Transcript Types of Problems

ACIDS and BASES (unit 11)

Notes start on slide 35 ***
ACIDS, BASES & SALTS
Unit 11
The Arrhenius Theory
of Acids and Bases
Arrhenius Theory of Acids and Bases:
an acid contains hydrogen and ionizes in
solutions to produce H+ ions:
HCl  H+(aq) + Cl-(aq)
Arrhenius Theory of Acids and Bases:
a base contains an OH- group and ionizes
in solutions to produce OH- ions:
NaOH  Na+(aq) + OH-(aq)
Neutralization



Neutralization: the combination of H+
with OH- to form water.
H+(aq) + OH-(aq)  H2O (l)
Hydrogen ions (H+) in solution form
hydronium ions (H3O+)
In Reality…
H+ + H2O  H3O+
Hydronium Ion
(Can be used
interchangeably with H+)
Commentary on Arrhenius Theory…
One problem with the Arrhenius
theory is that it’s not comprehensive
enough. Some compounds act like acids
and bases that don’t fit the standard
definition.
Bronsted-Lowry Theory
of Acids & Bases
Bronsted-Lowry Theory of Acids & Bases:

An acid is a proton (H+) donor

A base is a proton (H+) acceptor
for example…
Proton transfer
HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
Base
Acid
another
Water is a proton
donor, and thus an
example… acid.
ACID
CONJUGATE
BASE
NH3(aq) + H2O(l)  NH4+ (aq) + OH- (aq)
BASE
Ammonia is a proton
acceptor, and thus a
base
CONJUGATE
ACID
Conjugate acid-base pairs

Conjugate acid-base pairs differ by
one proton (H+)
A conjugate acid is the particle formed
when a base gains a proton.
A conjugate base is the particle that
remains when an acid gives off a proton.
Examples: In the following reactions,
label the conjugate acid-base pairs:


H3PO4 + NO2-  HNO2 + H2PO4acid
base
c. acid
c. base
CN- + HCO3-  HCN + CO32base
acid
c. acid c. base

HCN + SO32-  HSO3- + CNacid
base
c. acid c. base

H2O + HF  F- + H3O+
c. base c. acid
base acid
Amphoteric Substances
A substance that can act as both an
acid and a base (depending on what it is
reacting with) is termed amphoteric.
Water is a prime example.
Properties of Acids and Bases

ACIDS

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
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
Have a sour taste
Change the color of
many indicators
Are corrosive (react
with metals)
Neutralize bases
Conduct an electric
current
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BASES
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

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Have a bitter taste
Change the color of
many indicators
Have a slippery
feeling
Neutralize acids
Conduct an electric
current
Strength of Acids and Bases

A strong acid dissociates completely in sol’n:


A weak acid dissociates only partly in sol’n:


HNO2  H+(aq) + NO2-(aq)
A strong base dissociates completely in sol’n:


HCl  H+(aq) + Cl-(aq)
NaOH  Na+(aq) + OH-(aq)
A weak base dissociates only partly in sol’n:

NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
The Lewis Theory
of Acids and Bases
The Lewis Theory of Acids & Bases


Lewis acid: a substance that can accept
an electron pair to form a covalent
bond (electron pair acceptor).
Lewis base: a substance that can
donate an electron pair to form a
covalent bond (electron pair donor).
Neutralization (using Lewis)

Neutralization: the formation of a
coordinate covalent bond in which
both electrons originated on the
same (donor) atom.
Example 1:

Ionization of NH3:

NH3 + H2O  NH4+ + OH-
base
H
..
H N H + H
acid
..
O
..
H
H
+
 H N H +
H
Acid = electron pair acceptor, base = electron pair donor (to form the
covalent bond)
..
O
..
H
-
Example 2:

Auto-ionization of water:

..
O
..
H2O + H2O  H3O+ + OH-
base
H
H + H
acid
..
O
..
H
..
+
 H O H +
H
Acid = electron pair acceptor, base = electron pair donor (to form the
covalent bond)
..
O
..
H
-
Example 3:

Reaction of NH3 with HBr (a Lewis AND
a Bronsted-Lowry acid-base reaction):

base
H
..
NH3 + HBr  NH4+ + Br-
acid
H N H + H
..
Br
..
H
+
 H N H +
H
..
Br
..
-
SUMMARY OF ACID-BASE THEORIES
Theory
Acid Definition
Base Definition
Arrhenius
Theory
Any substance which
releases H+ ions in
water solution.
Any substance which
releases OH- ions in
water solution
BrǿnstedLowry Theory
Any substance which
donates a proton.
Any substance which
accepts a proton.
Lewis Theory
Any substance which
can accept an
electron pair.
Any substance which can
donate an electron pair.
Acid-Base Reactions


Neutralization reactions: reactions
between acids and metal hydroxide
bases which produce a salt and water.
H+ ions and OH- ions combine to form
water molecules:

H+(aq) + OH-(aq)  H2O(l)
Example 1: the reaction of HCl and NaOH
(there are 3 ways to write the chemical equation):

Balanced formula unit equation:


Total ionic equation:
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HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
H+ + Cl- + Na+ + OH-  H2O + Na+ + Cl-
Net ionic equation:

H+(aq) + OH-(aq)  H2O(l)
Example 2: Write the 3 types of equations for
the reaction of hydrobromic acid, HBr, with
potassium hydroxide, KOH.

