Download - Acid-Base Equilibria for Handout Chem 17
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Acids and Bases: Chem 16
• Arrhenius Definition ( ) OHOH + H l2
-
(aq)(aq) ←
→+
• ACIDS – donates H+
• HNO3, H3PO4, H2SO4, HCl, HI, HBr,
CH3COOH, organic-COOH, H2SO3
• BASES – donates OH-
• NaOH, KOH, LiOH, CsOH,
Mg(OH)2,Sr(OH)2,Ba(OH)2 , Ca(OH)2
• Bronsted-Lowry
Definition
++⇔+ 4(aq)
-
(aq)3(g)(l)2 NHOH NH OH
• ACIDS – donates H+
• (proton donor)
• HNO3, H3PO4, H2SO4, HCl, HI, HBr,
CH3COOH, organic-COOH, H2SO3
• BASES – accepts H+
• (proton acceptor)
• NH3, organic-NH2, NaOH, KOH, LiOH,
CsOH, Mg(OH)2,Sr(OH)2,Ba(OH)2 , Ca(OH)2
Acids and Bases: Chem 16 • Bronsted-Lowry Definition
Acids and Bases: Chem 16
• Amphoterism: – Species that can behave as an acid or base are called amphoteric.
– Proton transfer reactions in which a species behaves as either an acid or base is called amphiprotic
OH 2 ) Zn(NO HNO 2 Zn(OH) 22332 +→+
-2
4
-
2 Zn(OH)OH 2 Zn(OH) →+
HPO43- + H2O ���� H2PO4
2- + OH-
HPO43- + H2O ���� PO4
3- + H3O+
Acids and Bases: Chem 16 • Lewis Definition
• ACIDS – electron-pair acceptor
• H+ (∴ all molecules with H+)
• Electron deficient molecules (below-octet
atoms eg. Boron cmpds)
• BASES – electron-pair donor
• OH- (∴ all molecules with OH-)
• Molecules with lone e- pairs
acid base
OH NH OH NH -
423 ← +→+ +
base acid
BF Na BF NaF -
43 +←
→+ +
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Acids and Bases: Chem 16 � 17
• The Autoionization of water
← +→+ + -
(aq)(aq)3)(2)(2 OHOHOH OHll
Equilibrium-constant expression:2
2
-
3c
]O[H
]][OHOH[K
+
=
But concentration of water is constant
(and large) at 25oC, therefore: ]][OHOH[K K -
3wc
+=⇒
Experimental concentration H+ is
determined to be 1.00x10-7 at 25oC,
therefore: C25at 101.00x K
)10x 00.1)(10(1.00x K
o14-
w
-7-7
w
=
=
Acids and Bases: Chem 16 � 17
pH = -log [H3O+] or simply –log[H+]
pOH = -log[OH-]
Kw = 1.00 x 10-14 = [H+][OH-] at 25oC
∴ pOH + pH = 14.00
Proof:
-log Kw = -log [1.00 x 10-14] = -log ([H+][OH-])
-log Kw = 14.00 = -log[H+] + -log[OH]-
14.00 = pH + pOH
Acids and Bases: Chem 16 � 17• Calculate the [H3O+], pH, [OH-], and pOH for a 0.020 M HNO3
solution.
– Is HNO3 a weak or strong acid?
– What is the [H3O+] ?
[ ]( )
70.1pH
100.2-logpH
100.2OH
0.020 0.020 020.0
NOOHOHHNO
2
2
3
-
33
100%
23
=
×=
×=
+ →+
−
−+
+≈
M
M
MMM
• What is pH of water at its normal boiling point? Is it acidic or basic?– Given:
∆Hfo
H2O(l) = -285.83 kJ/mol
∆Hfo
OH-(aq) = -230.0 kJ/mol
� When concentration of acid reaches 1.00 x 10-5 and below,
the [H+] of H2O should be added, where [H+] = 1.00 x 10-7
(only used at 25oC)
� Example: What is the pH of a solution prepared by diluting
1.0 mL of 0.1 M HCl with 1000 liters of water?
