Transcript Ch. 19: Acids & Bases - Midland Park School District

Ch. 18: Acids & Bases
Sec. 18.4: Neutralization
Objectives
• Write chemical equations for neutralization
reactions.
• Explain how neutralization reactions are
used in acid-base titrations.
• Compare the properties of buffered &
unbuffered solutions.
Neutralization
• A neutralization reaction is a reaction
between an acid and a base that produces a
salt and water.
A salt is an ionic compound made up of a
cation from a base and an anion from an
acid.
Example: HCl + NaOH --> NaCl + H2O
Neutralization
• Neutralization is a double-replacement
reaction
replaces
2HCl + Mg(OH)2 --> MgCl2 + 2H2O
replaces
• Note: in order to write the correct formula
for the salt, you must “criss-cross” the ions,
Mg+2 and Cl-. Then you must BALANCE
the equation to conserve matter.
Practice Problems I
Write balanced chemical equations for the
following neutralization reactions:
1 HNO3 + CsOH -->
2 HBr + Ca(OH)2 -->
3 sulfuric acid + potassium hydroxide -->
4 HC2H3O2 + NH4OH -->
5 H3PO4 + Ba(OH)2 
Practice Problems II
Complete the following neutralization
reactions & then balance the equations.
1 ___ + ___ --> Mg(NO3)2 + H2O
2 ___ + ___ --> Ca3(PO4)2 + H2O
3 ___ + ___ --> lithium sulfate + water
4 ___ + ___ --> Na2CO3 + water
5 ___ + ____  Al2(SO4)3 + water
Net Ionic Equations
• It is important to realize that if the acid,
base, and salt completely dissociate in
aqueous solution, you can also write a
complete and net ionic equation for
neutralization reactions.
• HCl + NaOH --> NaCl + H2O
• H+ + Cl- + Na+ + OH- --> Na+ + Cl- + H2O
• H+ + OH- --> H2O
• Practice: Write complete and net ionic
equations for #1 on previous slide.
Titration
• The stoichiometry of acid-base reactions
provides the basis for a procedure known as
titration.
• Titration is a method for determining the
concentration of a solution by reacting a
known volume of the solution with a solution
of known volume AND concentration.
• To find the concentration of an acid, you
would titrate it with a base of known
concentration (and vice versa).
The Procedure for Titration
1 - Place a measured volume (or KNOWN volume) of a
acid or base of UNKNOWN concentration in a beaker.
Add an indicator. (See pg. 662)
Note: Any indicator can be used as long as it has a
color change at, or close to, a pH of 7. The color
change will indicate to you
when enough of the
solution of KNOWN
concentration has been added
to neutralize the solution of
UNKNOWN concentration.
The Procedure for Titration
• Phenolphthalein is generally used because there
is a clear indication of an acid state (colorless)
and a basic state (pink). In titration, a very pale
pink is the color sought.
The Procedure for Titration
2 - Fill a buret with a
KNOWN volume of
the solution of
KNOWN
concentration. This
solution is called the
standard solution or
titrant.
The Procedure for Titration
3 - Add measured volumes
of the standard solution
to a KNOWN volume of
the solution of
“unknown”
concentration, mixing
thoroughly. Continue
UNTIL the indicator
undergoes a color
change. This indicates
that the standard solution
has neutralized the
“unknown” solution.
Titration
• The end point is the point at which the
indicator in the titration changes color.
• The endpoint should closely approximate the
equivalence point.
• The equivalence point is the stoichiometric
point at which the moles of H+ ions from the
acid equal the moles of OH- from the base.
Titration Curves
• A graph like this
shows how the pH
changes during a
titration.
• In this titration, the
“unknown” is a
base. The pH
slowly decreases
when the standard
acid is added.
Titration Curves
• When nearly all the the OHions have been neutralized,
a dramatic decrease in pH is
seen with the addition of
very small volumes of the
standard acid.
• This abrupt change in pH
occurs at the equivalence
point.
• Additional amounts of acids
result in lowering the pH
even further.
Equivalence Points
• Not all titrations
have equivalence
points at a pH of 7.
• The pH at the
equivalence point
depends on the
relative strengths of
the reacting acid
and base.
Calculating the Unknown Molarity
• A chemist titrated 25 mL of an unknown
concentration of hydrochloric acid with 14.5
mL of 0.50 M NaOH. What is the
concentration of the acid?
1 Write the balanced equation for the reaction.
HCl + NaOH --> NaCl + H2O
2 Calculate the # of moles of the solution of KNOWN
molarity in the volume given. In this reaction, 14.5
mL of 0.50 M NaOH was used in the neutralization.
Remember: M = moles solute/liters of solution,
0.50 M = x moles/0.0145 L
x = 0.00725 mol NaOH
Calculating the Unknown Molarity
3 Use the moles of the KNOWN solution and a
mole ratio from the balanced equation to
calculate the moles of the UNKNOWN
solution needed for neutralization.
0.00725 mol NaOH x 1 mol HCl = 0.00725
1 mol NaOH mol HCl
4 Determine the molarity of the UNKNOWN by
using the moles (just found) & the volume of
the UNKNOWN solution used in the titration.
M = mol solute = 0.00725 = 0.29 M
liters sol’n
0.025
Practice Problems
1 What is the molarity of a Ca(OH)2 solution if
30.0 mL of the solution is neutralized by 26.4
mL of 0.250 M HBr?
2 What is the molarity of H2SO4 solution if 43.3
mL of 0.100 M KOH is needed to neutralize
20.0 mL of the unknown?
