acid-base equilibria 2

Acids and Bases II:

Acid-Base Equilbria

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

Acids and Bases: Chem 16 • 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

Lewis

Brønsted-Lowry

Arrhenius

Hehehe...

Acids and Bases: Chem 16 17

• The Autoionization of water -

(aq)(aq)3)(2)(2 OHOHOH OH

Equilibrium-constant expression:2

2

-3

c ]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 Ko14-

w

-7-7w

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

23

-33

100%23

M

M

MMM

• What is pH of water at its normal boiling point? Is it acidic or basic?– Given: ∆Hf

oH2O(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?

Acids and Bases: Chem 16 • 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

-33

c

But [H2O] = 55.6 M, very high and almost constant, therefore

COOH][CH

]COO][CHOH[KK

3

-33

ac

• 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

-3

a

COOH][CH

]COO][CHOH[KK

3

-33

ac

Ka is the acid-dissociation constant

• For weak bases,NH3(aq) + H2O(l) NH⇄ 4

+(aq) + OH-

(aq)

O]][H[NH

]][NHOH[K

23

4c

][NH

]][NHOH[K

3

4b

Kb is the base-dissociation constant. But why???Does the base dissociate, like acids?

[acid]

]base e][conjugatH[Ka

[base]

]acid e][conjugatOH[Kb

HA ⇄ H+ + A-

B- + H2O ⇄ OH- + 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


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

-33

a

-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

If Ka /[ ]initial is < 1.0x10-3 , 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 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.

• Combination of the anion of a weak acid with H3O+ ions from water to form nonionized weak acid molecules.


• 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


• 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 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 acidsBond strength has this periodic trend

HF >> HCl > HBr > HIAcid 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


• TERNARY ACIDS - hydroxides of nonmetals that produce H3O+ in water.– Consist of H, O, and a nonmetal. HClO4 H3PO4


• Acidity of ternary acids with same central element increase with increasing oxidation state of central element, and increasing O atomsHClO < 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 only!)

Hard and Soft Acids and BasesFrom 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 baseSOFT acid-SOFT base

>Dissociation constant

In general, molecules that involve

HARD acid-SOFT baseSOFT acid-HARD base

Hard Lewis acid

Which is the Hardest Lewis base? Which is the softest?

---3

-2

- CN ClO COOCH NO FHARDESTLewis base

SOFTESTLewis 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)

11-a2(aq)3

-23(aq)(l)2

-3(aq)

-7a1(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)

-4b1(aq)

-3(aq)(l)2

-23(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

-24(aq)

b1(aq)-2

4(aq)(l)2-3

4(aq)

b3a3

b2wa2

b1a1

KK

KKK

KK

13-a3(aq)3

-34(aq)(l)2

-24(aq)

8-a2(aq)3

-24(aq)(l)2

-4(aq)2

-3a1(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

-23(aq)(l)2

-3(aq)

-8b2(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

-8a2(aq)3

-24(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

-24(aq)

-13a3(aq)3

-34(aq)(l)2

-24(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?

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 acids

are basicExamples: NaCN (from NaOH and HCN)

K2C2O4 (from KOH and H2C2O4) Aqueous solutions of salts of weak bases and strong acids

are acidicExamples: 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


• 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 EquationSimplified equation for pH calculation involving

buffers

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

ICE 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- + H3O+ B + H2O ⇄ BH+ + 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 BuffersBuffers 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 = 2.12pKa2 = 7.21pKa3 = 12.38

Molarity H3PO4 (liquid) = 14.85 MFormula weightNaH2PO4•H2O = 137.99 g/moleNa2HPO4•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

-24

a2

componentsbuffer moles totalHPO moles POH moles -24

-42

M) (0.1)L 1

mL 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?

Acid-Base Neutralizations:Indicators, Titrations

andpH 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

-3

a

[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

Titration CurvesStrong 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

TotalVolume

(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

Titration CurvesWeak 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

TotalVolume

(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 violet

WA vs SB Titration Curve Regions

Initial Region: Use ICE and Ka

pH = ½ (pKa + pCHA)

Buffer Region: HH EquationpH = 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 EquationpH = pKa + log [moles

titrant/(moles analyte – moles titrant )]

Methyl orange

Phenolphthalein

Litmus

Methyl violet

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?

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 ICEpH = ½ (pKa1+ pCH2A)

Buffer Region: HH EquationpH = pKa1 + log (moles

HA- /moles H2A)

Amphiprotic RegionpH = ½ (pKa1 +

pKa2)

Buffer Region: HH EquationpH = 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 = 2.12pKa2 = 7.21pKa3 = 12.38

acid-base equilibria 2

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