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Acids and Bases
• Relative Strengths of Acids and Bases
Conjugate Acid-Base Pairs
� The weaker the acid or base, the stronger the conjugate partner.
� The stronger the acid or base, the weaker the conjugate partner.
2(g)(l)2(aq)32(aq)3(aq)3(aq)32
(aq)3(aq)(aq)(aq)3
COOHCOHOO2NaCHOOHCH2CONa
COOHCH NaCl HCl OONaCH
+⇒+→+
+→+
STRONGER
ACID and BASE
WEAKER
ACID and BASE
Ionization Constants for Monoprotic
Weak Acids and Bases• Consider an aqueous solution of acetic acid,
CH3COOH. What is the equilibrium constant expression?
CH3COOH(aq) + H2O(l) ⇄ CH3COO-(aq) + H3O+
(aq)
]OCOOH][H[CH
]COO][CHOH[K
23
-
33c
+
=
But [H2O] = 55.6 M, very high and almost constant, therefore
COOH][CH
]COO][CHOH[KK
3
-
33ac
+
=≈
• Because of its simplified form, we can write the equation for dissociation as
CH3COOH(aq) ⇄ CH3COO-(aq) + H+
(aq)
COOH][CH
]COO][CHH[K
3
-
3a
+
=
COOH][CH
]COO][CHOH[KK
3
-
33
ac
+
=≈
Ka is the acid-dissociation constant
• For weak bases,
NH3(aq) + H2O(l) ⇄ NH4+
(aq) + OH-(aq)
O]][H[NH
]][NHOH[K
23
4c
+−
=][NH
]][NHOH[K
3
4b
+−
=
Kb is the base-dissociation constant.
Does the base really dissociate, like acids?
[acid]
]base e][conjugatH[Ka
+
=
[base]
]acid e][conjugatOH[Kb
−
=
HA ⇄ H⇄ H⇄ H⇄ H++++ + A+ A+ A+ A----
B- + H2O ⇄ OH⇄ OH⇄ OH⇄ OH---- + BH+ BH+ BH+ BH
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� The ionization constant values for several acids are given below.
◦ Which acid is the strongest?
Acid Formula Ka value
Acetic CH3COOH 1.8 x 10-5
Nitrous HNO2 4.5 x 10-4
Hydrofluoric HF 7.2 x 10-4
Hypochlorous HClO 3.5 x 10-8
Hydrocyanic HCN 4.0 x 10-10
Ionization Constants for Monoprotic Weak Acids
and Bases
� The order of increasing acid strength for these weak acids is:
HCN>HClO>COOHCH>HNO>HF 32
� Calculate the concentrations of the various species in 0.15 M acetic acid, CH3COOH, solution.
� ALWAYS write down the ionization reaction and the ionization constant expression.
[ ][ ][ ]
5
3
-
33a
-
3323
108.1COOHCH
COOCHOHK
COOCHOH OHCOOHCH
−+
+
×==
← +→+
xMxM-x)M.(
xMxMxM
M
++
++
← +→+ +
150 ] [ mEquilibriu
- Change
0.15 ] [ Initial
COOCH OH OHCOOHCH -
3323 • Short-cut: Use the simplfying assumption-
Since x << 0.15, assume that 0.15 – x ≈ 0.15≈ 0.15≈ 0.15≈ 0.15
If Ka /[ ]initial is < 1.0x10-3 , the simplifying assumption is valid
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Percent Ionization of Weak
Acids/Bases
� Calculate the percent ionization of 0.15 M acetic acid. The percent ionization of acetic acid is
[ ] [ ][ ]
%1.1%10015.0
106.1ionization %
%100COOHCH
Hor COOCH= ionization %
3
original3
-
3
=××
=
×
−
+
M
M
equilequil
• Calculate the percent ionization of 0.15 M hydrocyanic acid, HCN. Ka = 4.0 x 10-10
– Compare the %ionization of HCN and HOAc.
