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


• 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


• TERNARY ACIDS - hydroxides of nonmetals

that produce H3O+ in water.

– Consist of H, O, and a nonmetal.

HClO4 H3PO4

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• 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


• 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

+− +←+


+− +←+

• 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.


+− +←

→+

� 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


• 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


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

7/25/2013

16

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)


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?

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17

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)


pH = pKa1 + log (moles HA- /moles H2A)

Amphiprotic Region

pH = ½ (pKa1 + pKa2)


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

acid-base equilibria for handout chem 17

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