15.1 Acids - Bases in Water

January 13 1 Acids - Bases in Water

15.1 Acids - Bases in Water …more equilibrium

Dr. Fred Omega Garces Chemistry 201 Miramar College


Acids (Properties) •Taste Sour •Dehydrate Substances •Neutralizes bases •Dissolves metals

Examples: •Juices: TJ, OJ, AJ •Wine •Banana •Coffee •Vitamin C •Soda

Acids-Bases Characteristics Base (Properties) •Taste Bitter •Denatures Proteins •Neutralizes acids •Turns metal g hydroxides

Examples: •Milk of Magnesia •Lime water •Lye, Drain •Ammonia •blood •Soap


Practice Naming Acids HNO2 HI HF H3PO4 H3PO3 HClO2 HClO H2CO3 HCN HC2H3O2 H2Cr2O7 HClO4

Nitrous Acid

Hydroiodic Acid

Hydrofluoric Acid

Phosphoric Acid

Phosphorous Acid

Chlorous Acid

Hypochlorous Acid

Carbonic Acid

Hydrocyanic Acid

Acetic Acid

Dichromic Acid

Perchloric Acid

Oxy-anions Oxy-acids

Add H+

Add H+

Add H+

Add H+


Svante Arrhenius (1859 - 1927)

Acid - increases H+ (H3O+) concentration

Bases - increases OH- concentration Arrhenius acids and bases are limited to aqueous solutions.

Behavior in water Acids are substances that have H in its formula and dissociates in water to yield yield , H+ (actually H3O+) H2SO4(aq) g H+ (aq) + HSO4 - (aq)

Bases are substances that have OH in its formula and dissociates in water to yield OH-. Hydroxide ions, OH-, are specie which readily react with H+ ions to form water: H+

(aq) + OH - (aq) g H2O (l)

Arrhenius Definition Svante Arrhenius, a Swedish scientist, was the first to recognize that acids, bases, and salts, when dissolved in water are dissociated into ions


Factors Affecting Acid/Base Strength What determines the strength of acids and Base? Dissociation property- Electrolyte - Substances which dissociate in water. Strong electrolyte completely dissociates to ions Weak electrolyte undergoes partial dissociation.

Acid Examples : HClO4 + H2O g H3O+ + ClO4

- 100% Dissociation (Strong Acid) HNO3 + H2O g H3O+ + NO3

- 100% Dissociation (Strong Acid) H2S + H2O D H3O+ + HS- Less 100% Dissociation (Weak Acid)

Base Examples : NaOH + H2O g OH- + Na+ 100% Dissociation (Strong Base) Ca(OH)2 + H2O g 2OH- + Ca+2 100% Dissociation (Strong Base) NH3 + H2O D OH- + NH4

+ Less 100% Dissociation (Weak Base)


Strong Acids Strong Acids Strong Electrolyte that ionizes (break up to H+ ions) in solution 100 % of time. Examples :

* H2SO4 , HClO4 , HClO3 , HNO3 , HX (X = Cl, Br, I)

Only strong acids dissociates 100% in water, Ka >>1

These are Strong Acids.

Rxn: HClO4 + H2O g H3O+ + ClO4- 100% Dissociation

HNO3 + H2O g H3O+ + NO3- 100% Dissociation

H2S + H2O D H3O+ + HS- Less 100% Dissociation

*All other acids under goes less than 100 % Dissociation.


Strong Acids; indicative by acid dissociation constant, Ka

The Ka values shown are the acid dissociation constant for monoprotic acids. The Ka is the equilibrium constant for the ionization of the acid, HA D H+ + A. Strong acids do not have meaningful Ka values and are left out from this table.


Strong Base Strong Base: Strong Electrolyte that ionizes (break up to OH-

ions) in solution 100 % of time. (Remember solubility Rules) Examples : Group-1 hydroxide: LiOH, KOH, NaOH, RbOH Heavy Grp-2 hydroxides: Ca(OH)2 , Sr(OH)2 , Ba(OH)2 , Ionic metal oxides: Li2O, Na2O, K2O, CaO

These are bases which dissociate 100% in water (or in the case of oxides,

react with water) to produce hydroxides.

Rxn: KOH g OH- + K+ 100% Dissociation Ca(OH)2 g 2OH- + Ca +2 100% Dissociation Li2O + H2O g 2OH- + Li+ 100% Dissociation CaO + H2O g Ca(OH)2 strong base produced NH3 + H2O D OH- + NH4

+ Less 100% Dissociation

*All other bases undergo less than 100 % Dissociation.


Chemistry of Acids and Bases Acids Base

Strong Acid Weak Acid


Hydronium ion: HCl(aq) g H+(aq) + Cl-(aq)

Proton ? H

Protons in water: Hydronium

e- p Proton in H2O

H+ is always associated with solvent. H+ attacks water

solvent to form hydronium H3O+ In reality:

H+ + H2O g H3O+

H+ protons & H3O+ Hydronium ion H+ (aq) D H3O+

considered the same. These terms are used interchangeably


Water: Acid-Base Properties Autoionization (Self-Ionization) of water Why does water have a pH of 7 ? Water is Amphoteric (it reacts with itself) 2 in 1 billion self-ionize.

H2O (l) + H2O (l) + E D H3O(aq) + OH- (aq)

Endothermic reaction ΔH°rxn = +55.9 kJ, Keq = Kw

Kw (ion-product constant) = 1•10-14

@ 25°C

Temperature Dependency of Kw

The equation Kw = [H3O+][OH-] is valid in pure water. In any aqueous solution. Kw is temperature-dependent, the auto-ionization rxn is endothermic, so Kw increases with temperature.

°C Kw

10 0.29•10-14

15 0.45•10-14

20 0.68•10-14

25 1.01•10-14

30 1.47•10-14

50 5.48•10-14


Consequences of Autoionization • A change in [H3O+] causes an inverse change in [OH-] & vice versa. Higher [H3O+] g lower [OH-] Higher [OH-] g lower [H3O+] • Both ions are present in all aqueous systems.

Acidic solution g [H3O+] > [OH-] Neutral solutions g [H3O+] = [OH-] Basic Solutions g [OH-] > [H3O+]

(as shown later)


Kw & [H3O+]

How does Kw dictate concentration of H+ and OH- ?

Kw = 1 ⋅10−14 = H3O+[ ] OH−[ ] at 25°C

For pure water, solve the iCe-table to determine the concentration of

[H3O]+ & [OH]-.

2H2O(l) ! OH-(aq) + H3O+

(aq )

i constant 0 0Δ - 2x + x + x[e] constant x x

Kw = 1 ⋅10-14 = x2

1 ⋅10-7M = x = [H3O]+ = [OH]-


For pure water [H3O+] = 1•10-7 M Since 1•10-7 M = 0.0000001M Concentration is so small, it is much more convenient to use a scale called “p-H” (power of hydrogen)

1•10-7 M g pH = 7

pH = - log [H3O+] or [H3O+] = 1•10-[pH]

Example: Calculate: pH for [H3O+] = 1.67•10-5 M [H3O+] of pH = 4.777 Calculator Sequence:

Answer : 4.777 1.67•10-5

pH Calculation

1.67 EE 5 + / - log + / - 4.78 + / - 10x


pH Scale pH and the Concentration of Acids

Conc [H3O+] Exp [H3O+] pH pOH

1M 1•100 0 14 0.1M 1•10-1 1 13 0.01M 1•10-2 2 12 0.001M 1•10-3 3 11 0.0001M 1•10-4 4 10 0.000001M 1•10-5 5 9 0.0000001M 1•10-6 6 8 0.00000001M 1•10-7 7 7 0.000000001M 1•10-8 8 6 0.0000000001M 1•10-9 9 5 0.00000000001M 1•10-10 10 4 0.000000000001M 1•10-11 11 3 0.0000000000001M 1•10-12 12 2 0.00000000000001M 1•10-13 13 1 0.000000000000001M 1•10-14 14 0


Other “p” Scale

“p” scale is the negative log

Convenient for expressing magnitude of small numbers i. [H3O+] pH = - log [H3O+] and [H3O+] = 1•10-[pH]

ii. [OH-] pOH = - log [OH-] and [OH-] = 1•10-[pOH]

ii. Keq i.e., Kw , Ka & Kb pKeq = - log [Keq] and [Keq] = 1•10-[pKeq]

pKw = - log [Kw] and [Kw] = 1•10-[pKw]

pKa = - log [Ka] and [Ka] = 1•10-[pKa]

pKb = - log [Kb] and [Kb] = 1•10-[pKb]


Determining pH, pOH, [OH-], [H3O+]

