_ in the swimming pool is 1.86X10-4
mol/L. What is the pH of this solution?
c. Ideally the [HOCl] = [OCl-] in the pool. To what pH should the pool be
adjusted so that this will be true?
d. A lower pH than normal value may be very irritating to the eyes of
swimmers. Using the equation and LeChatilier’s principle, explain why a
lower pH value might have this effect.