� Note that the [H+] (or %ionization) in 0.15 M acetic acid is 215 times greater than for 0.15 M HCN.
Solution Ka [H+] pH % ionization
0.15 M
HOAc
1.8 x 10-5 1.6 x 10-3 2.80 1.1
0.15 M
HCN
4.0 x 10-10 7.7 x 10-6 5.11 0.0051
Solvolysis: Reaction of Acid/Base with solvent
• Solvolysis is the reaction of a substance with the solvent in which it is dissolved.– Hydrolysis refers to the reaction of a substance with water
or its ions.
� Consider the acid HA
HA + H2O ⇄ A- + H3O+ Ka
Reverse form:
A- + H3O+ ⇄ HA + H2O Ka’ = 1/Ka
� Consider the conjugate base, A-
A- + H2O ⇄ HA + OH- Kb
Reverse form:
HA + OH- ⇄ A- + H2O Kb‘ = 1/Kb
Solvolysis: Reaction of Acid/Base with
solvent • How is Ka related to Kb?
HA + H2O ⇄ A- + H3O+ Ka
A- + H2O ⇄ HA + OH- Kb
H2O + H2O ⇄ H3O+ + OH- Kw
baw K x KK =a
wb
K
KK =
b
wa
K
KK =
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� The order of increasing acid strength for the weak acids is:
HCN>HClO>COOHCH>HNO>HF 32
� The order of increasing base strength of the anions
(conjugate bases) of the same acids is:
---
3
-
2
-CN<ClO<COOCH<NO<F
1.8 x 10-54.5 x 10-47.2 x 10-4 3.5 x 10-8 4.0 x 10-10
a
wb
K
KK =
5.6 x 10-101.4 x 10-112.2 x 10-11 2.9 x 10-7 2.5 x 10-5
The stronger the acid/base, the weaker is its conjugate
Strengths of Acids and Bases
• Strengths of BINARY Acids - acid strength increases with decreasing H-X bond strength.
– VIIA hydrohalic acids
Bond strength has this periodic trend
HF >> HCl > HBr > HI
Acid strength has the reverse trend.
HF << HCl < HBr < HI– VIA hydrides.
Bond strength has this trend.
H2O >> H2S > H2Se > H2Te
The acid strength is the reverse trend.
H2O << H2S < H2Se < H2Te
Strengths of Acids and Bases
• Down a group: ����size, ����energy to break H- bond (����electronegativity), ����acidity
Arrange in order of increasing acidity:
NH3, OH2, HF
���� NH3 < OH2 < HF
(Electronegativity trend: NH3 < OH2 < HF)
• Across a period: ����electronegativity, ����acidity
Strengths of Acids and Bases
• TERNARY ACIDS - hydroxides of nonmetals
that produce H3O+ in water.
– Consist of H, O, and a nonmetal.
HClO4 H3PO4
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Strengths of Acids and Bases
• Acidity of ternary acids with same central
element increase with increasing oxidation state
of central element, and increasing O atoms
HClO < HClO2 < HClO3 < HClO4
weakest strongest
Cl oxidation states
+1 +3 +5 +7
Strengths of Acids and Bases
• ternary acids of the same group and same number of O atoms increase in acidity with increase electronegativity of central atom
H2SeO4 < H2SO4
HBrO4 < HClO4
HBrO3 < HClO3
However for phosphorus
ternary acids:
H3PO2 > H3PO3 > H3PO4 –
relative position of H is
important (based on
structures)
Hard and Soft Acids and Bases
• From previous discussion, the STRENGTH of an acid or a base depends on the value of its dissociation constant Ka or Kb, respectively
• Hardness or Softness of an acid/base depends on POLARIZABILITY of molecule
– Recall that “Polarizability“ is used in describing the type of IMFA of molecules
– The concept of Hardness or Softness of acids/bases is Lewis-structure dependent, therefore it is a concept applied only in the LEWIS DEFINITION of acids/bases.
