weak conjugate base A buffer is resistant to changes in pH because it can neutralize any strong acid or base added to it 2A1 of 14 SOLUTION EQUILIBRIA A solution contains both HF and F ID: 710005
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Slide1
BUFFER – A solution of about equal amounts of a weak acid and its weak conjugate baseA buffer is resistant to changes in pH because it can neutralize any strong acid or base added to it
2A-1 (of 16)
SOLUTION EQUILIBRIASlide2
A solution contains both HF and F
-
Weak acid:
Weak conjugate base:
2 HF
2F
-
1 HCl molecule
If a strong acid is added to the buffer:
2A-2 (of 16)
1 H
3
O
+Slide3
A solution contains both HF and F
-
Weak acid:
Weak conjugate base:
2 HF
2 F
-
1 HCl molecule
If a strong acid is added to the buffer:
2A-3
(of 16)
it is reacted away completely by the WEAK CONJUGATE BASE (F
-
)
now 3 HF
now 1 F
-
1 H
3
O
+
still 1 H
3
O
+Slide4
A solution contains both HF and F
-
Weak acid:
Weak conjugate
base:
2 HF
2 F
-
If a strong base is added to the buffer:
it is reacted away completely by the WEAK ACID (HF)
1 OH
-
ion
2A-4
(of
16)
1 H
3
O
+
now 1 HF
now 3 F
-
still 1 H
3
O
+Slide5
The higher the concentrations of the weak acid and weak conjugate base in the buffer, the more strong
acid or strong base it can neutralizeThis buffer can neutralize
2 H
3
O
+
’s and 2 OH-’s
2A-5 (of 16)Slide6
Calculate the pH of a solution 0.25 M in hydrofluoric acid (Ka = 7.2 x 10-4) and 0.50 M in sodium fluoride
x
0.50 + x
HF
(aq
) + H2O (l) ⇆ H3O+ (aq) + F- (aq)
Initial M’sChange in M’sEquilibrium M’s
0.25~0
0.50
- x
+ x
+ x
0.25
- x
K
a
= [H
3
O
+
][F
-
]
_____________
[HF]
7.2
x
10
-4
=
x(0.50 + x)
______________
(0.25
– x)
7.2
x
10
-4
= x(0.50)
_________ 0.25 x = 3.6 x 10-4
2A-6
(of
16)
pH = 3.44
= [H
3
O
+
]
For a buffer: calculate its pH from either the
K
a
of the weak acid (HF) or the
K
b
of the weak conjugate base (F
-
)
CALCULATING THE pH OF A BUFFER SOLUTIONSlide7
[H3
O+] x [F-] = Ka _____ [HF]
-
log [
H
3
O
+] - log [F-] _____ [HF]
Calculate the pH of a solution 0.25 M in hydrofluoric acid (Ka = 7.2 x 10-4) and 0.50 M in sodium fluoride
2A-7 (of 16)
CALCULATING THE pH OF A BUFFER SOLUTION
[H
3
O
+
] x [F
-
]
=
K
a
_____
[HF]
pH - log [F-] = pKa _____ [HF]
=
- log
K
a
[H
3
O
+
][F
-
] =
K
a
____________
[HF]
pH
=
p
K
a
+
log
[F
-
]
_____
[HF]
log
[
H
3
O
+
] + log
[F
-
]
_____
[HF]
=
log
K
a
logSlide8
pH = pK
a + log [A-] _____ [HA]
HENDERSON-HASSELBALCH EQUATION
Calculate the pH of a solution 0.25 M in hydrofluoric acid (
K
a
= 7.2 x 10
-4
) and 0.50 M in sodium fluoride
CALCULATING THE pH OF A BUFFER SOLUTION2A-8 (of 16)Slide9
DAVID HASSELHOFF EQUATION
This can be used when a solution contains
both a weak acid and its weak conjugate base
Calculate the pH of a solution 0.25 M in hydrofluoric acid (
K
a
= 7.2 x 10
-4) and 0.50 M in sodium fluoride
CALCULATING THE pH OF A BUFFER SOLUTION
2A-9 (of 16)
pH = pKa + log [A-] _____ [HA]Slide10
pH
= pKa + log [weak conj base] ______________________ [weak acid]
pH
= p
K
a
+
log (nweak conj base) _________________ (nweak acid)
Calculate the pH of a solution 0.25 M in hydrofluoric acid (
Ka = 7.2 x 10-4) and 0.50 M in sodium fluoride
CALCULATING THE pH OF A BUFFER SOLUTION
Useful when buffers are produced from mixing different solutions together
pH
= p
K
a
+
log
(
n
w
eak
conj
base
) /
vol ________________________ (nweak ccid) / vol
2A-10
(of
16)Slide11
Calculate the pH of a solution 0.25 M in hydrofluoric acid (Ka = 7.2 x 10-4) and 0.50 M in sodium fluoride
CALCULATING THE pH OF A BUFFER SOLUTION
p
K
a
=
-log Ka
pH = pKa + log [F-]
_____ [HF]
= -log (7.2 x 10-4)
= 3.14
3
= 3.14
3
+
log (0.50 M)
__________
(0.25 M)
= 3.44
Determine the
p
K
a
of the weak acid
Use the David Hasselholf Equation
2A-11
(of
16)Slide12
Find the pH of a solution that is prepared by mixing 150. mL of 0.300 M nitrous acid (Ka = 4.0 x 10
-4) and 100. mL of 0.200 M sodium nitrite.