[H3O+] • [OH-] = Kw & pH + pOH = pKw

[H3O+] [OH-] Kw/

Kw/

-log [H3O+]

10-pH 10-pOH

-log

14.0 - pOH

14.0 - pH

[H3O+] • [OH -] = Kw

pH + pOH = pKw= 14

Use this chart to determine acid and base concentration at (25°C) Kw = 1.00 •10-14 & pKw = 14


Kw, pH and pOH

pH measures the concentration of [H3O+] Kw = 1•10-14 = [H3O+] [OH-] pKw = pH + pOH 14 = pH + pOH

Example: What is the pH of a solution having a [OH-] = 3.0•10-13 M

Way of Multiplying Way of adding / subtracting Using the formula: Using the formula: 1•10-14 = [H3O+] [OH-] 14 = pH + pOH

1•10-14 = [H3O+] [OH-] 14 = pH + pOH 1•10-14 =[H3O+] [3.0•10-13] 14 = pH + 12.52 [H3O+] =3.33•10-2M pH = 1.48 pH = 1.48


pH Values pH Values for various samples:

• Smaller the pH value the greater the acidity

• larger the pH value the lower the acidity

Acid: pH < 7 Neutral: pH = 7 Base: pH > 7


Rainwater pH (H20) not 7.0

Rain water isn’t neutral

Other sources of Acidic Rain: SO2(g) + O2(g) g 2 SO3(g) & N2 g) + O2 g) g 2 NO2(g) from burning fossil from car combustion fuels and from volcanoes

2 SO3(g) + H2O(l) g H2SO4(aq) sulfuric acid

2 NO2(g) + H2O(l) g HNO3(aq) + HNO2 (aq) nitric acid nitrous acid

Dissolving CO2 CO2(g) + H2O(g) g H2CO3 (aq ) carbonic acid

H2CO3 (aq) ionization: H2CO3 (aq) g H+ (aq) + HCO3 - (aq)

Overall: CO2(g) + H2O(l) g H+ (aq) + HCO3 - (aq)


Rainwater and Acid Rain Affects of Acid Rain on

Historical Artifacts and the Environment b) The same statue some 50 years later (June 1994) after the destructive effect of acid rain.

a) Details of the marble that is part of the Washington Square Arch in Washington Square Park completed on July 17, 1944.

The combustion of fossil fuels produces CO2; automobile contain NO2; and burning low-grade coal releases SO2. All these chemicals contribute to acid rain.

A forest damaged by acid rain


Measuring pH in the Laboratory pH range for various indicators

Mechanism for phenolphthalein indicator. At Low pH phenolphthalein is colorless and has a structure in which there is a five membered ring. In the presence of excess acid the five membered ring is broken and the resulting change in conformation gives rise to a compound which is pink.

Some common acid-base indicators. The color changes occur over a range of pH values. Notice that a few indicators have two color changes over two different pH ranges.


Acid Base in Water

Summary Pure water has a low conductivity because it autoionizes to a small extent.

This process is described by an equilibrium reaction whose equilibrium

constant is the ion-product constant for water, Kw (1•10-14 at 25°C). Thus,

[H3O+] and [OH-] are inversely related. In acidic solution, [H3O+] is

greater than [OH-], the reverse is true in basic solution, and the two are

equal in neutral solution. To express small values of [H3O+] more simply, we

use the pH scale (pH = -log [H3O+]). A high pH represents a low [H3O+].

Similarly, pOH = -log [OH-], and pK = - log K. At 25°C, in acidic solutions,

pH < 7.00, in basic solutions, pH > 7.00; and in neutral solutions, pH = 7.0.

The sum of the pH and pOH equals pKw (14.00 at 25°C)

January 13 1 Acids - Bases; Proton Transfer

15.2 Acids - Base Proton Transfer


Important Notes:

Ka when H3O+ is produced, Kb when OH- is produced


Brønsted-Lowry Acids-Bases

Brønsted - Lowry definition

Acid - Proton H+ (H3O+) donor Base - Proton H+ (H3O+) acceptor.

example: acids: HCl (aq) g H+(aq) + Cl- (aq) Bases: NH3 (aq) D NH4

+ (aq) HCl (aq) + NH3 (aq) g NH4

+ (aq) + Cl- (aq

In an acid - base reaction, H+ & OH- always combine together to form water and an ionic compound (a salt):

HCl(aq) + NaOH(aq) g H2O(l) + NaCl(aq)


Neutralization reaction is a reaction between acids and base

NH3 + HCl g NH4+ + Cl-

Base NH3 : Bronsted-Lowery: proton acceptor (Conjugate Base1) Cl-: Bronsted-Lowery: proton accept (Conjugate Base2)

Acid HCl: Bronsted-Lowery: proton donor (Conjugate Acid2) NH4

+: Bronsted-Lowery: proton donor (Conjugate Acid1)

Consider a Neutralization Reaction

Base1 Conj Base2

Conj Acid1

Acid2 Conjugates differ by a proton (H+)


… more Conjugate Acid-Base pairs Other Examples of Conjugate Pairs:

1. HCO3- (aq) + H2O(l) D H3O+

(aq) + CO3 2- (aq) Acid Base Conj.. acid Conj. base

of H2O of HCO3-

2. HCO3- (aq) + HS-

(aq) D H2S(aq) + CO3-2

(aq) Acid Base Conj. acid Conj. base

of HS- of HCO3-

3. CH3CO2H(aq) + CN-(aq) D HCN(aq) + CH3CO2

- (aq)

Acid Base Conj. acid Conj. base

of CN- of CH3CO2H

4. HSO3-(aq) + NH3 (aq) D NH4

+(aq) + SO3

-2(aq)

Acid Base Conj. acid Conj. base of NH3 of HSO3

-


Conjugate (Example)

Q: Give the conjugates of NH3, IO-, C2H3O2-, HAsO4

2-

A: NH4 , HIO, HC2H3O2 , H2AsO4

-

Q: Identify the acids and bases on both sides of the following equations and show the conjugate acid-base pairs.

HNO2 + H2O g NO2- + H3O+

A:

HPO42- + NH4

+ g H2PO4- + NH3

B:

Acid1 ConjAcid2 ConjBase1 Base2

Base1 ConjAcid1 Acid2 ConjBase2


Relative Strengths and Net Directions The net direction of an acid-base reaction depends on the relative strengths of the acids and bases involved. A reaction proceeds to a greater extent in the direction in which a stronger acid and stronger base form a weaker acid and weaker base. The strength of the conjugate acid-base pairs are shown as the stronger the acid, the weaker the conjugate base. The strongest acid appears on the top left and the strongest base on the bottom right. When an acid reacts with a base farther down the list, the reaction proceeds in the forward (right) direction.


…more Acids and Bases conjugate pairs Conjugate Acid Conjugate Base

Name Formula Formula Name Perchloric acid HClO4 ClO4 - Perchlorate ion Sulfuric acid H2SO4 HSO4

- Hydrogen sulfate ion Hydrochloric acid HCl Cl - Chloride ion Nitric acid HNO3 NO3

- Nitrate ion Hydronium ion H3O + H2O Water Hydrogen sulfate ion HSO4

- SO4 2- Sulfate ion Phosphoric acid H3PO4 H2PO4 - Dihydrogen phosphate ion Acetic acid CH3CO2H CH3CO2 - Acetate ion Carbonic acid H2CO3 HCO3

- Hydrogen carbonate ion Hydrogen sulfide H2S HS - Hydrogen sulfide ion Dihydrogen phosphate ion H2PO4 - HPO4

2- Hydrogen phosphate ion

Ammonium ion NH4 + NH3 Ammonia Hydrogen cyanide HCN CN- Cyanide ion Hydrogen carbonate ion HCO3 - CO3

2- Carbonate ion Water H2O OH - Hydroxide ion Ethanol C2H5OH C2H5O - Ethoxide ion Ammonia NH3 NH2

- Amide ion Hydrogen H2 H - Hydride ion Methane CH4 CH3 - Methide ion

Stro

ng A

cids

Strong Base


Concentration Calculations for Strong Acids/Bases- Since strong acids or base undergo 100% ionization, the concentration of H+ or OH- is the analytical concentration of these acid or base solution. That is [HClO4] = [H3O+] yields pH of solution [HNO3] = [H3O+] yields pH of solution

and [KOH] = [OH-] yields pOH then pH [Ca(OH)2] = 2 [OH-] yields pOH then pH

Consider 1. Acid: HClO4 + H2O g H3O+ + ClO4

-

0.0010 M

2. Base: Ca(OH)2 g 2OH- + Ca +2 0.0010 M


Concentration Calculations for Strong Acids/Bases Since strong acids or base undergo 100% ionization, the concentration of H+ or OH- is the analytical concentration of these acid or base solution. That is [H2SO4] = [H3O+] yields pH of solution [HNO3] = [H3O+] yields pH of solution

and [KOH] = [OH-] yields pOH then pH [Ca(OH)2] = 2 [OH-] yields pOH then pH

Consider 1. Acid: HClO4 + H2O g H3O+ + ClO4

-

0.0010 M Since this proceeds 100%, then [H3O+ ] also equals 0.0010 M The pH of the solution = - log[0.0010] = 3.00

2. Base: Ca(OH)2 g 2OH- + Ca +2 0.0010 M Since this proceeds 100%, then [OH-] equals 2 • 0.0010 M The pOH of the solution is - log[0.0020] = 2.70 and pH = 11.30


Weak Acids / Bases

Weak acids and bases ionizes less than 100%? Dissociation properties- Electrolyte - Substances which dissociate in water. Strong electrolyte completely dissociates to ions Weak electrolyte undergoes partial dissociation.

