Transcript Acids and Bases

ACIDS AND BASES
SOME PROPERTIES OF ACIDS
 Produce H+ (as H3O+) ions in water (the hydronium ion is a
hydrogen ion attached to a water molecule)
 Taste sour
 Corrode metals
 Electrolytes
 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”
SOME PROPERTIES OF BASES

Produce OH- ions in water

Taste bitter, chalky

Are electrolytes

Feel soapy, slippery

React with acids to form salts and water

pH greater than 7

Turns red litmus paper to blue
“Basic Blue”
Acid/Base definitions
Definition 1: Arrhenius
Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
4.3
ACID/BASE DEFINITIONS

Definition #2: Brønsted – Lowry
Acids – proton donor
Bases – 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
H
Cl H
acid
+
O
H
base
+
H
HO
+
Cl
H
conjugate acid conjugate base
conjugate acid-base pairs
A Brønsted-Lowry acid is a proton donor
A Brønsted-Lowry base is a proton acceptor
base
acid
conjugate
acid
conjugate
base
ACID-BASE THEORIES
The Brønsted definition means NH3 is a BASE in
water — and water is itself an ACID
NH3 + H 2O
Base
Acid
NH4+ + OH Acid
Base
CONJUGATE PAIRS
LEARNING CHECK!
Label the acid, base, conjugate acid, and
conjugate base in each reaction:
HCl + OH-  Cl- + H2O
H2O + H2SO4  HSO4- + H3O+
The ph scale is a way of
expressing the strength of acids
and bases. instead of using very
small numbers, we just use the
negative power of 10 on the
molarity of the H+ (or OH-) ion.
UNDER 7 = ACID
7 = NEUTRAL
OVER 7 = BASE
CALCULATING THE PH
pH = - log [H+]
(Remember that the [ ] mean Molarity)
Example: If [H+] = 1 X 10-10
pH = - log 1 X 10-10
pH = - (- 10)
pH = 10
Example: If [H+] = 1.8 X 10-5
pH = - log 1.8 X 10-5
pH = - (- 4.74)
pH = 4.74
TRY THESE!
Find the pH of these:
1) A 0.15 M solution of
Hydrochloric acid
pH = - log [H+]
pH = - log 0.15
pH = - (- 0.82)
pH = 0.82
2) A 3.00 X 10-7 M
solution of Nitric
acid
pH = - log 3 X 10-7
pH = - (- 6.52)
pH = 6.52
PH CALCULATIONS – SOLVING FOR H+
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
MORE ABOUT WATER
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
MORE ABOUT WATER
Autoionization
OH-
H3O+
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
In a neutral solution [H3O+] = [OH-]
and so [H3O+] = [OH-] = 1.00 x 10-7 M
POH
 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
+
[H3O ],
[OH ]
AND PH
What is the pH of the 0.0010 M NaOH solution?
[OH-] = 0.0010 (or 1.0 X 10-3 M)
pOH = - log 0.0010
pOH = 3
pH + 3 = 14
pH = 11
; pH + pOH = 14
Strong and Weak Acids/Bases

Generally divide acids and bases into STRONG or
WEAK ones.
STRONG ACID: HNO3 (aq) + H2O (l) 
H3O+ (aq) + NO3- (aq)
HNO3 is about 100% dissociated in water.

Weak acids are much less than 100% ionized in
water.
*One of the best known is acetic acid = CH3CO2H
Strong and Weak Acids/Bases

Strong Base: 100% dissociated in water.
NaOH (aq)  Na+ (aq) + OH- (aq)
•Strong bases are the group I hydroxides
•Calcium, strontium, and barium hydroxides are strong, but
only soluble in water to 0.01 M
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)
Strong Acid
Weak Acid
15.4
What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation.
Start 0.002 M
HNO3 (aq) + H2O (l)
End 0.0 M
0.0 M
0.0 M
H3O+ (aq) + NO3- (aq)
0.002 M 0.002 M
pH = -log [H+] = -log(0.002) = 2.7
What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?
Ba(OH)2 is a strong base – 100% dissociation.
Start 0.018 M
Ba(OH)2 (s)
End 0.0 M
0.0 M
0.0 M
Ba2+ (aq) + 2OH- (aq)
0.018 M 0.036 M
pOH = -log (0.036) =1.44
; pH + pOH = 14
pH = 14-1.44 = 12.56
15.4
EQUILIBRIA INVOLVING
WEAK ACIDS AND BASES
Consider acetic acid, HC2H3O2 (HOAc)
HC2H3O2 + H2O ↔ H3O+ + C2H3O2 Acid
Conj. base
[H3O+ ][OAc - ]
-5
Ka 
 1.8 x 10
[HOAc]
(K is designated Ka for ACID)
IONIZATION CONSTANTS FOR
ACIDS/BASES
EQUILIBRIUM CONSTANTS
FOR WEAK ACIDS
Weak acid has Ka < 1
EQUILIBRIA INVOLVING A WEAK ACID
You have 1.00 M HOAc. Calculate the
equilibrium concentration of HOAc, H3O+,
OAc-, and the pH.
Ka= 1.8 x 105
Step 1. Define equilibrium concentration in ICE
table.
[HOAc]
[H3O+]
[OAc-]
initial
1.00
0
0
change
-x
+x
+x
equilib
1.00-x
x
x
EQUILIBRIA INVOLVING A WEAK ACID
Step 2. Write Ka expression
+
2
[H
O
][OAc
]
x
3
Ka  1.8 x 10-5 =