• More polarizable molecules (greater # of π e-
or lone pairs) – SOFT Lewis acids/bases
• Less polarizable molecules – HARD Lewis
acids/bases
HARD acid-HARD base
SOFT acid-SOFT base >Dissociation
constant
In general, molecules that involve
HARD acid-SOFT base
SOFT acid-HARD base
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Hard Lewis acid
Which is the Hardest Lewis base? Which is the softest?
---
3
-
2
-CN ClO COOCH NO F
HARDEST
Lewis base
SOFTEST
Lewis base
• In a 0.12 M solution of a weak monoprotic acid, HY, the acid is
5.0% ionized. Calculate the dissociation constant for the weak
acid.
� The pH of a 0.10 M solution of a weak monoprotic acid, HA,
is found to be 2.97. What is the value of its dissociation
constant?
Polyprotic Acids/Bases• Many weak acids contain two or more acidic hydrogens.
– Examples include H3PO4 and H3AsO4.
• The calculation of equilibria for polyprotic acids is done in a stepwise fashion.
– There is a dissociation constant for each step
• Consider arsenic acid, H3AsO4, which has three ionization constants.
1 Ka1 = 2.5 x 10-4
2 Ka2 = 5.6 x 10-8
3 Ka3 = 3.0 x 10-13
• Calculate the concentration of all species in 0.100 M
arsenic acid, H3AsO4, solution.
You may apply the simplifying assumption in each step (1 ICE table/
dissociation)
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11-
a2(aq)3
-2
3(aq)(l)2
-
3(aq)
-7
a1(aq)3
-
3(aq)(l)23(aq)2
10 x 4.7 K OH COOH HCO
10 x 4.4K OH + HCOOHCOH
=+↔+
=↔++
+
8-
b2(aq)3(aq)2(l)2
-
3(aq)
-4
b1(aq)
-
3(aq)(l)2
-2
3(aq)
10 x 2.3 K HO COHOH HCO
10 x 1.2K HO + HCOOHCO
=+↔+
=↔+−
−
b2a2
w
b1a1
KK
K
KK K HO POHOH POH
K HO POHOH HPO
K HO + HPOOHPO
b3(aq)4(aq)3(l)2
-
4(aq)2
b2(aq)
-
4(aq)2(l)2
-2
4(aq)
b1(aq)
-2
4(aq)(l)2
-3
4(aq)
−
−
−
+→+
+→+
→+
b3a3
b2wa2
b1a1
KK
KKK
KK
13-
a3(aq)3
-3
4(aq)(l)2
-2
4(aq)
8-
a2(aq)3
-2
4(aq)(l)2
-
4(aq)2
-3
a1(aq)3
-
4(aq)2(l)24(aq)3
10 x 3.60 K OH POOH HPO
10 x 6.20 K OH HPOOH POH
10 x 50.7K OH + POHOHPOH
=+→+
=+→+
=→+
+
+
+
• The behaviour of an amphiprotic species (acting as base or acid) depends on its dissociation constants
11-
a2(aq)3
-2
3(aq)(l)2
-
3(aq)
-8
b2(aq)3(aq)2(l)2
-
3(aq)
10 x 4.80 K OH COOH HCO
10 x 2.38 K OH COHOH HCO
=+→+
=+→++
−
12-
b3(aq)4(aq)3(l)2
-
4(aq)2
-8
a2(aq)3
-2
4(aq)(l)2
-
4(aq)2
10 x 1.33 K OH POHOH POH
10 x 6.20 K OH HPOOH POH
=+→+
=+→+−
+
7-
b2(aq)
-
4(aq)2(l)2
-2
4(aq)
-13
a3(aq)3
-3
4(aq)(l)2
-2
4(aq)
10 x 1.61 K OH POHOH HPO
10 x 3.6 K OH POOH HPO
=+→+
=+→+−
+
• What is the pH of the resulting solution obtained
by dissolving 1.52 g of NaH2PO4•2H2O in 50.00
mL water? If the salt added was Na2HPO4•2H2O
instead, will the solution be basic or acidic?