Acid Examples : H2S + H2O D H3O+ + HS- Less 100% Dissociation (Weak Acid)

Base Examples : NH3 + H2O D OH- + NH4

+ Less 100% Dissociation (Weak Base) Concentration of Acid / Base depends on the percent dissociation.

Both of these are an equilibrium Problem !!!


Weak Acids and Bases, Ka values (revisited)

The % ionization is dependent of the Keq constant. For acids, this is called K-acid or Ka.


Solving Weak-Acid Equilibria As in equilibrium type problems, there are also two type of acid base equilibrium problems. • Given the equilibrium concentrations, find Ka or Kb • Given the Ka (or Kb) and concentrations information, find the other equilibrium concentrations.

Assumptions

• [H30+] from water, ~ 1•10-7 is negligible compare to x.

[H3O+]eq = 1•10-7 M + x = x

• A weak acid has a very small Ka, therefore x is very small compared to the original acid concentration.

[HA]eq = [HA]int - x = [HA]int

Problem Solving Strategy:

• Balance eqn & Mass action Expression

• Construct iCe table

• define x as the unknown

• Make simplifying assumptions

• Solve Mass action expression for x

• Check assumption using the 5% rule.

…do second iteration if error is greater than 5%.


Weak Acids : pH Calculation

Consider HF: HF(aq) + H2O (l) D H3O+

(aq) + F- (aq)

100 molc’ Ka = 6.8•10-4

100 molc’ HF g 97molc’ HF & 3F- & 3H+

(Analytical Conc) (Solution at Equilb)

HF(aq) + H2O (l) D H3O+ (aq) + F-

(aq) i 0.10M lots 1•10-7 0 C e


Weak Acids : pH Calculation

Consider HF: HF(aq) + H2O (l) D H3O+

(aq) + F- (aq)

100 molc’ Ka = 6.8•10-4

100 molc’ HF g 97molc’ HF & 3F- & 3H+

(Analytical Conc) (Solution at Equilb)

HF(aq) + H2O (l) D H3O+ (aq) + F-

(aq) i 0.10M lots 1•10-7 0 C -x -x +x +x e 0.10 - x lots 1•10-7+x x

Amount Dissociation: Determine conc. Ka = [H3O+] [F- ] << 1 by Ka [HF]

Since Ka << 1 , The major component in solution is the HF


The pH value of weak acids is higher than those of strong acids for solutions with the same concentration. Example: Determine the pH of 0.10 M HF and of 0.10 M HCl

Weak Acids : pH comparison with strong acid -

HCl(aq) ! H+(aq) + Cl-(aq) 100% dissociation

i 0.10 M 1 ⋅10-7 0.0 C - 0.10M + 0.10M + 0.10Me 0 0.10M 0.10M

pH = -log 0.10[ ]= 1.00


The pH value of weak acids is higher than those of strong acids for solutions with the same concentration. Example: Determine the pH of 0.10 M HF and of 0.10 M HCl

Weak Acids : pH comparison with strong acid

HCl(aq) ! H+(aq) + Cl-(aq) 100% dissociation

i 0.10 M 1 ⋅10-7 0.0 C - 0.10M + 0.10M + 0.10Me 0 0.10M 0.10M

pH = -log 0.10[ ]= 1.00

HF(aq) ! H+(aq) + F-

(aq) ka = 6.8 ⋅10-4

i 0.10 M 1 ⋅10-7 0.0 less than 100% dissociationC -x +x +x

e 0.10 -x 1 ⋅10-7 + x x

Assumption method:

ka = 6.8 ⋅10-4 = x ⋅ 1 ⋅10-7 + x( )

0.10-x assume 1 ⋅10-7 + x ≅ x

0.10-x ≅ .010

6.8 ⋅10-4 = x ⋅x0.10

after applying simplifying assumption

x2 = 6.8 ⋅10-5

solving for x Verifyinging assumption

x =[H+] = 8.25 ⋅10-3M, 8.25 ⋅10-3

.10 = 8.24% > 5% rule

2nd iteration:

6.8 ⋅10-4 = x ⋅ x( )0.10-x

assume 1 ⋅10-7 + x ≅ x 0.10 - 8.25 ⋅10-3 ≅ 0.0918

6.8 ⋅10-4 = x ⋅ x( )

0.10-8.25 ⋅10-3

6.8 ⋅10-4 = x ⋅ x( ).0918

solving for x Verifyinging assumption

x =[H+] = 7.90 ⋅10-3M, pH = 2.10 8.25 ⋅10-3 −7.90 ⋅10-3( )

0.0918 = .38% < 5% rule

∴ Two iteration of the simplifying assumption method leads to the correct solution

The Quadratic Formula can also be used to solve this problemQuadratic Formula method:

ka = 6.8 ⋅10-4 = x ⋅x0.10-x

solving for x, x =[H+] = 5.8 ⋅10-3M, 5.8 ⋅10-3 > 5% of 0.05

Using quadratic equation : x2 + 6.8 ⋅10-4 x - 6.8 ⋅10-5 = 0,

x =[H+] = 7.9 ⋅10-3M, pH = 2.10


Find Ka, Given Concentrations

Type problem: HA + H2O (l) D H3O+ (aq) + A-

(aq)

Mass Action Expression: Ka = [H3O+] [A- ] [HA]


Find Concentrations, Given Ka

Type problem: Concentrations provided for reaction HPr + H2O (l) D H3O+

(aq) + Pr- (aq)

What is the [H3O+] of 0.10M, Propanoic acid, Ka = 1.3•10-5 Use iCe table to express equilibrium concentrations in terms of x Solve Mass action expression and solve for concentrations.


Find Concentrations, pH and Ka, Given percent dissociation or ionization (α)-

A 0.200 M solution of a weak acid HX is 9.40 percent ionized (α). Using this info., calculate [H+], [X-], [HX] and Ka for HX. Percent dissociation (α) = [HA]Dissoc = X

[HA]Init

[HA]Init = 0.200M * 9.40% of 0.200 M is 0.0188 M (x = 0.0188)


Weak Acids: percent dissociation (α) Lower concentration of weak acids will dissociate to a greater extent because according to LeChatelier, the system will go in the direction to offset the lower number of particles (concentrations) by breaking up to more particles.* For a weak acid the percent ionization varies as the inverse square-root of the acid concentration.

HF(aq) ! H+(aq) + F -

(aq) less than 100% dissociation i [HF]o 0.0 0.0 C - x +x +x e [HF]o - x x x

ka = x ∗x[HF]o - x

=x2

[HF]o

assume x << [HF]o

x = ka[HF]o[ ]1 /2

% α =x

[HF]o

•100

% α = ka[HF]o[ ]1 /2

[HF]o

•100 = ka[ ]1 /2

[HF]o1/2 •100

Note: % Ionization is inversely related to [HF]o concentration.

A graph illustrating that the percent ionization, a, is

inversely proportional to the original acid concentration.


Behavior of Polyprotic Acids Many acids have more than one ionizable H atom. Polyprotic acids are acids with multiple ionizable protons

Consider the following Acids: 1.  HNO2 + H2O g H3O+ + NO2

- Ka1 = 7.1•10 -4 Monoprotic

2. H2SO3 + H2O g H3O+ + HSO3- Ka1 = 1.7•10 -2

HSO3- + H2O D H3O+ + SO3

2- Ka2 = 6.4•10 -8 Diprotic

3. H3PO4 + H2O D H3O+ + H2PO4- Ka1 =7.3•10 -3

H2PO4- + H2O D H3O+ + HPO4

2- Ka2 = 6.2•10 -8

HPO42- + H2O D H3O+ + PO4

3- Ka3 = 4.2•10 -13 Triprotic

Note that the proceeding Ka’s are always smaller than the previous. This makes sense in that the

removal of the next H+ is from a negatively charge specie. The proceeding Ka’s (for successive

H+ ionization) are generally smaller by factors of 1000.