[HOAc]
1.00 - x
This is a quadratic. Solve using quadratic
formula.
EQUILIBRIA INVOLVING A WEAK ACID
Step 3. Solve Ka expression
[H3O+ ][OAc - ]
-5
Ka  1.8 x 10 =

[HOAc]
x2
1.00 - x
First assume x is very small because Ka
is so small.
Ka  1.8 x 10-5 =
x2
1.00
Now we can more easily solve this
approximate expression.
EQUILIBRIA INVOLVING A WEAK ACID
Step 3. Solve Ka approximate
expression
2
Ka  1.8 x 10-5 =
x
1.00
x2 = 1.8 x 10-5
x = [H3O+] = [OAc-] = 4.2 x 10-3 M
pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37
EQUILIBRIUM CONSTANTS
FOR WEAK BASES
Weak base has Kb < 1
EQUILIBRIA INVOLVING A WEAK BASE
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O ↔ NH4+ + OHKb = 1.8 x 10-5
Step 1. Define equilibrium concs. in ICE table
[NH3]
[NH4+]
[OH-]
initial
0.010
0
0
change
-x
+x
+x
equilib
0.010 - x
x
x
EQUILIBRIA INVOLVING A WEAK BASE
Step 2. Solve the equilibrium expression
[NH 4+ ][OH- ]
x2
-5
Kb  1.8 x 10 =
=
[NH3 ]
0.010 - x
Assume x is small, so
x2 = 1.8 x 10-5 x 0.010 = 1.8x10-7
x= 4.2 x 10-4
EQUILIBRIA INVOLVING A WEAK BASE
Step 3. Calculate pH
[OH-] = 4.2 x 10-4 M
so pOH = - log [OH-] = 3.37
Because pH + pOH = 14
pH = 14 – pOH
pH = 10.63
TYPES OF ACID/BASE REACTIONS:
SUMMARY
15.4
Ionized acid concentration at equilibrium
percent ionization =
x 100%
Initial concentration of acid
For a monoprotic acid HA
Percent ionization =
[H+]
[HA]0
x 100%
[HA]0 = initial concentration
15.5
Ionization Constants of Conjugate Acid-Base Pairs
HA (aq)
A- (aq) + H2O (l)
H2O (l)
H+ (aq) + A- (aq)
OH- (aq) + HA (aq)
H+ (aq) + OH- (aq)
Ka
Kb
Kw
KaKb = Kw
Weak Acid and Its Conjugate Base
Kw
Ka =
Kb
Kw
Kb =
Ka
15.7
EXERCISE

Calculate the pH of 1 moldm-3 ethanoic acid
with Ka is 1.7x10-5
CH3CO2H (aq)  CH3CO2- (aq) + H+ (aq)

Calculate the pH of 1 moldm-3 ethanoic acid with Ka
is 1.7x10-5
CH3CO2H (aq)  CH3CO2- (aq) + H+ (aq)
Before
Equi
1
1-x
0
x
1.7x10-5 = x.x
1-x ~ 1
So, x = 4.12 x 10-3
pH = -log (4.12 x 10-3 ) ; pH = 2.4
\
0
x
EXERCISE

Calculate the pH of 0.1 mol dm-3 ethanoic acid.
CH3CO2H (aq)  CH3CO2- (aq) + H+ (aq)

Calculate the pH of 0.1 mol dm-3 ethanoic
acid.
Using the same method, we get:
Ka = x.x
0.1-x ~ 0.1
So, 1.7 x 10-5 = x.x
0.1
x= 1.30 x 10-3
pH = -log 1.30x10-3 ; pH = 2.9
EXERCISE

What are the concentrations of H3O+ and OHions in 0.05 mol dm-2 HCl?
HCl (aq) + H2O  H3O+ (aq) + Cl - (aq)
EXAMPLE

What are the concentrations of H3O+ and OHions in 0.05 mol dm-2 HCl?
HCl (aq) + H2O  H3O+ (aq) + Cl - (aq)
[H3O+] = [HCl]
= 0.05 mol dm-2 (complete dissociation)
Kw = [H3O+] [OH- ]
1.0 x 10-14 = [0.05 mol dm-2 ] [OH- ]
[OH- ] = 2.0 x 10-13mol dm-3
THANK YOU..