• You accidentally spilled muratic acid (2.0 M HCl)
on the rubber slippers of your room mate. To
neutralize the acid, you looked for a base in your
dorm stock room, and you found two salts –
sodium bicarbonate and sodium phosphate.
Which of the two salts will you use?
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Inorganic Lewis Acids – Hydrolysis of
Metal Ions• Because metal ions are positively charged, they
attract the electrons of oxygen atoms in water.– Depending on the strength of electron interacting
with the cation, the water molecule can turn into hydroxide anion and release H+
– The acid strength of these ion-complexes acting as Lewis acids depend on size and charge of cation center
3-
a(aq)
2
(aq)52
3
)aq(62 10 x 2.0 K H (OH)O)Fe(H )OFe(H =+←
→ +++
Na(NO3) Ca(NO3)2 Zn(NO3)2 Al(NO3)3
7.0 6.9 5.5 3.5
Salts of acids and bases
� Aqueous solutions of salts of strong acids and strong basesare neutral
Examples: NaCl (from HCl and NaOH)
K2SO4 (from KOH and H2SO4)
� Aqueous solutions of salts of strong bases and weak acidsare basic
Examples: NaCN (from NaOH and HCN)
K2C2O4 (from KOH and H2C2O4)
� Aqueous solutions of salts of weak bases and strong acidsare acidic
Examples: NH4Cl (from NH3 and HCl)
(CH3)3NHBr ((CH3)N and HBr)
How about – KHC2O4? NaHSO4? LiHSO3?
Salts of acids and bases
• Aqueous solutions of salts of weak bases and weak acids can be neutral, basic or acidic.
The values of Ka and Kb determine the pH.
NH4CH3COO?
Compare Ka of NH4+ vs Kb of OAc-
MgNH4PO4?
Compare Ka of NH4+ and Mg2+ vs Kb of OAc-
NH4(HCO3)?
Compare Ka of NH4+ vs Kb/Ka of amphiprotic HCO3
-
Common Ion Effect
and
Buffers/Buffer Capacity
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Common Ion Effect –
A special name for a Le Chatelier-based shift
• Consider a solution of 0.05 M acetic acid, CH3COOH (50.00
mL)
5-
a(aq)3(aq)3)l(2(aq)3 10 x 1.8 K OH COOCH OH COOHCH =+←
→+ +−
OH COOCH OH COOHCH (aq)3(aq)3)l(2(aq)3
+− +←+
OH COOCH OH COOHCH (aq)3(aq)3)l(2(aq)3
+− +←+
• Describe the direction of equilibrium shift
After adding 10.00 mL of 0.5 M HCl
After adding 10.00 mL of 0.5 M NaCH3COO
In COMMON-ION effect, the direction of shift of equilibrium is always towards the
side that diminishes the added common/similar ion
Common Ion Effect –
A special name for a Le Chatelier-based shift
• Consider a solution of 0.05 M acetic acid,
CH3COOH (50.00 mL)
• Describe the pH of the final mixture
– After adding 10.00 mL of 0.5 M HCl
– After adding 10.00 mL of 0.5 M NaCH3COO
DECREASE pH, more ACIDIC
INCREASE pH, less ACIDIC
Buffers
• Solutions that contain BOTH acid component and its conjugate base
– Conjugate base is directly added, not the one calculated from equilibrium
– Examples :
• Acetic acid added with sodium acetate
• Ammonium chloride added with aqueous ammonia solution
• Solutions that resist drastic pH changes
Henderson-Hasselbalch Equation• Simplified equation for pH calculation involving
buffers
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Buffers• What is pH of a buffer that is 0.12 M in lactic acid, HC3H5O5,
and 0.10 M in sodium lactate?