Ka’s of Polyprotic Acids The pH estimate of a polyprotic acid can be estimated by only the pKa1 for polyprotic acids if pKa1 differs by a factor of 103 or greater to pKa2


Series Equation for Polyprotic

Consider the ionization of Carbonic acid: The reaction is never: H2CO3 + H2O g 2 H3O+ + CO3

2-

rather it is a two step process:

step i ) H2CO3 + H2O g H3O+ + HCO3- Ka1 =4.3•10 -7

Ka1 = [H3O+] [HCO3- ] / [H2CO3] g [H3O+](1) g pH (1)

step ii) HCO3- + H2O g H3O+ + CO3

2 - Ka2 = 5.6 • 10 -11

Ka1 = [H3O+] [CO3-2 ] / [HCO3

- ] g [H3O+](2) g pH (2)

Note that when Ka1 >> Ka2 ,

then [H3O+](1) >> [H3O+](2) and therefore

[H3O+](T) = [H3O+](1) + [H3O+](2) = [H3O+](1)

pH soln = pH (1)

When solution can be treated as a mixture of strong acid and weak acid, the contribution from the weaker specie is negligible.

negligible


Example: pH Estimate for Polyprotic Acids

Phosphoric acids is a triprotic acid. Calculate the pH of a 0.100 M solution of phosphoric acid.

H3PO4 + H2O g H3O+ + H2PO4- Ka1 =7.3•10 -3

H2PO4- + H2O g H3O+ + HPO4

2- Ka2 = 6.2•10 -8

HPO42- + H2O g H3O+ + PO4

3- Ka3 = 4.2•10 -13

Ka1 = [H3O+] [H2PO4

-] = x2 g 7.3 •10 -3 (0.100 ) = x2 a x = [H3O+] = 2.70•10 -2 M [H3PO4 ] 0.100

Ka2 = [2.7•10-2+x] [HPO42-] = x = 6.2•10-8 a x = [H3O+] = 6.2•10 -8 M

[2.7•10-2-x ]

Ka3 = [2.7•10-2+x] [PO43-] = 4.35•105 x = 4.2•10-13 a x = [H3O+] = 9.64•10 -19 M

[6.2•10-8-x ]

[H3O+ ] = 2.7•10-2 M + 6.2•10 -8 M + 9.64•10 -19 M ~ 2.7•10 -2 M pH = 1.57

Name Formula Ka1 Ka2 Ka3

Phosphoric H3PO4 7.31⋅10-3 6.2 ⋅10-8 4.2 ⋅10-13

Acid


In Class Assignment: pH Estimate for Polyprotic Acids

Sulfuric acid is a diprotic acid. Its two stages of ionization is shown below. H2SO4 (aq) g H+ + HSO4

-(aq) pKa = ???

HSO4

-( (aq) D H+ + SO4

2- pka = 1.991 p766 Silberberg

Use your text to find relevant information to determine the following:

a)  Calculate the concentration of HSO4- & SO4

2- ion in a 0.5 M H2SO4 solution

b)  Determine the pH of the solution.


Weak Bases

In a base equilibrium process, hydroxides are formed. When the hydroxide formation is not stoichiometric, the strategy of equilibrium analysis must be used.

Weak bases produces OH- at less than 100%?

Base Examples :

NH3 + H2O D OH- + NH4+

HS- + H2O D OH- + H2S

Less 100% Dissociation (Weak Base)


Types of Weak Bases Two type of behavior for hydroxide formation. 1) Molc’ with atom with nonbonding pair of electrons serving as proton acceptor

site. i.e., amines such as NH3, CH3NH2, pyridine. (Lone pair)

2) Anions of weak acids (conjugate base of a weak acid). i.e., acetate ion (C2H3O2

-) which is the conjugate of acetic acid HC2H3O2. (Lone pair via anion)


Example: pH for weak Base soln’ Calculate [OH -] and pH for a 0.075 M solution of ethylamine, C2H5NH2. (Kb= 6.4•10-4)

C2H5NH2 + H2O D C2H5NH3+ + OH -

i 0.075 M lots 0 1•10-7

C -x -x + x + x [e] 0.075 - x lots x 1•10-7 + x

Kb = 6.4•10-4 = [OH -] [C2H5NH3

+ ] = x2

[C2H5NH2 ] 0.075

6.4•10-4 (0.075 ) = x2 g x = [OH -] = [C2H5NH3 ] = 6.93•10 -3 M, pOH = 2.16, pH = 11.84

Verify assumption method, % error = 9.2% > 5% necessitate 2nd iteration

Second iteration Kb = 6.4•10-4 = x2 = x2

0.075 - 0.00693 0.0681 yields x = [OH -] = [C2H5NH3 ] = 6.60•10 -3 M, pOH = 2.18, pH = 11.82

Using the Quadratic Equation, pH = 11.82, confirms second iteration answer.


Relationship between Ka and Kb We note that weak acid have conjugate strong base. This suggest there is a relationship between the ka of the weak acid and the Kb of the strong base.

That is the strength of an acid and the strength of its conjugate base is expressed by: Ka • Kb = Kw

Consider (1) HF + H2O D H3O+ + F- Ka = 3.5•10-4

(2) F- + H2O D OH- + HF Kb = ?

ka =H3O+[ ]• F−[ ]• OH−[ ]

kb • F−[ ]kb =HF[ ]• OH−[ ]

F-[ ] ka =

H3O+[ ] • F−[ ]HF[ ]

kb • F-[ ]OH−[ ]

= HF[ ]Plug

ka• kb = H3O+[ ]• OH−[ ] = Kw

ka• kb = Kw


Amphoteric Specie which can act as either a base or an acid by donating or accepting protons are considered amphoteric substances. Under certain conditions, chemicals can either give up a proton or accept a proton.

i.e., H2O, HCO3- , HPO4

2 -, H2PO4-

Consider Hydrogen carbonate (bicarbonate) :

Acid process:

H2CO3 + H2O g H3O+ + HCO3- 1st H+, Ka1 =4.3 •10 -7

HCO3- + H2O g H3O+ + CO3

-2 2nd H+, Ka2 = 5.6 • 10 -11

Base process:

CO3-2 + H2O g OH - + HCO3

- 1st OH-, Kb1 = 1.79•10 -4

HCO3 - + H2O g OH - + H2CO3

2nd OH-, Kb2 =2.33 •10 -8

Note from the above reaction that Ka1 competes against Kb2

these two equilibrium constant are related by Kw: Kb2 = kw / Ka1

To determine acidity or basicitiy of an amphoteric solution, the Ka’s and Kb’s must be weighted against each other.


Is the solution Acidic or Basic?

Other salts that pose the same problem-

NaHSO3 sodium bisulfate

NaH2PO4 sodium dihydrogen phosphate

K2HPO4 Potassium biphosphate

KHC2O4 Potassium oxalate

KH2C6H5O7 Potassium citrate

NH4HCO3 Ammonium bicarbonate

From polyprotic acids

Amphoteric Behavior of Carbonic Acid

H2CO3 HCO3- CO3

-2 Ka1 = 4.3•10-7 Ka2 = 5.6•10-11

Kb1 = 1.79•10-4 Kb2 = 2.33•10-8


From the magnitude of the equilbrium constant for bicarbonate ion - For the HCO3

- : |Kb2| > |Ka2| In this process, the HCO3

- is a better base than HCO3- is an acid.

Therefore is a solution basic or acidic, if HCO3

- is introduced in a solution i.e., NaHCO3 ? i.e., 42 g NaHCO3 is added to 500 ml of water, What is the pH ?

Amphoteric Behavior of Carbonic Acid

H2CO3 HCO3- CO3

-2 Ka1 = 4.3•10-7 Ka2 = 5.6•10-11

Kb1 = 1.79•10-4 Kb2 = 2.33•10-8

HCO3-

NaHCO3

Kb2 = 2.33•10-8 Ka2 = 5.6•10-11 OH- H+

To determine the pH, solve the equilb rxn for both reaction. Kb2 = [OH -] [H2CO3

] g [OH-] & Ka2 = [H3O+] [CO32 ] g [H3O+]

[HCO3- ] [HCO3

- ]

Excess of OH- and H3O+ will determine the pH of the solution.

From inspection, Kb2 >> Ka2, therefore this solution is basic.


Solubility of salts - influenced by acid, base properties of the salt’s cation & anion.