2A-12
(of
16)
HNO
2
: Weak Acid
NaNO
2
: Salt w/ Weak Conj. Base
Solutions are mixed
calculate
the
moles
of all
species present
x
0.150
L solution
= 0.0450
mol
HNO
2
0.300
mol
HNO
2
____________________
L
solution
0.200 M NaN
O
2
0.200 M Na
+
and 0.200 M NO
2
-
x
0.100
L solution
= 0.0200
mol
N
O
2
-
0.200
mol
N
O
2
-
___________________
L
solutionSlide13
pK
a = -log Ka
pH
=
p
K
a + log nNO2- _______ nHNO2
= -log (4.0 x 10-4
)
= 3.398
= 3.39
8
+
log (0.0200
mol
)
_______________
(0.0450
mol
)
= 3.05
Determine the pKa of the weak acid
Use the David Hasselholf Equation
Find the pH of a solution that is prepared by mixing 150. mL of 0.300
M
nitrous acid (
K
a
= 4.0 x 10
-4
) and 100. mL of 0.200 M sodium nitrite.
HNO
2
: Weak Acid
NaNO
2
: Salt w/ Weak Conj. Base
2A-13
(of
16) Slide14
Find the pH of a buffer solution that is prepared by mixing 300. mL of 2.50 M acetic acid (Ka = 1.8 x 10-5) and 200. mL of 2.50
M sodium acetate.
2A-14
(of
16)
HC
2
H3O2
: Weak Acid
NaC
2H3O2
: Salt w/ Weak Conj. Base
Solutions are mixed
calculate
the
moles
of all
species present
x
0.300
L solution
= 0.750
mol
HC
2
H
3
O
2
2.50
mol
HC
2
H
3
O
2
_______________________
L
solution
2.50
M Na
C
2
H
3O2
2.50 M Na
+
and 2.50 M C
2
H
3
O
2
-
x
0.200
L solution
= 0.500
mol
C
2
H
3
O
2
-
2.50
mol
C
2
H
3
O
2
-
______________________
L
solutionSlide15
pK
a = -log Ka
pH
=
p
K
a + log nC2H3O2- __________ nHC2H3O2
= -log (1.8 x 10
-5)
= 4.745
= 4.74
5
+
log (0.500
mol
)
______________
(0.750
mol
)
= 4.57
Determine the pK
a
of the weak acid
Use the David Hasselholf Equation
Find the pH of a buffer solution that is prepared by mixing 300. mL of 2.50
M
acetic acid (
K
a
= 1.8 x 10
-5
) and 200. mL of 2.50 M sodium acetate.
HC
2
H
3
O
2
: Weak Acid
NaC
2
H
3
O
2
: Salt w/ Weak Conj. Base
2A-15
(of
16) Slide16
What must be the [HNO2]/[NO2-] to make a buffer with a pH = 3.50?
2A-16 (of 16)
pH =
pK
a
+ log
[NO2-] _________ [HNO
2]
3.50 = 3.398 + log [NO2-] _________
[HNO2]
0.10
2
= log [
NO
2
-
]
_________
[HNO
2]
1.2
6 = [NO2-] _________ [HNO2]
[HNO
2
] = 0.79
_________
[NO
2
-
]
HNO
2
: Weak Acid
NO
2
-
: Weak Conjugate Base
A good buffer has close to a 1:1 ratio of weak acid to weak conjugate base (but no more than a 10:1
or 1:10 ratio
)
This will happen if the
pK
a
of the weak acid is within 1 pH unit of the desired pH of the bufferSlide17Slide18
Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and 1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added.