Ka = 1.40 x 10-4
The most important aspect of buffer solutions is that they resist drastic
changes of pH upon adding strong acids or bases!
OH COOOHC OH COOHOHC (aq)3(aq)344)l(2(aq)344
+− +←
→+
0.10 M0.12 M
+ x + x - x
I
∆
E 0.12 - x 0.10 + x x
(0.12)
(0.10) pK pH a +=
x) (0.12
x) (x)(0.10 K a −
+=
Common Ion Effect – Buffers• Non-buffer Case - Consider the a solution of 50.00 mL of 0.05 M acetic
acid, CH3COOH. Calculate the pH of the final mixture after adding 10.00
mL of 0.05 M HCl.
OH COOCH OH COOHCH (aq)3(aq)3)l(2(aq)3
+− +←
→+
� Because HCl is a strong acid, it directly contributes to the initial concentration of
H3O+ in the equilibrium calculation
� Set-up ICE table in MOLE basis, then convert to molarity when
calculating/equating to Ka.
(10.00 mL) •
(0.05 M)
(50.00 mL) •
(0.05 M)
+ x + x - x
I
C
Common Ion Effect – Buffers� Buffer CASE - Consider a solution of 50.00 mL 0.05 M acetic acid
(CH3COOH) and 0.05 M sodium acetate (NaCH3COO). Calculate the pH of
the final mixture after adding 10.00 mL of 0.05 M HCl.
Step 1: STOICHIOMETRIC calculation
Which has the lowest change in pH (∆pH)?
� Try calculating the pH after adding 10.00 mL of 0.05 M NaOH (instead HCl) for the 2 cases.
Step 2: EQUILIBRIUM calculation, HH
mL) 60.00 volume(total
M) mL)(0.05 (60.00
mL) 60.00 volumetotal(
M) mL)(0.05 00.40(
log pKpH a
=
=
+=
Henderson-Hasselbalch Equation
HA + H2O ⇄ A⇄ A⇄ A⇄ A---- + H+ H+ H+ H3333OOOO++++B + H2O ⇄ BH⇄ BH⇄ BH⇄ BH++++ + HO+ HO+ HO+ HO----
� HH equation only used when salt is present – that is, present separately, not the
[salt] from ICE calculation
� The salt-component must be added separately, or generated by neutralizing the
main component
� In calculations involving buffers, ICE table must be in terms of MOLES especially if
volumes are not same. However, equating with Ka must be in MOLARITY.
� HH equation is allowed only when “simplifying assumptions” are valid
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Preparation of Buffers• Buffers can be prepared in three ways
– Adding a solid component to a liquid component• Ex. NaCH3COO solid added to a solution of acetic acid (acetic-acetate
buffer)
– Neutralizing a liquid component with a strong opposite component • Ex. Aqueous ammonia added with liquid HCl (ammonia-ammonium
buffer)
Aqueous phosphoric acid added with solid NaOH (phosphate buffer)
– Mixing two solid components in the same volume of water• Ex. Solid NaH2PO4•H2O and solid Na2HPO4•7H2O dissolved in water
(phosphate buffer)
Preparation of Buffers
• How many grams of NH4Cl must be added to 2.0 L
of 0.10 M NH3 to form a buffer of pH 9.00?
Kb NH3 = 1.8 x 10-5
� What volume of 0.5 M NaOH must be added to 50 mL
of 0.1 M benzoic acid (C6H5COOH) to make 100.0 mL of
0.05 M benzoate buffer that is pH 4.5?
Ka benzoic acid = 6.3 x 10-5
Preparation of Buffers• Prepare a 100.0 mL 0.1 M phosphate buffer, pH 8.00.