Idea - LeChatelier’s Principle is obeyed. Consider Solubility of BaF2 in acidic solution: BaF2(s) D Ba+2

(aq) + 2F- (aq) Eqn(1) but F - + H2O D OH-

(aq) + HF

Factors Affecting Solubility: Acidic Soln’

F- reacts further decreasing the amt. of F - in eqn 1

As HF is produced, more BaF2 dissolves to maintain Ksp equilibrium.

F - can be removed further by addition of H3O+

BaF2(s) + H3O+ (in acidic conditions, soluble)

D Ba+2 (aq) + 2F - (aq)

+ H3O+ D HF

Solubility increases because H3O+ removes F - and converts it to HF

BaF2(s) Ba+2 (aq) + 2F- (aq)

Acidic Solution, Solubility increase


Factors Affecting Solubility: Basic Soln’

F- reacts further to produce HF from eqn 1.

According to LeChatelier’s principle, less BaF2 will not dissociate.

Solubility of salts - influenced by acid, base properties of the salt’s cation & anion.

Idea - LeChatelier’s Principle is obeyed. Consider Solubility of BaF2 in basic c solution: BaF2(s) D Ba+2

(aq) + 2F- (aq) Eqn(1) but F - + H2O D OH-

(aq) + HF

BaF2(s) Ba+2 (aq) + 2F- (aq)

Basic Solution, Solubility decrease

Solubility decreases because F- is generated when HF reacts with excess OH-. Increase F- shift equilibrium back to solid.

BaF2(s) + OH- (in basic conditions, insoluble)

D Ba+2 (aq) + 2F -(aq)

+ H2O D HF (aq) + OH- D F - (aq) + H2O


Acid Base Reaction, which chemical dominates Consider the following reactions below. Determine which chemical in each equation is the dominant specie in solution. Which direction will the equilibrium favor?

H2S + H2O D H3O+ + HS-

H2PO4- + NH4

+ D H3PO4 + NH3

HBrO4 + BrO3- D HBrO3 + BrO4

-

HCH3CO2 + CN- D CH3CO2- + HCN

HIO

HF

H2SO3

H2C2O4 NaHCO3

KH2PO3

H2S

NaCN


In Class Assignment:

1  Calculate the pH if 20.0 grams of sodium hydroxide is dissolved in enough water to form 1.00L solution.

pH =

2.  Calculate the pH if 52.0 grams of sodium bisulfite is dissolved in enough water to form 1.00L solution.

pH =

3.  Calculate the pH if 35.5 grams of sodium phosphate is dissolved in enough water to form 1.00L solution

pH =

pH of Sodium hydroxide, sodium bisulfite and sodium phosphate NaOH, NaHSO3, Na3PO4


Summary

Strength of Acid or Base Keq or Ka, kb value.

The competition for amphoteric substance to give

OH- or H3O+

Solubility of salt is directly influence by LeChatelier

Principle and how reactant or product reacts with

the acid.

January 131 Acids - Base i. Salt Solution ii. Characteristics of Acid-Base

15.3 Acids - Bases i. Characteristics of Acid-Base

ii. Salt Solution

Acids-Base in solution and structural features of acids



Acid-Base Strength

Factors Affecting Acid Strength

Polarity: A molecule will transfer H+ only if the H-X bond is polarized

towards the X atom.

Bond Strength: Weak H-X bonds will lead to greater dissociation;

stronger acid.

Stability of conjugate base X-: Greater stability of X-, the stronger

the acid.

Solvation of ions: The extent in which the ions are stabilized by

water, leads to a stronger acid.

Acid Strength involves the breaking of H-A and the formation of H-OH2. Factors that

affect the ease with which bonds are broken and formed will influence the strength of

the acid.


I. Acid-Base Strength: Binary Hydrides

Bond Strength; Stronger the bond strength the weaker the acid character. 1. Size increase of X, leads to longer and weaker H-X bond, so H+ is release more

easily. (predominantly down PT) 2. EN of X: As X become more electronegative, the H-X becomes more polar, and H

+ is more easily release. (predominantly across PT)

H-F H-Cl H-Br H-I

569 kJ/mol 431 kJ/mol 368 kJ/mol 297 kJ/mol

Bond Strength from atomic radius (orbital overlap)

Na-H CH4 NH3 HF 1.0 2.5 3.1 4.1

Electro-negativity


I. Acid-Base Strength: Binary Hydrides

Bond Strength; Stronger the bond strength the weaker the acid character. 1. Size increase of X, leads to longer and weaker H-X bond, so H+ is release more

easily. (predominantly down PT) 2. EN of X: As X become more electronegative, the H-X becomes more polar, and H+

is more easily release. (predominantly across PT)

H-F H-Cl H-Br H-I

569 kJ/mol 431 kJ/mol 368 kJ/mol 297 kJ/mol

Bond Strength from atomic radius (orbital overlap)

Increasing atomic size, Increasing acid strength

Na-H CH4 NH3 HF 1.0 2.5 3.1 4.1

Electro-negativity

Increasing EN, Increasing Acid Strength


Acid-Base Strength: Binary Hydrides

Trend of Acidity for Binary compounds Bond Strength: Atomic size (Down PT) Bond polarity: Electronegative (across PT)


II. Acid-Base Strength: Oxyacids Oxyacids HOXOn: HNO3 , H2SO4 , H3PO4

Compounds with hydrogen, oxygen and third element Hypo-halous acid, HOCl, HOBr, HOI

H-O-I H-O-Br H-O-Cl 2. Electro negativity

1. Size: (No effect since H is bonded to O H-O-I H-O-Br H-O-Cl

H-O-Cl H-O-ClO H-O-ClO2 H-O-ClO3

3. Number of oxygen atom


2. Acid-Base Strength: Oxyacids and EN (Mechanism)

Variation of the third element Mechanism: The greater the electro-negativity (EN) of the X atom, the more the O-atom has to compensate for the e- density loss by removing e- density from H-atom thereby weakening the H-O bond.

H

O

X

e- density

In general, acidity increases as electronegativity of the third element increases.


2. Acid-Base Strength: Oxyacids and EN (Example)

Variation on the identity of third atom Electronegativity

Which is the stronger acid ? H2SO4 because EN (S) > EN (Se).

(HO)mXOn

(HO)2SO2

(HO)2SeO2OO Se

HO

HO

OO S

HO

HO


3. Acid-Base Strength: Oxyacids and Variation of Oxygen Atoms

Variation on the number of oxygen Polarity

Which is a stronger acid in terms of polarity ?

(HO)mXOn

(HO)2SO2OO S

HO

HO

OS

HO

HO

(HO)2SO

OO S

HO

HO

OS

HO

HOe- density is attracted to oxygen to a greater extent in H2SO4 compared to H2SO3. This weakens the H-O bond to a greater extent.


3. Acid-Base Strength: Oxyacids

Variation on the number of oxygen Stability of conjugate (resonance energy)

Hypochlorous acid HOCl (OH)Cl n=0 pKa= 7.5

chlorous acid HClO2 (OH)ClO n=1 pKa= 2.0

chloric acid HClO3 (OH)ClO2 n=2 pKa= S.A.

perchloric acid HClO4 (OH)ClO3 n=3 pKa= -10

(HO)mXOn


Acid-Base Strength: Organic Acids

Organic Acids Carboxylic acids are the largest category of organic acids. These acids contain the carboxyl group (blue). Carboxylic group contains the acidic proton on the O-H group and additional O withdrawing electron density on the carbon

Examples:

Acetic Acid Formic Acid Benzoic acid

CO

O HR

CO

O HCH

HH C

O

O HH C

O

O HC


Acid-Base Strength: Organic Acids Factors affecting acid behavior: 1. Additional oxygen on C result in polarizing O-H bond. This also stabilize the conjugate base. Acetic acid vs methanol

2. Strength of these acids result from the stability of their anions through resonance.

In addition, increasing the number of electronegative atoms in the acid increases acid character, trifluoric acetic acid Ka = 5.0 •10-1

Super acids have negative pH.

CO

O HCH

HH C

H

HH O H

CO

OCH

HH C

O

OCH

HH

CO

O HCF

FF


Summary of Acid Strength

Acid Type Trend

Binary 1. The more polar the H-X bond, the stronger the acid. This effect is dominant for acids of the same period.

2. The weaker the H-X bond, the stronger the acid.

This effect is dominant for acids of the same group.

Oxyacid 1. The greater the number of O atoms attracted to the central atom (the greater the oxidation number of the central atom), the stronger the acid.

2. For the same number of O atoms attached to the central atom, then the greater the electronegativity of the central atom, the stronger the acid.