Balanced formula unit equation:


Total ionic equation:


HBr(aq) + KOH(aq)  H2O(l) + KBr(aq)
H+ + Br- + K+ + OH-  H2O + K+ + Br-
Net ionic equation:

H+(aq) + OH-(aq)  H2O(l)
Example 3: Write the 3 types of equations for
the reaction of nitric acid, HNO3, with calcium
hydroxide, Ca(OH)2.

Balanced formula unit equation:


Total ionic equation:
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
2HNO3(aq) + Ca(OH)2(aq)  2H2O(l) + Ca(NO3)2(aq)
2H+ + 2NO3- + Ca2+ + 2OH-  2H2O + Ca2+ + 2NO3-
Net ionic equation:

H+(aq) + OH-(aq)  H2O(l)
Demos…
DEMO: Sponge
How does that work?...



The sponge is soaked in Congo red.
Congo red is a dye, a biological stain, and a
pH indicator. It has been used as a direct
fabric dye for cotton to produce a bright red
color.
Scientists use Congo red as a pH indicator (a
substance that will change color in the
presence of different ion concentrations,
[H+])
Variety of pH indicators…


There are many different types of pH
indicators, such as universal indicator
and litmus paper.
Litmus paper comes in red Litmus paper
and blue Litmus paper. 

Red litmus paper in an acids turns…

BLUE
Blue litmus paper in a base turns …
RED
Demo: tap water vs. dH2O


Both waters have Universal indicator in them
(= pH indicator (changes color in the
presence of ions), which is a type of weak
acids)
The water will change pH, and therefore
COLOR (which helps us determine if a
solution is acidic or basic) with the addition of
HCl (acid) and NaOH (base)
Universal Indicator Color Chart
pH scale
0
Acid
7
Neutral
14
Base
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Why does it take more drops of acid or
base to make the tap water change
color than it does for the distilled
water?
What is distilled water made of? What is
tap water made of?
Buffered Solutions
A solution of a weak acid
and a common ion is called a
buffered solution.
Thus, the solution maintains
it’s pH in spite of added acid or
base.
pH and pOH
Ionization of water

Experiments have shown that pure
water ionizes very slightly:
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
2H2O  H3O+ + OH-
Measurements show that:
[H3O+] = [OH-]=1 x 10-7 M
Pure water contains equal
concentrations of H3O+ + OH-, so it is
neutral.
pH

pH is a measure of the
concentration of hydronium
ions in a solution.

pH = -log [H3O+]
or

pH = -log [H+]
Example: What is the pH of a solution
where [H3O+] = 1 x 10-7 M?
pH = -log [H3O+]
-7
 pH = -log(1 x 10 )
 pH = 7

Example: What is the pH of a solution
where [H3O+] = 1 x 10-5 M?
pH = -log [H3O+]
-5
 pH = -log(1 x 10 )
 pH = 5


When acid is added to water, the [H3O+]
increases, and the pH decreases.
Example: What is the pH of a solution
where [H3O+] = 1 x 10-10 M?
pH = -log [H3O+]
-10)
 pH = -log(1 x 10
 pH = 10


When base is added to water, the [H3O+]
decreases, and the pH increases.
The pH Scale
0
Acid
7
Neutral
14
Base
pOH

pOH is a measure of the
concentration of hydroxide
ions in a solution.

pOH = -log [OH-]
Example: What is the pOH of a solution
where [OH-] = 1 x 10-5 M?
pOH = -log [OH-]
-5
 pOH = -log(1 x 10 )
 pOH = 5

How are pH and pOH related?

At every pH, the following relationships
hold true:

[H3O+] • [OH-] = 1 x 10-14 M

pH + pOH = 14
Example 1: What is the pH of a solution
where [H+] = 3.4 x 10-5 M?
pH = -log [H+]
-5
 pH = -log(3.4 x 10
M)
 pH = 4.5

Example 2: The pH of a solution is
measured to be 8.86. What is the [H+] in
this solution?
pH = -log [H+]
+
 8.86 = -log [H ]
+
 -8.86 = log [H ]
+
 [H ] = antilog (-8.86)
+
-8.86
 [H ] = 10
+
-9 M
 [H ] = 1.38 x 10

***you may have to put your calculator into sci mode
to get the decimals
Example 3: What is the pH of a solution
where [H+] = 5.4 x 10-6 M?
pH = -log [H+]
-6
 pH = -log(5.4 x 10 )
 pH = 5.3

Example 4: What is the [OH-] and pOH
for the solution in example #3?
[H3O+][OH-]= 1 x 10-14
-6
-14
 (5.4 x 10 )[OH ] = 1 x 10
-9 M
 [OH ] = 1.9 x 10