– Given:
H3PO4
pKa1 = 2.12
pKa2 = 7.21
pKa3 = 12.38
Molarity H3PO4 (liquid) = 14.85 M
Formula weight
NaH2PO4•H2O = 137.99 g/mole
Na2HPO4•7H2O = 268.07 g/mole
7.21 pK OH HPOOH POH a2(aq)3
-2
4(aq)(l)2
-
4(aq)2 =+→+ +
)POH (moles
)HPO (moleslog pK pH
-
42
-2
4a2 +=
componentsbuffer moles totalHPO moles POH moles -2
4
-
42 =+
M) (0.1)
L 1mL 1000(
mL) (100.0 HPO moles POH moles -2
4
-
42 =+
Buffer Capacity
• Amount of acid or base (usually in mL) needed to change the pH of a buffer solution by 1 degree.
Compare the two buffers:
- 100 mL of 1.0 M NaCH3COO and 1.0 M CH3COOH
- 100 mL of 0.1 M NaCH3COO and 0.1 M CH3COOH
- Which has the highest buffer capacity relative to 1.0 M NaOH?
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Acid-Base Neutralizations:
Indicators, Titrations
and
pH curves
Acid-Base Indicators
• The point at which chemically equivalent amounts of acid and
base have reacted is called the equivalence point.
• The point at which a chemical indicator changes color is called
the end point.
2color 1color
OH In OH HIn (aq)3(aq))l(2(aq)
+− +←
→+
bcolor acolor
OH HIn OH In (aq)(aq))l(2(aq)
−− +←
→+
Acidic indicator
Basic indicator
Acid-Base Indicators
• The equilibrium constant expression for an indicator
would be expressed as:
[HIn]
]][InO[H K
-
3a
+
= [HIn]
][In
]O[H
K -
3
a =+
pH range when indicator
changes its color depends
largely on Ka of the indicator
Acid-Base Indicators
Color change ranges of some acid-base indicators
Indicator
Color in
acidic
range pH range
Color in
basic range
Methyl violet Yellow 0 - 2 Purple
Methyl orange Pink 3.1 – 4.4 Yellow
Litmus Red 4.7 – 8.2 Blue
Phenolphthalein Colorless 8.3 – 10.0 Red
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Titration Curves
• Strong Acid titrated with Strong Base
Given: 25.00 mL of 0.5 M HClO4, calculate the pH of the resulting
solution after adding the following volumes of 0.5 M NaOH:
Volume of
0.5 M NaOH
(mL)
Mmoles
NaOH
Mmoles
HClO4
remaining
Total
Volume
(mL)
[H+]final pH
0.00 0.0 12.5 25.0 0.5000 0.301
5.00 2.5 10.0 30.0 0.3333 0.477
10.00 5.0 7.5 35.0 0.2143 0.669
15.00 7.5 5.0 40.0 0.1250 0.903
20.00 10.0 2.5 45.0 0.0555 1.256
25.00 12.5 0 50.0 1x10-7 7.000
30.00 15.0 0 55.0 2.2x10-13 12.657
Methyl orange
Phenolphthalein
Litmus
Methyl violet
• Strong Acid/ Strong Base Titration Curve
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
0 5 10 15 20 25 30 35 40
pH
Volume Titrant (0.5 M NaOH)
SA-SB curve
Titration Curves• Weak Acid titrated with Strong Base
Given: 25.00 mL of 0.5 M CH3COOH, calculate the pH of the resulting
solution after adding the following volumes of 0.5 M NaOH:
Volume of
0.5 M
NaOH (mL)
Mmoles
NaOH
Mmoles
CH3COOH
remaining
Mmoles
CH3 COO-
produced
Total
Volume
(mL)
pH
0.00 0.0 12.5 0 25.0 0.301
5.00 2.5 10.0 2.5 30.0 4.143
10.00 5.0 7.5 5.0 35.0 4.569
15.00 7.5 5.0 7.5 40.0 4.921
20.00 10.0 2.5 10.0 45.0 5.348
25.00 12.5 0 12.5 50.0 9.071
30.00 15.0 0 12.5 55.0 12.658
Methyl orange
Phenolphthalein
Litmus
Methyl violet0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
0 5 10 15 20 25 30 35 40
pH
Titrant volume