Carboxylic 1. The greater the electronegativity of the groups attached (Organic) to the carboxyl group, the stronger the acid.


Acid-Base and Salt solutions Conjugates:

Strong Acid - Base CA g C+ + A - (100%)

Weak Acids - Base CA D C+ + A - < (100%) A- is a base (proton acceptor) A- + H2O D AH + OH- C+ is an acid (proton donor) C+ + H2O D *C + H3O+

C = NH4+ or polyatomic with H in formula i.e., H2PO3

-2, HSO4-, HCO3

-

(i) Salts from Strong base and strong acid:

Consider NaOH and HCl g the salt NaCl no pH = 7 (ii) Salts from strong base and weak acid:

Consider NaOH and HClO g the salt NaClO pH > 7

(iii) Salts from weak base and a strong acid:

Consider NH4Cland HNO3 g the salt NH4NO3 pH < 7

(iv) Salts from a weak base and a weak acid:

Consider NH4Cland HClO g the salt NH4ClO pH = ?


• Neutral: No affect of pH Salts form from strong acids and Bases

Na from NaOH and Cl from HCl form NaCl Ca from Ca(OH)2 and NO3 from HNO3 for Ca(NO3)2

______________________________________ • Basic: pH above 7 Salts are from strong base and weak acid

Na from NaOH and ClO- form HClO form NaClO Ba from from Ba(OH)2 and C2H3O from HC2H3O2 form Ba(C2H3O2)2

______________________________________ • Acidic: pH below 7 Salts from a weak base and strong acid

NH4+ from NH3 and Cl- from HCl form NH4Cl

Al+3 from AlCl3 and NO3- from HNO3 form Al(NO3)3

______________________________________

• Acidic or Basic: Hydrolysis Calculation Salts from weak base and weak acids

NH4C2H3O2 from NH4+ and HC2H3O2

NH4CN from NH4+ and HCN

FeCO3 from Fe+2 and H2CO3

Acid-Base Properties of Salt Solution


Salts in Water, pH Properties


Hydrolysis: Salts from weak base and weak acids

What is the pH of a salt solution when 37 g of NH4F is dissolved in 0.50L of water.

H3O+ Calculation :

NH4+ ! NH3 + H3O

+

i 2.0M 0 1⋅10-7

C -x +x +xe 2.0M- x +x 1⋅10-7 +x

NH4F → NH4+ + F−

NH4+ ! NH3 + H3O+ Ka = 5.56 ⋅ 10-10

F− ! HF + OH- Kb = 1.47 ⋅ 10-11

OH− Calculation : F− + H2O ! HF + OH−

i 2.0M Excess 0 1⋅10-7

C -x - x +x +xe 2.0M- x Excess x 1⋅10-7 +x


Hydrolysis of Amphoteric substance What is the [H3O+],[OH-],[NH3],[HCO3

-] and pH of a salt solution when 3.95 g of NH4HCO3 is dissolved in 1.00L of water.

H3O+ from NH4+ Calculation:

NH4+ ! NH3 + H3O+

i .05M 0 1 ⋅10-7

C -x +x +x

e .05M-x +x 1 ⋅10-7 +x

NH4HCO3 ! NH4+ + HCO3

−

NH4+ ! NH3 + H3O+ Ka = 5.56 ⋅10-10

HCO3− + H2O ! H3O+ + CO3

− Ka = 4.7 ⋅10-11

HCO3− + H2O ! OH- + H2CO3 Kb = 2.2 ⋅10-8

OH− from HCO3- Calculation:

HCO3− + H2O ! OH− + H2CO3

i .05M Excess 0 1⋅10-7

C -x -x +x +x

e .05M-x Excess x 1 ⋅10-7 +x

H3O+ from HCO3

- Calculation : HCO3

- + H2O ! H3O+ + CO3

-

i .05M Excess 0 1⋅10-7

C -x - x +x +x e .05M- x Excess x 1⋅10-7 +x

Solve for H3O+ based on dissociation of NH4

+.

Solve for H3O+ based dissociation of HCO3

-

Solve for OH- based association of HCO3

-

Strategy: Calculate the H3O+ from NH4

+, H3O+ from HCO3

- and the OH- from HCO3

-. These must then be added to determine which is excess in order to determine the pH of the solution.

Answer


Generalizing Bronsted-Lowry Concept: Leveling Effect When dealing with Bronsted-Lowry Acid-Base systems, all reactions taking place in water with acid systems yield H3O+, where as base systems yield OH-.

The question that comes to mind is-

“Why are strong acids and strong bases equally strong in water?”

Consider putting HCl or HNO3 in water, as soon as it goes into water, H3O+ forms. In fact all strong acid dissociate completely and react with water to form H3O+, The strong acid, i.e., HCl and HNO3 no longer exist. Any strong acid considered will simply donate it’s proton to H2O and form H3O+. The same is true for strong base, any strong base will accept a proton from H2O to form OH-. In water, the strongest acid is H3O+ and in base the strongest base is OH-. This phenomena is called the Leveling effect.

Water equalizes (level) the strength of all strong acids by reacting with the acid to form products of water’s autoionization.

To rank strong acids in terms of relative strength require using a different solvent system. i.e., dissolve HCl, HNO3 in acetic acid.


Lewis Acids-Bases Acid: Electron pair acceptor / Base: Electron pair donor Lewis Base, specie with lone pair electron Lewis Acid - Specie with empty valence orbital

i) Reaction between ammonia and boron trihydride

H

N

H

H

H

B

HH

Lewis Base Lewis Acid H

N

H

HH

BHH

Coordinate covalent bond

ii) Metal ion and ligand; Cu(NH3)42+ : a) Cu+2 ion is electron acceptor: acid / NH3 ion is electron donor: base

Cu+2(aq)

HN

HH

H

N

H

H

H

N

H H

Cu

H

N

HH

HN

H

HH

N

H

H

2+

b) H2O + CO2 D H2CO3: CO2 is the Lewis Acid, H2O is the Lewis base

HN

HH

HN

HH


Lewis Acid/Base: Metal Hydrolysis

iii) Metal ion hydrolysis:

High positive small cations attracts polar water. The ion polarizes the O-H bond

which weakens the O-H bond in water. (Stronger Lewis acid)

a)  Fe(H2O)63+ (aq) D Fe(H2O)5(OH)2+

(aq) + H+ (aq)

b)  Be(H2O)42+ (aq) D Be(H2O)3(OH)2+

(aq) + H+ (aq)

c) Al(OH)3 (s) + OH - D Al(OH)4 - (aq)


Summary

Acid Type Trend

Binary 1. The more polar the H-X bond, the stronger the acid. This effect is dominant for acids of the same period.

2. The weaker the H-X bond, the stronger the acid.

This effect is dominant for acids of the same group.

Oxyacid 1. The greater the number of O atoms attracted to the central atom (the greater the oxidation number of the central atom), the stronger the acid.

2. For the same number of O atoms attached to the central atom, then the greater the electronegativity of the central atom, the stronger the acid.

Carboxylic 1. The greater the electronegativity of the groups attached to the carboxyl group, the stronger the acid.

January 13 1 Common Ion Effect

16.1 Common Ion Effect Buffer Solutions

The resistance of pH change



Common Ion Effect Ionization of an electrolyte, i.e., salt, acid or base is decreased when a common ion is added to that solution.

i) What is the % ionization for 0.100 M acetic acid ? (Pure) HC2H3O2 + H2O D C2H2O2

- + H3O+ Ka =1.8•10-5 M Solving the iCe problem: ka = =1.8•10-5 M = [C2H3O2

-] [H3O+] /0.10 M g [H3O+]=1.34•10-3 M % α = (1.34•10-3 / 0.10 ) * 100 = 1.34 % pH = 2.87

ii) What is % α if 0.100 M HC2H3O2 is mix w/ 0.100M NaC2H3O2 ? (Buffer) HC2H3O2 + H2O D C2H2O2

- + H3O+ Ka =1.8•10-5 M i 0.100 Lots 0.100 1•10-7

C -x -x +x +x [c] 0.100-x Lots 0.100+x 1•10-7+x ka = 1.8•10-5 M = [0.100+ x ] [x] /( 0.100 -x) ∼ [0.100] [x] /( 0.100 ) x = [H3O+]= 1.8•10 -5 M pH = 4.74 % α = (1.8•10-5 / 0.10 ) * 100 = 0.0180 % Ionization % decrease in presence of common ion !!