2B-1 (of 14)
Solutions mixed, so calculate
the moles of all species present
HC
2
H
3
O2
: Weak Acid
NaC
2
H
3
O
2
: Salt
w/
Weak Conj. Base
NaOH
: Strong Base
x
0.500
L solution
= 0.750
mol
HC
2
H
3
O
2
1.50
mol
HC
2
H
3
O
2
_______________________
L
solution
0.500
mol
C
2
H
3
O
2
-
x
0.500
L
sol’n
= 0.500
mol
Na
C
2
H
3
O
2
1.00
mol
Na
C
2
H
3
O
2
________________________
L
solution
0.100
mol
NaOH
0.100
mol
OH
-Slide19
Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and 1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added.
2B-2 (of 14)
STRONG
bases react
COMPLETELY
with acids
OH- (aq) + HC2H3O
2 (aq) → H2O (l) + C2H3O2- (aq)
HC
2H3O2
: Weak Acid
NaC
2
H
3
O
2
: Salt
w/
Weak Conj. Base
NaOH
: Strong Base
Initial
moles
Reacting moles
Final moles
0.100
0.500
0
0.600
– 0.100
+ 0.100
0.750
0.650
– 0.100
an amount of weak acid is reacted away that is equal to the amount of strong base
an amount of weak base is produced that is equal to
the amount of strong baseSlide20
Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and 1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added.
2B-3 (of 14)
Amend the
David Hasselholf Equation
pH
=
p
Ka + log nC
2H3O2- __________ nHC2H3O2
+
n
OH-
____________
–
n
OH
-
= 4.74
5
+
log (0.500
mol
+ 0.100
mol) ______________________________ (0.750 mol – 0.100 mol)
= 4.71
HC
2
H
3
O
2
: Weak Acid
NaC
2
H
3
O
2
: Salt
w/
Weak Conj. Base
NaOH
: Strong BaseSlide21
200. mL of a solution 0.40 M in ammonia (Kb = 1.8 x 10-5) and 0.30 M in ammonium chloride are mixed with 100. mL of 0.45 M hydrochloric acid. Find the pH.
2B-4 (of 14)
NH
3
: Weak Base
NH
4
Cl
: Salt
w/
Weak Conj. Acid
HCl
: Strong Acid
Calculate
the moles of all species present
x
0.200
L solution
= 0.080
mol
NH
3
0.40
mol
NH
3
_________________
L
solution
x
0.200
L solution
= 0.060
mol
NH
4
Cl
0.30
mol
NH
4
Cl
___________________
L
solution
x
0.100
L solution
= 0.045
mol
HCl
0.45
mol
HCl
_________________
L
solution
0.045
mol
H
3
O
+
0.060
mol
NH
4
+Slide22
200. mL of a solution 0.40 M in ammonia (Kb = 1.8 x 10-5) and 0.30 M in ammonium chloride are mixed with 100. mL of 0.45 M hydrochloric acid. Find the pH.
2B-5 (of 14)
H
3
O
+ (aq) + NH3
(aq) →
STRONG acids react COMPLETELY with basesH2O (l) + NH4+ (aq)
NH
3
: Weak Base
NH
4
Cl
: Salt
w/
Weak Conj. Acid
HCl
: Strong Acid
Initial moles
Reacting moles
Final moles
0.080
0.060
0.035
0.105
– 0.045
+ 0.045
0.045
0
– 0.045
an amount of weak
base
is reacted away that is equal to the amount of strong
acid
an amount of weak
acid
is produced that is equal to
the amount of strong
acidSlide23
200. mL of a solution 0.40 M in ammonia (Kb = 1.8 x 10-5) and 0.30 M in ammonium chloride are mixed with 100. mL of 0.45 M hydrochloric acid. Find the pH.
2B-6 (of 14)
Determine the
p
K
a
of the weak acid
Kw =
Ka Kb
Kw = Ka ____Kb
= 5.5
6
x 10
-10
= 1.00 x 10
-14
______________
1.8 x 10
-5
p
K
a
= -log
K
a
= -log (5.5
6
x 10
-10
)
= 9.25
5
NH
3
: Weak Base
NH
4
Cl
: Salt
w/
Weak Conj. Acid
HCl
: Strong AcidSlide24
200. mL of a solution 0.40 M in ammonia (Kb = 1.8 x 10-5) and 0.30 M in ammonium chloride are mixed with 100. mL of 0.45 M hydrochloric acid. Find the pH.