WA-SB curve
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WA vs SB Titration Curve Regions
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
0 5 10 15 20 25 30 35 40
pH
Titrant volume
WA-SB curve
Initial Region: Use ICE
and Ka
pH = ½ (pKa + pCHA)
Buffer Region: HH Equation
pH = pKa + log (moles A- /moles HA)
At half equivalence point:
pH = pKa
Equivalence Region: Use ICE
and Kb
pOH =
½ (pKb + p[moles HA/total
volume])
Excess base region:
pOH = -log (moles excess base/ total volume)
Buffer Region: HH Equation
pH = pKa + log [moles titrant/(moles analyte –
moles titrant )]
Methyl orange
Phenolphthalein
Litmus
Methyl violet
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
0 5 10 15 20 25 30 35 40
acetic curve
lactic curve
HClO curve
Titration
Curve of
Different
Acids vs
Strong Base
• A 0.1044-g sample of an unknown monoprotic acid required
22.10 mL of 0.0500 M NaOH to reach the endpoint. (a) What
is the molecular weight of the acid? (b) As the acid is titrated,
the pH of the solution after the addition of 11.05 mL of the
base is 4.89. What is the Ka of the acid?
� A biochemist needs 750 mL of an acetic acid-sodium acetate
buffer with pH 4.50. Solid sodium acetate, NaC2H3O2, and
glacial acetic acid, HC2H3O2, are available. Glacial acetic acid is
99% pure by mass and has a density of 1.05 g/mL. If the
buffer is to be 0.20 M in HC2H3O2, how many grams of the salt
and how many milliliters of glacial acetic acid must be used?
7/25/2013
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Titration Curves of Polyprotic Acids vs SB
H2A + H2O ⇄ HA- + H3O+ Ka1
HA- + H2O ⇄ A2- + H3O+ Ka2
pH Curve will show:
� 2 Equivalence points
� Buffer region similar to weak monoprotic acid region
� Amphiprotic region
First Equivalence Point: H2A + NaOH ⇄ Na+ + HA- + H2O
Second Equivalence Point: HA- + NaOH ⇄ Na+ + A2- + H2O
Amphiprotic region starts at the 1st equivalence point region. The
region between the 1st equivalence point and 2nd equivalence point is
a buffer region composed of HA- and A2-. In this region, the buffer HA-
and A2- predominates the amphiprotic species HA-. �
Weak H2A vs SB
Initial Region: Use ICE
pH = ½ (pKa1+ pCH2A)
Buffer Region: HH Equation
pH = pKa1 + log (moles HA- /moles H2A)
Amphiprotic Region
pH = ½ (pKa1 + pKa2)
Buffer Region: HH Equation
pH = pKa2 + log (moles A2- /moles HA-)
Equivalence Region: Use ICE and Kb1
pOH = ½ (pKb1 + p[moles H2A/total volume])
Excess base region:
pOH = -log (moles excess base/ total volume)
Weak H3A vs Strong Base Titration • Suppose you want to do a physiological experiment that requires a pH 6.5 buffer. You find that the organism with which you are working is stable to a certain acid and its sodium salts (H2X: Ka1 = 2x10-2, Ka2 = 5.0x10-7). You have available 1.0 M of this acid and 1.0 M NaOH in the lab. How much of the NaOH should be added to 1.0 L of the acid to make a buffer at pH 6.50?
� What is the pH of a solution made by mixing 0.30 mole
NaOH, 0.25 mole Na2HPO4 , and 0.20 mole H3PO4 with
water and diluting with 1.00 L?
H3PO4
pKa1 = 2.12
pKa2 = 7.21
pKa3 = 12.38