acid-base equilibria chem 17 handout new
TRANSCRIPT
Review of Acid and Base Definition
• 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
Acids and Bases II:
• 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
• Bronsted-Lowry Definition: Correct assignment of acid-base
conjugate pairs is based on properly drawn Lewis structures
• Amphoterism:
– Species that can behave as an acid or base are
called amphoteric.
– Amphiprotic: special term for an amphoteric
species which involves proton transfer reactions
to show behavior as either an acid or base
OH 2 ) Zn(NO HNO 2 Zn(OH) 22332 +→+
-2
4
-
2 Zn(OH)OH 2 Zn(OH) →+
HPO42- + H2O ���� H2PO4
- + OH-
HPO42- + H2O ���� PO4
3- + H3O+
• 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 ← +→+ +
acid base
BF Na BF NaF -
43 +←
→+ +
Acids and Bases: Chem 16 � 17
• The Autoionization of water
← +→+ + -
(aq)(aq)3(l)2(l)2 OHOHOH OH
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
=
=
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
• 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 the 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
� NOTE: When concentration of H+ coming from an acid source
(e.g. HA) 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?
� 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
Relative Strengths of Acids and Bases
Conjugate Acid-Base Pairs
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
-
33
ac
+
=≈
• We can simply 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
� 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 decreasing 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 ��
�������
1.0 � 10��, the simplifying assumption is valid
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
Percent Ionization of Weak
Acids/Bases• 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 - 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 =
� The order of decreasing 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
• 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?
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
Down a group: ����size, ����energy to break H- bond
(����electronegativity), ����acidity
Strengths of Acids and Bases
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
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)
Base Strength of Amines
� The electronic properties of
the substituents (alkyl groups
enhance the basicity, aryl
groups diminish it).
� Steric hindrance offered by
the groups on nitrogen.
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)
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 flip-flops of your roommate. To neutralize the acid, you looked for a base in the dorm stock room, and you found two salts –sodium bicarbonate and sodium phosphate. Which of the two salts will you use to quickly and effectively neutralize the acid?
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 bases
are neutralExamples: 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
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 present in the initial concentration
of components
– 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
Henderson-Hasselbalch Equation• 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
C
E 0.12 - x 0.10 + x x
(0.12)
(0.10) pK pH a +=
x) (0.12
x) (x)(0.10 K 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
Preparation of Buffers• Buffers can be prepared in three ways
– Adding a solid salt component to a liquid• 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 salts 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 ×=+
Common Ion Effect – Buffers
• Non-buffer Case: Consider the a solution of 50.00 mL of 0.05 M acetic acid, CH3COOH. – Determine the pH of the solution.
– Calculate the pH of the final mixture after adding 10.00 mL of 0.05 M HCl.
– What is the ∆pH?
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 + x + x + x + x + x ---- x x x x
IIII
CCCC
• Buffer CASE - Consider a solution of 50.00 mL 0.05 M acetic acid
(CH3COOH) and 0.05 M sodium acetate (NaCH3COO).
– Determine the pH of the solution.
– Calculate the pH of the final mixture after adding 10.00 mL of 0.05 M HCl.
– What is the ∆pH?
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
=
=
+=
Buffer Capacity
�Amount of acid or base (usually in mL) needed to
change the pH of a buffer solution by 1 degree.
� ∆B – number of moles of acid/base added per 1 L of buffer solution
� ∆pH – change in pH associated with the addition of acid/base
� ↑∆B, ↓ ∆pH, ↑β,
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?
Buffer capacity is highest when pH = pKa
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
-
3In
+
= ][In
[HIn]K ]O[H
-In3 =+
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
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
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?
� 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