Common Ion Effect Equation Consider the previous problem in which a common ion is in the same solution. HC2H3O2 + H2O D C2H3O2

- + H3O+ Ka =1.8•10-5 M i 0.100 Lots 0.100 1•10-7

C -x -x +x +x [c] 0.100-x Lots 0.100+x 1•10-7+x

or [c] [HC2H3O2 ] Lots [C2H3O2- ] [H3O+ ]

ka = [C2H3O2- ] [H3O+] rearrange the equation [H3O+] = ka • [HC2H3O2

]

[HC2H3O2] [C2H3O- ] Taking the - log of both side -

- log [H3O+] = - log (ka • [HC2H3O2 ] / [C2H3O2

-] ) or pH = -log ka - log( [HC2H3O2] / [C2H3O2

- ] ) let Ca = [HC2H3O2] and Cb = [C2H3O2 - ] ) therefore pH = pKa - log Ca / Cb

or pH = pKa + log Cb / Ca

This is the Henderson Hasselbach Equation:

pH = pKa + log Cb / Ca or pOH = pKb + log Ca / Cb


Henderson-Hasselbach Equation

pH of a solution can be calculated using a useful equation: pH = pKa + log [A-] / [HA] Where HA & A- are the weak acid and its conjugate and Ka is for HA

Similarly,

pOH = pKb + log [HA] / [A-] Where HA & A- are the weak base and its conjugate and Kb is for A-


Henderson-Hasselbach Equation: Example

Consider the common ion effect problem and lets see how the Henderson-Hasselbach equation can be used to simplify this problem.

What is pH if 0.100 M HC2H3O2 is mix w/ 0.100M NaC2H3O2 ?

HC2H3O2 + H2O D C2H2O2- + H3O+ Ka =1.8•10-5 M

i 0.100 Lots 0.100 1•10-7

C -x -x +x +x [c] 0.100-x Lots 0.100+x 1•10-7+x

Using the Henderson-Hasselbach equation:

pH = - log (4.3•10-7 ) + log (0.100 / 0.100) pH = 4.74 + log 1 pH = 4.74 + 0 pH = 4.74

Note: When a common ion is present in the same solution, the strategy to solve the problem requires a Buffer Type of calculation.


Henderson-Hasselbach Equation and Buffer Problems (sRF)

A buffer 0.100 M acetate and 0.200 M acetic acid is prepared (Ka = 1.8 •10-5). i) What is the pH of the buffer? ii) Calculate the initial pH, final pH, and change in pH that result when 1.00 mL of 0.100 M HCl is added to 100.0 mL of the buffer. iii) Calculate the initial pH, final pH, and change in pH that result when 1.00 mL of 0.100 M HCl is added to 100.0 mL of water. Note: HCl = 0.100 M • 1.00mL = 0.1 mmol.

C2H3O2- =0.100 M •100mL = 10 mmol and HC2H3O2

=0.200 M •100mL = 20 mmol

i) pH = pKa + log Cb/Ca = -log(1.8•10-5) + log ( 0.10 / 0.20) g pH = 4.44

ii) C2H3O2- + H3O+ D HC2H2O2 + H2O

s 10mmol 0.1 mmol 20 mmol Lots R -0.1 -0.1 +0.1 - f 9.9 0 20.1 Lots [c] 9.9/101 0 20.1/101 VT = 101 mL

pH = -log (1.8•10-5)+log [(9.9/101) / [20.1/101)] = 4.74 - 0.31 → pH = 4.43

pH (initial) = 4.44, pH (final) 4.43, ΔpH (change) = -0.01


...Continue: Henderson-Hasselbach Equation and Buffer Problems

...continue A buffer 0.100 M acetate and 0.200 M acetic acid is prepared (Ka = 1.8 •10-5). Reger 14.19

iii) Calculate the initial pH, final pH, and change in pH that result when 1.00 mL of 0.100 M HCl is added to 100.0 mL of water. Note: HCl = 0.100 M • 1.00mL = 0.100 mmol.

iii) HCl + H2O D H3O+ + Cl- s 0.100 mmol 1•10-7M - R -0.100mmol +0.100mmol - f 0 0.100 mmol [c] 0 0.100mmol / 101 mL

[H3O+] = 9.9•10-4 M g pH = 3.00

pH (initial) = 7.00 , pH(final) 3.00, ΔpH(change) = -4.00


Essential Feature of Buffer Systems A buffer solution exhibits very small change in pH changes when H3O+ and OH- is added. A buffer solution consists of relatively high concentration of the components of a conjugate weak acid-base pair. The buffer-components concentration ratio determines the pH, and the ratio and pH are related by the Henderson-Hasselbalch equation. A buffer has an effective range of pKa + 1 pH unit.


Blood Buffer System

Buffer - A solution whose pH is resistant to change

Your body uses buffers to maintain the pH of your blood

Blood pH 7.35 - 7.45 Buffer system in body - 1. Proteins 2. Phosphates HPO4

2- / H2PO4- : 1.6 / 1

3. Carbonates H2CO3 / HCO3

- : 10 / 1 Reaction: H3O+ + HCO3

- D H2CO3 + H2O H2CO3 g H2O + CO2 (exhale)


Acidosis Blood pH i 7.35 (ACIDOSIS)

Depression of the acute nervous symptom. Or respiratory center in the medulla of the brain is affected by an accident or by depressive drugs.

Symptoms: •Depression of the acute nervous system

•Fainting spells

•Coma •RIP

Causes: 1. Respiratory Acidosis Difficulty Breathing (Hypo-ventilation) Pneumonia, Asthma anything which diminish CO2 from leaving lungs.

2. Metabolic Acidosis Starvation or fasting Heavy exercise

Mechanism: 1. Respiratory Acidosis CO2 doesn’t leave lungs which result in the build up of H2CO3 in the blood

2. Metabolic Acidosis If body doesn’t have enough food then Fatty acids (Fat) are used. Fatty Acids g Acidic. Furthermore, exercise leads muscle to produce lactic acid.


Alkalosis Blood pH h 7.45 (ALKALOSIS)

Hyperventilation during extreme fevers or hysteria. Excessive ingestion of basic antacids and severe vomiting

Symptoms: •Over simulation of the nervous system

•Muscle cramps

•Convulsion •Death

Causes: 1. Respiratory Alkalosis Heavy rapid breathing (hyperventilation). Results from - fear, hysteria, fever, infection or reaction with drugs.

2. Metabolic Alkalosis Metabolic irregularities or by excess vomiting

Mechanism: 1. Respiratory Alkalosis Excessive loss of CO2 lowers H2CO3 and raise HCO3

- level (Can be remedied by breathing in a bag)

2. Metabolic Alkalosis Vomiting removes excess acidic material from stomach. (pH of stomach equals one).


Buffer System at Work Buffer - System that resists change in pH when H3O+ or OH- is added. Buffer solution may be prepared by a weak acid and its conjugate base. How it Works: A- g HA H3O+ Buffer H2O

Remember pH = Conc. of H3O+

Your blood Rxn: HCO3- D H2CO3

Acidosis Excess H3O+ + HCO3- g H2CO3 + H2O

H3O+ CO2 + H2O Alkalosis Excess OH- + H2CO3 D HCO3

- + H2O

OH-


Summary

Equation / Concept Function1 [H+] [OH-] = Kw Permits the calculation of [H+] or [OH-] when the other is known.2 p X = - log X This equation is the basis of the p-scale.3 pH + pOH = 14.00 This equation shows the relationship between the pH and the pOH4 HA ! H++ A-

Ka =[H+ ] [A −]

[HA]

This is the Mass Action Equation for the ionization of a weak acid inwater. This equation yields the ka given the equilibriumconcentration of all specie. The equation also yields the [H3O+]given the initial concentration of the weak base [HA] and the ka.

5 B + H2O ! HB + OH-

Kb =[HB] [OH−]

[B]

This is the Mass Action Equation for the ionization of a weak basein water. This equation yields the kb given the equilibriumconcentration of all specie. The equation also yields the [OH-]given the initial concentration of the weak base [B] and the kb.

6 Percent ionization (α)

α =amount ionizedinitial amount

×100%

The percent ionization can be calculated from the initialconcentration of the acid (or base) and the change in theconcentration of the ions. Given the percent ionization (α) and thepH, the ka (or kb) can be determined.

7 Ka • Kb = Kw This equation relates Ka and Kb for conjugate pairs in aqueoussolution,

8 ID of the solute as :i) only a weak acidii) only a weak baseiii) a mixture of a weak acidand its conjugate base

Identification of the function of the solute leads to the correct MassAction expression and thereby leading to the correct equilibriumlaw. This is a critical first step to solve any acid-base equilibria

9 Identification of acidiccations and basic anions

Identification of function of cation and anion of a salt lead to pH ofthe salt solution. Given the ka or kb of the conjugates of these ionsleads to the calculation of the pH or pOH

10 Assumption whichsimplifies Mass Action

In order to simplify the math calculation of a Mass Actionexpression, assumption can be made base on the ka or kb value.

