Transcript Acids and Bases

Produce H+ (as H3O+) ions in water (the hydronium
ion is a hydrogen ion attached to a water molecule)
Taste sour
Corrode metals (react to  H2 (g))
Are electrolytes (conduct electricity)
React with bases to form a salt and water
pH is less than 7
Turns blue litmus paper to red “Blue to Red A-CID”
Produce OH- ions in water
Taste bitter, chalky
Are electrolytes (conduct electricity)
Feel soapy, slippery
React with acids to form salts and water
pH greater than 7
Turns red litmus paper to blue
“Basic Blue”
HCl
Hydrochloric Acid
Stomach acid
HNO3
H2SO4
H3PO4
Nitric Acid
Sulfuric Acid
Phosphoric Acid
Jewelry making
Paper making; Car batteries
Preservative in Coca-Cola
NaOH
Sodium hydroxide
lye
KOH
Potassium hydroxide
liquid soap
Ba(OH)2
Barium hydroxide
stabilizer for plastics
Mg(OH)2
Magnesium hydroxide “MOM” Milk of magnesia
Al(OH)3
Aluminum hydroxide
Maalox (antacid)
 Definition #1: Arrhenius (traditional)
Acids – produce H+ ions (or hydronium ions H3O+)
Bases – produce OH- ions
(problem: some bases don’t have hydroxide ions!)
Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
 Strong acid- ionizes completely in aqueous solution
 Strong electrolytes
 HCl, HNO3
 Weak acid- releases few hydrogen ions in aqueous
solution
 HCN and acetic acid (-COOH)
 Strong base- ionizes completely in aqueous solution
 Strong electrolytes
 Weak base- releases few hydroxide ions in aqueous
solution
Strong and Weak Acids/Bases
The strength of an acid (or base) is
determined by the amount of
IONIZATION. STRONG=100% IONIZED
HNO3, HCl, H2SO4 and HClO4 are among the
only strong acids.
Strong and Weak Acids/Bases
 Weak acids are much less than 100% ionized in water.
One of the best known is acetic acid = CH3COOH
Strong and Weak Acids/Bases
 Strong Base: 100% dissociated in water.
NaOH (aq) ---> Na+ (aq) + OH- (aq)
Other common strong
bases include KOH and
Ca(OH)2.
CaO (lime) + H2O -->
Ca(OH)2 (slaked lime)
CaO
Strong and Weak Acids/Bases
 Weak base: less than 100% ionized in
water
One of the best known weak bases is
ammonia
NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq)
 Definition #2: Brønsted – Lowry
Acids – molecule or ion that is a proton
donor
Bases – molecule or ion that is a proton
acceptor
A “proton” is really just a hydrogen atom
that has lost it’s electron!
A Brønsted-Lowry acid is a proton donor
A Brønsted-Lowry base is a proton acceptor
base
acid
conjugate
acid
conjugate
base
The Brønsted definition means NH3 is
a BASE in water — and water is itself
an ACID
NH3
Base
+
H2O
Acid
NH4+ + OHAcid
Base
Definition #3 – Lewis
Lewis acid - a substance
that accepts an electron
pair
Lewis base - a
substance that
donates an electron
pair
Formation of hydronium ion is also an excellent
example.
H
+
ACID
•• ••
O—H
H
BASE
••
H O—H
H
•Electron pair of the new O-H bond
originates on the Lewis base.
 NO, NO2, CO2, SO2, and SO3 gases from industrial
processes can dissolve in atmospheric water to
produce acidic solutions.
 Burning of fossil fuels by coal-burning power
plants, factories, and automobiles
example:
SO3 (g) + H2O(l)  H2SO4 (aq)
Very acidic rain is known as acid rain.
Acid rain can erode statues and affect ecosystems.
Amphoteric Compounds
Any species that can react as either an acid or a base
is described as amphoteric.
example: water
water can act as a base
H2SO4 (aq) + H2O(l)  H3O (aq) + HSO4Ğ(aq)
acid1
base2
acid2
base1
water can act as an acid

Ğ



NH3 (g) + H2O(l) 
NH
(aq)

OH
(aq)

4
base1
acid2
acid1
base2
H2O can function as both an ACID and a BASE.
In pure water there can be AUTOIONIZATION
Equilibrium constant for water = Kw
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
 In the self-ionization of water, two water
molecules produce a hydronium ion and a
hydroxide ion by transfer of a proton.
-