3 What is the concentration of household
ammonia (NH3) if 49.9 mL of 0.59 M HCl is
required to neutralize 25.9 mL of ammonia?
Salt Hydrolysis
• Recall the definition of a salt . . . .
• When a salt is added to water, will any
reaction occur?
• Why then does an indicator added to a salt
and water solution show that some salts are
neutral, some are basic, and some are
acidic?
Bromthymol blue and salts
• The ammonium
chloride solution
turns yellow,
indicating it is
acidic.
• Sodium nitrate
(blue) is neutral.
• Potassium
fluoride (green) is
basic.
Salt hydrolysis
• Many salts react with water in salt
hydrolysis.
• When dissolved, the ions of a salt separate.
• The anions of the salt may accept H+ from
water molecules OR the cations may donate
H+ to water.
• It all depends on the acid & base used to
produce the salt.
An Example
• Potassium fluoride (KF) was produced from
?
_____ + _____  KF + H2O
• Note that KF results from the neutralization
of a weak acid (HF) by a strong base
(KOH).
• When the KF produced is then dissolved,
the salt dissociates: KF  K+ + F-
Example Continued
• Once dissolved, this what happens next: the
“weak” ion of the salt reacts with water. In
this example, the anion of the salt (F-)
accepts H+ from water molecules.
F- + H2O  HF + OH-
• Fluoride ions act as a Bronsted base and
OH- is produced! The solution is basic.
Another example
• NH4Cl is formed from a weak base (NH3)
and a strong acid (HCl)
• In solution, NH4Cl dissociates and the weak
NH4+ reacts with water. When the cation
donates H+ to water, an acidic solution is
formed.
NH4Cl  NH4+ + ClNH4+ + H2O  NH3 + H3O+
• NH4Cl produces an acidic solution.
Conclusions
• Have you noticed?
• The salt of a strong base produces a basic
solution.
• The salt of a strong acid produces a acidic
solution.
• Little or no hydrolysis occurs when a salt of
a strong acid and a strong base dissociates
in water. These salt solutions are neutral.
Predictions
• Therefore, in order to predict whether a salt
will produce a acidic, basic, or neutral
solution, we must determine the acid and
base that produced the salt.
• In addition, we must know which acids are
considered strong acids and which bases are
strong bases.
Strong Acids (pg. 603)
Strong acid ionize completely in water.
•
•
•
•
•
•
Hydrochloric (HCl)
Hydrobromic (HBr)
Hydroiodic (HI)
Perchloric (HClO4)
Nitric (HNO3)
Sulfuric (H2SO4)
Strong bases (pg. 606)
Strong bases dissociate entirely in water.
•
•
•
•
•
•
Sodium hydroxide (NaOH)
Potassium hydroxide (KOH)
Rubidium hydroxide (RbOH)
Cesium hydroxide (CsOH)
Calcium hydroxide (Ca(OH)2)
Barium hydroxide (Ba(OH)2)
Practice Problems
First, write the equation for the
neutralization reaction producing the
following salts. Then, write equations for
their salt hydrolysis reactions and classify
their solutions as acidic, basic, or neutral.
• Magnesium sulfate
• Calcium carbonate
• Rubidium acetate
Buffered Solutions
• pH is an
important
abiotic factor
for all living
things.
• Control of pH is
as important in
an aquarium as
it is in your
body.
Buffered Solutions
• A buffer is a solution that resists a change
in pH when small amounts of an acid or
base are added.
• Adding small amounts of HCl to pure
water, for example, may lower its pH from
7 to 2. But adding the same amount of HCl
to a buffered solution, the pH may decrease
only to 6.
How do buffers work?
• A buffer is a mixture of a weak acid and its
conjugate base OR a weak base and its
conjugate acid.
• This solution reacts with any H+ or OHadded to it.
Example
• Suppose a buffer solution contains HF (the
weak acid) and F- (its conjugate base).
• Since it is a weak acid, an equilibrium is
established when the acid dissociates:
HF
H+ + F• Any small addition of H+ shifts the
equilibrium to the left, forming more HF.
• The pH will still remain fairly constant
because the [H+] does not change that
much.
HF
+
H
+
F
• If a base (OH-) is added to the buffering
solution, the H+ ions in the solution react
with the OH- to form water.
• The equilibrium will shift to the right but
pH will still remain fairly constant. The
[H+] does not change that much.
Buffered Solutions
• Great amounts of acid or base added to a
buffered solution will change the pH.
• The amount of acid or base a buffer can
absorb without a large change in pH is
called the buffer capacity.
• The greater the concentration of substances
in the buffering solution, the greater the
buffer capacity.
Buffering Systems in Blood
• The pH of human blood must be kept within
a narrow range – 7.1 to 7.7.
• Outside of this range, among other things,
proteins will denature (and not function
properly).
• There are a number of buffering systems in
your blood that maintain pH.
Blood Buffers
CO2 + H2O
H2CO3
H+ + HCO3This buffer is the most important.
• If H+ ions enter the blood, the equilibriums
shift left. The kidneys remove more water
from your blood and the lungs expel greater
amounts of CO2. Your rate of breathing
increases.
CO2 + H2O
H2CO3
H+ + HCO3-
• If OH- ions enter the blood, H+ ions react
with them and the equilibriums shift to the
right. The kidneys remove more HCO3ions and the lungs expel less CO2. Your
rate of breathing slows.
Practice Problems
In the following situations, predict whether
the blood pH will rise or fall, and which
way the H2CO3/HCO3- equilibrium will
shift.
• A person becomes overexcited and
hyperventilates.
• A person takes an overdose of the antacid
NaHCO3.
• A person has the flu and vomits a lot.