Acids and Bases
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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..