Ğ



H2O(l) + H2O(l) 
H
O
(aq)
+
OH
(aq)
 3
Autoionization
OH-
H3O+
Kw = ionization constant of water
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
In a neutral solution [H3O+] = [OH-]
so Kw = [H3O+]2 = [OH-]2
and so [H3O+] = [OH-] = 1.00 x 10-7 M
Kw = [H+1][OH-1] = 1E-14
Used to find unknown [H+1] or [OH-1] if the other is
known or given
If [H+1] = .0003M, find the [OH-1]
1E-14 = [.0003M] [OH-1]
[OH-1]=3.33E-11M
 Base 10 logarithms (log)
 The log of a number is the power to which 10 must be
raised to get that number
 log 1000 = 3 because 103 = 1000
 log 0.1 = -1 because 10-1 = 0.1
 Use your calculator to solve
 log (1.25*10-3) = -2.90
 -log (3.64*10-12) = 11.44
pH = - log
Example: If [H+] =1E-10
pH = - log( 1E-10)
pH = - (- 10)
pH = 10
Example: If [H+] = 1.8E-5
pH = - log 1.8E-5
pH = - (- 4.74)
pH = 4.74
+
[H ]
Find the pH of these:
1) A 0.15 M solution of Hydrochloric acid
(answer: .8239)
2) A 3.00 X 10-7 M solution of Nitric acid
(answer: 6.5229)
STOP
If the pH of Coke is 3.12, [H+] = ???
Because pH = - log [H+] then
- pH = log [H+]
Take antilog (10x) of both
sides and get
10-pH = [H+]
[H+] = 10-3.12 = 7.6 x 10-4 M
*** to find antilog on your calculator, look for “Shift” or “2nd function”
and then the log button
 A solution has a pH of 8.5. What is the
Molarity of hydrogen ions in the solution?
pH = - log [H+]
8.5 = - log [H+]
-8.5 = log [H+]
Antilog -8.5 = antilog (log [H+])
10-8.5 = [H+]
3.16E-9 M = [H+]
 Since acids and bases are opposites,
pH and pOH are opposites!
 pOH does not really exist, but it is
useful for changing bases to pH.
 pOH looks at the perspective of a base
pOH = - log [OH-]
Since pH and pOH are on opposite
ends,
pH + pOH = 14
[OH-]
[H+]
pOH
pH
The pH of rainwater collected in a certain region of the
northeastern United States on a particular day was
4.82. What is the H+ ion concentration of the
rainwater?
(answer: 1.51E-5M)
The OH- ion concentration of a blood sample is
2.5 x 10-7 M. What is the pH of the blood?
(answer: 7.3979)
Calculating [H3O+], pH, [OH-], and pOH
Problem 1: Calculate the [H3O+], pH, [OH-], and
pOH of a solution of 0.0024 M hydrochloric acid
at 25°C.
[H3O+]= 0.0024 M
pH= 2.62
[OH-]= 4.17 E^-12M
pOH= 11.38
End
 Problem 2: What is the [H3O+], [OH-], and pOH of a
solution with pH = 3.67? Is this an acid, base, or
neutral?
 [H3O+]= 10^ -3.67 = 2.14E-4 M
 [OH-]= 10^-14 / (2.14E-4) = 4.67E-11 M
 pOH= 14 – 3.67 = 10.33
 Acid
 Tests the voltage of the
electrolyte
 Voltage changes as
hydronium ion
concentration changes
 Converts the voltage to pH
 Very cheap, accurate
 Must be calibrated with a
buffer solution
 Indicators are compounds that will change
color in the presence of an acid or base.
 Indicators are either weak acids or weak bases
 Indicators only work in a specific range of pH
 Some dyes are natural, like radish skin or red
cabbage
 In aqueous solutions, neutralization is the reaction of
hydronium ions and hydroxide ions to form water
molecules.
 H3O+(aq) + OH(aq) 
2H2O(l)
 A salt is an ionic compound composed of a cation from a
base and an anion from an acid.
HCl(aq) + NaOH(aq)  NaCl(aq)  H2 O(l)
 Titration is the controlled addition and
measurement of the amount of a solution of
known concentration required to react completely
with a measured amount of a solution of unknown
concentration.
 The point at which the two solutions used in a
titration are present in chemically equivalent
amounts is the equivalence point.
 The point in a titration at which an indicator
changes color is called the end point of the
indicator.
Setup for titrating an acid with a base
1. Add solution from the buret to
the flask.
2. Reagent (base) reacts with
compound (acid) in solution in
the flask.
3. Indicator shows when exact
stoichiometric reaction has
occurred. (Acid = Base)
This is called the END POINT
where NEUTRALIZATION has
occurred.
Titration Math(short version)
M1 • V1 = M2 • V2
Moles H3O+ = Moles OH-
BURET
Standard Solution
= NaOH
Solution of
Unknown
Concentration =
HCl
 Set up the buret and the chemicals the same except no
Phenolphthalein is added
 Set up the Nova with a pH sensor attached and insert the
probe into the Unknown solution. Proceed with the
addition of the base solution until the graph looks like the
one attached
 The volume where the large
jump takes place is the End
Point, moles Base=moles Acid
 Begin calculations
Molarity and Titration
1. Start with the balanced equation for the
neutralization reaction, and determine the
chemically equivalent amounts of the acid and
base.
2. Determine the moles of acid (or base) from the
known solution used during the titration.
3. Determine the moles of solute of the unknown
solution used during the titration using the balanced
equation.
4. Determine the molarity of the unknown solution.
Problem: Determine the molarity of an acidic solution,
10 mL HCl, by titration.
(HCl of unknown molarity in the flask, 5.0 x 10-3 M
NaOH in the buret)
Titrate the acid with a standard base solution
20.00 mL of 5.0 × 10−3 M NaOH was titrated
1. Write the balanced neutralization reaction
equation.
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
Determine the chemically equivalent amounts of
HCl and NaOH.
1 mol HCL reacts with 1 mol NaOH
3. Calculate the number of moles of NaOH used in
the titration. Then calculate the number of moles of
HCl initially in the flask.
20.0 mL of 5.0 × 10−3 M NaOH is needed to reach the
end point
5.0  10-3 mol NaOH
1L