***you may have to put your calculator into sci mode
to get the decimals
pH + pOH = 14
 pOH = 14 – 5.3 = 8.7

Extra Practice

Classify each solution as acidic, basic, or
neutral
***MUST SOLVE FOR pH (write down
this #) and use the pH scale
a. [H+] = 6.0 x 10-10 M
b. [OH-] = 3.0 x 10-2 M
c. [H+] = 2.0 x 10-7 M
d. [OH-] = 1.0 x 10-7 M
pH = 9.2, basic
pOH = 1.5, pH = 12.5, basic
pH = 6.7, acidic
pOH = 7 = pH, neutral
Example #5


Which is the MOST basic from
question #4?
B, pH = 12.7
Acids and bases: Titrations


The amount of acid or base in a
solution is determined by carrying out a
neutralization reaction;
an appropriate acid-base indicator
(changes color in specific pH range)
must be used to show when the
neutralization is completed.
DEMO of lab…
Read
the
buret
Buret
Solution with
Indicator
correctly
=2
decimal
places
Acid Base Titration: the addition of a known amount of
solution to determine the volume or concentration of
another solution
A very accurate method to
measure concentration.
Acid + Base  Salt + Water
H+ + OH- 
H2O
Moles H+ = Moles OH-
3 Steps:
(SHOW DEMO OF TITRATION)
1.
2.
3.
Add a measured amount of an acid of
unknown concentration to a flask.
Add an appropriate indicator to the flask
Add measured amounts of a base of known
concentration using a buret. Continue until
the indicator shows that neutralization has
occurred. This is called the end point of the
titration
(*** look at page 620 fig. 19-20)
Example:


A 25 mL solution of H2SO4 is neutralized
by 18 mL of 1.0 M NaOH using
phenolphthalein as an indicator. What is
the concentration of the H2SO4
solution?
Equation:
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
 (steps)
Steps
1)
2)
3)
How many mol of NaOH are needed
for neutraliztion?
How many moles of H2SO4 were
neutralized?
Calculate the concentration of the
acid:

1) How many mol of NaOH are
needed for neutraliztion?

Mol NaOH = 1.0 mol x 0.018 L = 0.018 mol
1L
2) How many moles of H2SO4
were neutralized?
Mol H2SO4 = 0.018 mol NaOH x 1 mol H2SO4 =
2 mol NaOH
9.0 x 10-3 mol H2SO4
3) Calculate the concentration of
the acid:
M = mol = 9.0 x 10-3 mol = 0.36 M
L
0.025 L
Titration Curve
PAGE 619 fig 19-17


A graph showing how the pH changes
as a function of the amount of added
titrant in a titration.
Data for the graph is obtained by
titrating a solution and measuring the
pH after EVERY drop of added titrant.
Equivalence point =



The point on the curve where the moles
of acid equal the moles of base;
the midpoint of the steepest part of the
curve is a good approximation of the
equivalence point.
PAGE 619 fig 19-17
PAGE 620 fig. 19-19


Knowledge of the equivalence point can
then be used to choose a suitable
indicator for a given titration;
the indicator must change color at a
pH that corresponds to the equivalence
point.
Calculations of Titrations
The mole method and molarity
Calculate the molarity of a sulfuric acid solution
if 23.2 mL of it reacts with 0.212 g of
Na2CO3.
Na2CO3 + H2SO4  Na2SO4 + CO2 + H2O
****use steps 1-3 that we just did.
 answer
ANSWER
Molarity = of a sulfuric acid solution if 23.2
mL of it reacts with 0.212 g of Na2CO3.
Na2CO3 + H2SO4  Na2SO4 + CO2 + H2O
M
mol 0.212 g Na 2CO3 1000mL
1mol Na 2CO3
1mol H 2SO4

x
x
x
 0.0862M H 2SO 4
L
23.2mL
1L
105.99 g Na 2CO3 1mol Na 2CO3
Normality


The Normality (N) of a solution is
defined as the molarity x the total
positive oxidation number of the solute.
EXAMPLES
HCl  H+ + ClH2SO4  2 H+ + OHNaOH  Na+ + OHBa(OH)2  Ba2+ + 2 OH-
Example

Calculate the molarity and normality of
a solution that contains 34.2 g of
Ba(OH)2 in 8.00 L of solution.

M=?

N=?
 answers
Answer (molarity)
M = 34.2 g of Ba(OH)2 x 1 mol = 0.0249 M
8.00 L
171.35g
 answer for normality
Answer (normality)

N = M x +2 (for Ba2+) = 0.0249 M x 2 =
0.0499 N
Using Normality

1 equivalent of an acid reacts with 1
equivalent of a base, so in titration
problems, you can use this equation
NaVa = NbVb
Example

30.0 mL of 0.0750 N HNO3 required
22.5 mL of Ca(OH) 2 for neutralization.
Calculate the normality and molarity of
the Ca(OH)2 solution.
2 HNO3 + Ca(OH)2  Ca(NO3)2 + 2 H2O
 answers
ANSWER (normality)

N of Ca(OH)2 = ?
NaVa = NbVb
(0.0750 N)(30.0 mL) = Nb(22.5 mL)
= 0.100 N
Answer (molarity)

N = M x (+ oxidation #)

M = N / (+ oxidation #)

M = 0.100 N / (2) because Ca2+

= 0.0500 M