2B-7 (of 14)
Amend the David Hasselholf Equation
pH
=
p
K
a + log n
NH3 ______ nNH4+
– nH3O
+ ____________
+ n
H
3
O
+
= 9.25
5
+
log (0.080
mol
–
0.045
mol) _____________________________ (0.060 mol + 0.045 mol)
= 8.78
NH
3
: Weak Base
NH
4
Cl
: Salt
w/
Weak Conj. Acid
HCl
: Strong AcidSlide25
A buffer can be made by mixing solutions of:(1) a weak acid and a salt containing its weak conjugate base(2) a weak acid and a lesser amount of a strong base(3)
a weak base and a lesser amount of a strong acid2B-8
(of
14)Slide26
0.50 mol NaNO2____________________ L solution
(1) If 600. mL of 0.50 M nitrous acid (Ka = 4.0 x 10-4
) and 400. mL of 0.50 M sodium nitrite are mixed, a buffer is produced. Find the
pH.
2B-9
(of
14)
HNO
2
: Weak Acid
NaNO
2
: Salt w/ Weak Conj. Base
Calculate the moles of all species present
x 0.600
L solution
= 0.30
mol
HNO
2
0.50
mol
HNO
2
___________________
L
solution
x 0.400
L solution
= 0.20
mol
NaN
O
2
0.20
mol
N
O
2
-Slide27
pH =
pKa + log nNO2- ________ nHNO2
(1) If
600. mL of 0.50 M nitrous acid (
K
a
=
4.0 x 10-4) and 400. mL of 0.50 M sodium nitrite are mixed, a buffer is produced. Find the pH.
2B-10 (of 14)
pKa = -log Ka
= -log (4.0 x 10
-4
)
= 3.39
8
= 3.39
8
+
log (0.20
mol
)
_____________
(0.30 mol)
= 3.22
Determine the p
K
a
of the weak acid
Use the David Hasselholf Equation
HNO
2
: Weak Acid
NaNO
2
: Salt w/ Weak Conj. BaseSlide28
(2) If 600. mL of 0.50 M nitrous acid and 400. mL of 0.50 M sodium hydroxide are mixed, a buffer is produced. Find the pH.
2B-11 (of 14)
Calculate the moles of all species
x
0.600
L solution
= 0.30
mol
HNO2
0.50 mol HNO2___________________ L solution
x
0.400
L solution
= 0.20
mol
NaOH
0.50
mol
NaOH
___________________
L
solution
HNO
2
: Weak Acid
NaOH
: (Lesser Amount of) Strong Base
0.20
mol
OH
-Slide29
(2) If 600. mL of 0.50 M nitrous acid and 400. mL of 0.50 M sodium hydroxide are mixed, a buffer is produced. Find the pH.
2B-12 (of 14)
Use
the
Amended David
Hasselholf Equation
pH
=
pKa + log nNO2
- _______ nHNO2
+ nOH- __________
– nOH-
= 3.39
8
+
log (0
mol
+ 0.20
mol
)
__________________________ (0.30 mol – 0.20 mol)
= 3.70
HNO
2
: Weak Acid
NaOH
: (Lesser Amount of) Strong BaseSlide30
(3) If 500. mL of 0.60 M sodium nitrite and 250. mL of 0.80 M hydrochloric acid are mixed, a buffer is produced. Find the pH.
2B-13 (of 14)
Calculate the moles of all species
x
0.500
L solution
= 0.30
mol
NaNO2
0.60 mol NaNO2____________________ L solution
x
0.250
L solution
= 0.20
mol
HCl
0.80
mol
HCl
_________________
L
solution
NaNO
2
: Salt w/ Weak Conj. Base
HCl
: (Lesser Amount of) Strong Acid
0.30
mol
NO
2
-
0.20
mol
H
3
O
+Slide31
(3) If 500. mL of 0.60 M sodium nitrite and 250. mL of 0.80 M hydrochloric acid are mixed, a buffer is produced. Find the pH.