 20 mL  1.0  10-4 mol NaOH used
1L
1000 mL
Amount of HCl = mol NaOH = 1.0 × 10−4 mol
4. Calculate the molarity of the HCl solution
1.0  10-4 mol HCl 1000 mL

 1.0  10-2 M HCl
10.0 mL
1L
Sample Problem F
In a titration, 27.4 mL of 0.0154 M Ba(OH)2 is added
to a 20.0 mL sample of HCl solution of unknown
concentration until the equivalence point is
reached. What is the molarity of the acid solution?
Sample Problem F Solution
Given: volume and concentration of known solution
= 27.4 mL of 0.0154 M Ba(OH)2
Unknown: molarity of acid solution
Solution:
1.
balanced neutralization equation 
chemically equivalent amounts
Ba(OH)2 + 2HCl
1 mol
2 mol
BaCl2 + 2H2O
1 mol 2 mol
Sample Problem F Solution, continued
2. volume of known basic solution used (mL)
amount of base used (mol)
mol Ba(OH)2
1L
 mL of Ba(OH)2 solution 
 mol Ba(OH)2
1L
1000 mL
3. mole ratio, moles of base used
moles of acid used from unknown solution
2 mol HCl
 mol of Ba(OH)2 in known solution  mol HCl
mol Ba(OH)2
Sample Problem F Solution, continued
4. volume of unknown, moles of solute in unknown
 molarity of unknown
amount of solute in unknown solution (mol) 1000 mL

volume of unknown solution (mL)
1L
 molarity of unknown solution
Sample Problem F Solution, continued
1.
1 mol Ba(OH)2 for every 2 mol HCl.
0.0154 mol Ba(OH)2
 24.7 mL of Ba(OH)2 solution
2.
1L
1L
3.80  10-4 mol Ba(OH)

 4.22
2
1000 mL
2 mol HCl
Ğ-4
4
3.80

4.22

10
mol of Ba(OH)2
3. 1 mol Ba(OH)
2
 7.61
8.44  10
Ğ-4
4
mol HCl
 Sample Problem F Solution, continued
 4.
-4
7.61  10 mol HCl 1000 mL
8.44
3.8

 4.22
 10-2 M HCl
20.0 mL
1L
End