2B-14 (of 14)
Use
the
Amended David
Hasselholf Equation
pH
=
pKa + log nNO2
- _______ nHNO2
– nH3O+
_
___________ + n
H
3
O
+
= 3.39
8
+
log (0.30
mol
– 0.20
mol) ___________________________ (0 mol + 0.20 mol)
= 3.10
NaNO
2
: Salt w/ Weak Conj. Base
HCl
: (Lesser Amount of) Strong AcidSlide32Slide33
ACID-BASE TITRATIONS2C-1 (of 17)Acid-base titrations use a STRONG BASE to completely neutralize an acid, or a STRONG ACID to completely neutralize a base……because strong acids or bases react to completionWhen the moles of base added equals the moles of acid in the flask, the acid has been neutralizedNeutralization turns acid molecules into their conjugate bases
OH- + HA → A- + H2OWhen all of the acid has been neutralized (all converted into its conjugate base), this is called the EQUIVALENCE POINTAn indicator can be used to signal when all of the acid has been neutralizedSlide34
20.0 mL of an HA solution titrated with a 0.400 M NaOH solution.(a) Calculate the molarity of the HA solution if 50.0 mL of the NaOH solution are needed to reach the equivalence point.are
x 1 ____________ 0.0200 L NaOH + HA →
NaA
+ H(OH)
0.400 M
50.0 mL
x M20.0 mL1 mol1 molx 1 mol HA ________________ 1 mol NaOH
0.400 mol NaOH x 0.0500 L _____________________ L =
1.00 M HA2C-2 (of 17)Stoichiometric calculation: strong bases react to completionisSlide35
20.0 mL of an HA solution are titrated with a 0.400 M NaOH solution.(b) Calculate the pH of the 1.00 M HA solution before the titration if the Ka of HA is 1.0 x 10
-5.2C-3 (of 17)xx HA
(
aq
) +
H
2O (l) ⇆ H3O+ (aq
) + A- (aq) Initial M’s Change in M’s Equilibrium M’s1.00~00
- x+ x+ x1.00 - xThe pH of a weak acid is an equilibrium calculation determined from an ICE table for the ionization reaction of the weak acidSlide36
20.0 mL of an HA solution are titrated with a 0.400 M NaOH solution.(b) Calculate the pH of the 1.00 M HA solution before the titration if the Ka of HA is 1.0 x 10-5.
2C-4 (of 17)Ka = [H3O+][A-] ____________
[HA]
1.0
x 10
-5 = x2 ____________ (1.00 – x)
3.16 x 10-3 M = x pH = -log (3.16 x 10-3)= 2.50
The pH of a weak acid is an equilibrium calculation determined from an ICE table for the ionization reaction of the weak acidSlide37
20.0 mL of an HA solution are titrated with a 0.400 M NaOH solution.(c) Calculate the pH of the solution after 10.0 mL of the NaOH was added.2C-5 (of 17) This solution will contain both a weak acid and its conjugate base
A buffer with solutions mixed: calculate moles of all species x 0.0200 L solution
= 0.0200
mol
HA
1.00
mol
HA_______________ L solution
x 0.0100 L solution
= 0.00400 mol OH-
0.400
mol
OH
-
_________________
L
solutionSlide38
20.0 mL of an HA solution are titrated with a 0.400 M NaOH solution.(c) Calculate the pH of the solution after 10.0 mL of the NaOH was added.2C-6 (of 17)
Use the Amended David Hasselholf Equation
pH
=
p
K
a + log nA- ______ nHA
+
nOH- __________ – nOH-
= 5.00
+ log (0
mol + 0.00400 mol
)
___________________________________
(0.0200
mol – 0.00400 mol
)
= 4.40Slide39
20.0 mL of an HA solution are titrated with a 0.400 M NaOH solution.(d) Calculate the pH of the solution after 25.0 mL of the NaOH was added.2C-7 (of 17) A buffer with solutions mixed: calculate moles of all species
x 0.0200 L solution
= 0.0200
mol
HA
1.00
mol HA________________ L solution
x
0.0250 L solution
= 0.0100 mol OH-
0.400
mol
OH
-
__________________
L
solutionSlide40
+ n
OH- __________ – nOH-20.0 mL of an HA solution are titrated with a 0.400 M NaOH solution.(d) Calculate the pH of the solution after 25.0 mL of the NaOH was added.
2C-8
(of
17)
Use
the
Amended David Hasselholf Equation
pH
= pKa + log nA- _____ nHA
= 5.00
+
log (0
mol
+ 0.0100
mol
)
_________________________________
(0.0200 mol – 0.0100
mol)
= 5.00
At the half equivalence point, pH =
pKaSlide41
20.0 mL of an HA solution are titrated with a 0.400 M NaOH solution.(e) Calculate the pH of the solution at the equivalence point. (remembering 50.0 mL of base are needed) Calculate moles of all species2C-9 (of 17)
x 0.0200 L solution
= 0.0200
mol
HA
1.00
mol HA________________ L solution
x
0.0500 L solution
= 0.0200 mol OH-
0.400
mol OH-
__________________
L
solution
At the equivalence point, the solution only contains the weak conjugate baseSlide42
20.0 mL of an HA solution are titrated with a 0.400 M NaOH solution.(e) Calculate the pH of the solution at the equivalence point.2C-10 (of 17)The pH of a weak base is determined from an ICE table for the reaction of the weak base with water (its Kb reaction)
At the equivalence point, the solution only contains the weak conjugate base 0.0200 mol of HA is converted to 0.0200 mol A-
K
b
=
Kw ____ Ka
= 1.0 x 10-9= 1.00 x 10-14 ______________ 1.0 x 10-5
Calculate the Kb of the weak base A-Slide43
20.0 mL of an HA solution are titrated with a 0.400 M NaOH solution.(e) Calculate the pH of the solution at the equivalence point.2C-11 (of 17)The pH of a weak base is determined from an ICE table for the reaction of the weak base with water (its Kb reaction)
At the equivalence point, the solution only contains the weak conjugate base 0.0200 mol of HA is converted to 0.0200 mol A-
Calculate the molarity of the A
-
at the equivalence point
= 0.285
7
M A
-
0.0200 mol A-______________________0.0200 + 0.0500 LSlide44
A- (aq) + H2O (l) ⇆ HA (aq) + OH- (aq) Initial M’s Change
in M’s Equilibrium M’s0.2857x x0~0
- x
+ x
+ x
0.285
7
- x20.0 mL of an HA solution are titrated with a 0.400 M NaOH solution.(e) Calculate the pH of the solution at the equivalence point.
Kb = [HA][OH-] _____________ [A-]
1.0 x 10-9 = x2 _______________ (0.2857 – x)
1.69 x 10-5 = x 4.77 = pOH
= [OH
-
]
9.23 = pH
2C-12
(of
17)Slide45
20.0 mL of an HA solution are titrated with a 0.400 M NaOH solution.(f) Calculate the pH of the solution after 60.0 mL of the NaOH was added.2C-13 (of 17)
Calculate moles of all species
x
0.0200
L solution
= 0.0200
mol
HA
1.00 mol HA________________ L solution
x 0.0600 L solution
= 0.0240 mol OH-
0.400
mol
OH
-
__________________
L
solution
There are 0.0240 – 0.0200 = 0.0040 moles of
EXCESS
strong baseSlide46
20.0 mL of an HA solution are titrated with a 0.400 M NaOH solution.(f) Calculate the pH of the solution after 60.0 mL of the NaOH was added.2C-14 (of 17)
Calculate the molarity of the
EXCESS
OH
-
in the solution
= 0.050 M
OH-
0.0040 mol OH-______________________0.0200 + 0.0600 LpOH = 1.30
pH = 12.70 Slide47
At the half-equivalence point :half HA, half A- pH = pKa
At the start : pure HAAt the equivalence point : pure A-A weak acid-strong base titration produces an equivalence point pH > 7 because:Weak Acid-Strong Base Titration Curve (HA and
NaOH
)
2C-15
(of
17)NaOH + HA →
NaA + H(OH)from NaOH (a strong base)
Na+ is a non acid
from HA (a weak acid) A- is a weak baseSlide48
pH =
pKa + log [Ind-] ________ [HInd]INDICATORS
Phenolphthalein is a weak organic acid (
p
K
a
= 9.75)
colorless pink HInd (aq) + H2O (l) ⇆ H3O+ (aq)
+ Ind- (aq)The color change technically occurs when [Ind
-] = [Hind]A solution is colorless if it has more HInd
2C-16 (of 17)Phenolphthalein has a pKa = 9.75, so it will change color at pH = 9.75
good for titrations in which the equivalence point pH is slightly basic
so when …
pH
=
p
K
a
, pink
if
it has more
Ind
-Slide49
A strong acid-strong base titration produces an equivalence point pH = 7 because:Strong Acid-Strong Base Titration Curve (HZ and NaOH)An indicator with a pKa = 7 should be used for this titration
2C-17 (of 17) NaOH + HZ → NaZ + H(OH)
from
(add OH
-
)
NaOH
which is a strong base ∴ Na+ is a non acid
from (add H+’s) HZwhich is a strong acid∴ Z- is a non base