11 Reactions when H+ or OH-are added to a buffersolution.

Understanding the buffer reaction permits the determination of theeffect of a strong acid or strong base on the pH of the solution.Adding H+ lowers the [A-] and raises [HA], adding OH- lowers[HA] and raises [A-].

The following summary lists the important tools needed to solve problems dealing with acid-base equilibria.

January 13 1 Strong-Acids / Strong Bases Titration

16.2 Strong-Acids / Strong-Bases Titration



Reaction Between Acid - Base What is the pH of a solution when an acid is mixed with a base? Stoichiometry Problem:

HA g H3O+ + A- MOH g OH- + M+

MOH + HA g H2O + MA

Stoichiometry Problem:

The amount of H3O+ or OH- remaining after a portion is neutralize determines the pH of the solution.

In an acid - base reaction, H+ & OH- always combine together to form water and an ionic compound (a salt): Neutralization Reaction.

HCl(aq) + NaOH(aq) g H2O(l) + NaCl(aq)

Analysis is a Stoichiometry problem only if a strong acid is combined with a strong base.


Titration Acid/Base Titration A technique of chemical analysis to determine the amount of a substance in a sample. i.e., What is the acidic content of Lake Elsinor?

A sample can be tested by titration. In a titration experiment, a known volume of a standard concentrated solution (the titrant) is used to analyze a sample (the analyte). One is usually an acid, the other a base. An indicator is added to the analyte to signal when the titration is complete. This is called the endpoint. When the moles of acid(H3O+) and moles of base (OH-) are equal in a titration experiment, the stoichiometric equivalent point is reached. This is called the equivalent point. Indicator changes color @ endpoint moles titrant = moles analyte.

@ equivalent point.


Acid-Base Indicator

Some common acid-base indicators. The color changes occur over a range of pH values. Notice that a few indicators have two color changes over two different pH ranges.

Mechanism for phenolphthalein indicator. At Low pH phenolphthalein is colorless and has a structure in which there is a five membered ring. In the presence of excess acid the five membered ring is broken and the resulting change in conformation gives rise to a compound which is pink.


S Acid- S Base Analysis Indicator changes color @ endpoint. Indicator is chosen so that endpoint occurs at equivalent pt. The following is true at the equivalence point.

moles H3O+ = moles OH-

Macid • Vacid = Mbase • Vbase For monoprotic A & B Macid • Vacid = Mbase • Vbase For polyprotic A & B

but Nacid•Vacid = Nbase•Vbase for monoprotic and polyprotic A & B

eq H3O+ • L H3O+ = eq OH- • L OH-

L Soln L Soln

For Acid base calculation at Equivalence pt: Sometimes: Macid • Vacid = Mbase •Vbase but always: Nacid•Vacid = Nbase•Vbase


S Acid- S Base Calculation

Titration of HCl with NaOH (Reger 14.1) A titration is used to detm’ conc. of HCl solution. Exactly 20.00mL of the acid solution was placed in a flask, with phenolphthalein added. 18.34mL of 0.0982 M NaOH was needed to reach the endpoint. What is the conc. of the HCl ?

20.00 mL HCl

18.34 mL NaOH @ end pt.

Rxn: NaOH g OH 100% HCl g H+ 100%

Net H+ + OH- g H2O

Conc H3O+:

= 0.0982 mol OH- • 18.34 mL • 1 mol H+ • 1

1 L 1mol OH- 20.00 ml

[HCl] = 0.09005 M


Titration: Thought Experiment

Consider the titration of a Strong acid with a Strong Base. What is the pH after incremental addition of some moles of base to the acid.

Amt Amt Amt Remaining pH

NaOH HCl H2O (mol) or [OH- ] [H2O+] pOH

0 100 H+ - 100H+ pH = - log [100 / vT]

20 100-20 20 H2O 80 H+ - log [80 / vT]

50 100-50 50 H2O 50 H+ - log [50 / vT]

99 100-99 99 H2O 1 H+ - log [1 / vT]

100 100-100 100 H2O 0 H+ - log [0 / vT] pH = ?*

101 100-101 100 H2O 1 OH- pOH = - log [1 / vT]

200 100-200 100 H2O 100 OH- pOH = - log [100 / vT]

* Neutral pH is determine by the Autoionization of water pH = 7.0


Titration S Acid - S Base: Example A 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH. What is the pH after addition of 0.00, 20.00, 49.00, 50.00, 51.00 and 60.00 mL of base.

Titration curve between strong acid and strong base. Analyte is HCl and titrant is NaOH.

Rxn: NaOH g OH 100% H Cl g H+ 100%

Net H+ + OH- g H2O

?

Click for simulation


i) 0.00 mL Base 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH 0.00mL of base.

pH based on the [H3O+] of HCl HCl is a strong acid, therefore [HCl] = [H3O+] [HCl] = [H3O+] = 0.100 M = 1•10-1 M g pH = 1.00

ii) 25.00 mL Base: Vtotal = 75 ml solution

0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH 25.00mL of base mol HCl = 0.100M • (50mL) =5.00 •10-3 mol HCl or 5.0 mmol = H3O+

mol NaOH = 0.100M • (25mL) = 2.5 •10-3 mol NaOH or 2.5 mmol = OH -

pH based on the excess HCl remaining. Since HCl is a strong acid.

HCl + NaOH g H2O + Na+ + Cl- s 5 mmol 2.5 mmol - - - R 2.5 2.5 2.5 2.5 2.5 f 2.5 0 2.5 2.5 2.5 [c] 2.5 mmol / 75.00 ml = 3.3•10-2 M

[HCl] = [H3O+] = 3.3•10-2 M g pH = 1.48

Titration Strong Acid - Strong Base: Example ( 0.0 & 25.0 ml)


iii) 49..00 mL Base: Vtotal = 99 ml solution 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH 49.00mL of base.

mol HCl = 0.100M • (50mL) = 5.0 mmol HCl = H3O+

mol NaOH = 0.100M • (49mL) = 4.9 mmol NaOH = OH -

HCl + NaOH g H2O + Na + Cl- s 5.0 mmol 4.9 mmol - - - R -4.9 -4.9 - - - f 0.1 mmol 0 - - - [c] 0.1 mmol / 99.0 ml = 1.01 •10-3 M

[HCl] = [H3O+] = 1.01 •10-3 M g pH = 3.0

Titration S Acid - S Base: Example ( 49.0 & 50.0 mL)

iv) 50.00 mL Base: Vtotal = 99 ml solution 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH 50.00mL of base.



HCl + NaOH g H2O + Na + Cl- s 5.0 mmol 5.0 mmol - - - R -5.0 -5.0 - - - f 0.1 mmol 0 - - -

[c] 0.1 mmol / 100.0 ml = 0 M Is the pH = zero ?

No, Autoionization Water has [H3O+] = 1.00 •10-7 M g pH = 7.0


v) 51.00 mL Base: Vtotal = 101 ml solution 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH 51.00mL of base.



HCl + NaOH g H2O + Na + Cl- s 5.0 mmol 5.1 mmol - - - R -5.0 -5.0 - - - f 0 mmol 0.1 - - - [c] 0.1 mmol / 101.0 ml = 9.90 •10-4 M

[NaOH] excess = [OH-] = 9.90 •10-4 M g pOH = 3.00 g pH = 11.0

Titration S Acid - S Base: Example ( 51.0 & 60.0 mL)

vi) 60.00 mL Base: Vtotal = 110 ml solution 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH 60.00 mL of base.

mol HCl = 0.100M • (50 mL) = 5.0 mmol HCl = H3O+

mol NaOH = 0.100M • (60 mL) = 6.0 mmol NaOH = OH -

HCl + NaOH g H2O + Na + Cl- s 5.0 mmol 6.0 mmol - - - R -5.0 -5.0 - - - f 0 mmol 1.0 - - - [c] 1.0 mmol / 110.0 ml = 9.90 •10-3 M

[NaOH] excess = [OH-] = 9.90 •10-3 M g pOH = 2.04 g pH = 11.96


Summary of the titration of 0.100 M HCl with 0.100 M NaOH. What is the pH after incremental addition of some moles of base to the acid.

Vol Amt Amt Net Conc pH NaOH NaOH HCl-NaOH H3O+ H+ or OH- or (mL) (mmol) (mmol) (mmol) (M) pOH

0 0 5mmol - 0 5 5 / 50=0.1 pH = 1.0

25 2.5 5 - 2.5 2.5 2.5/75=3.3•10-2 pH = 1.48

49 4.9 5 - 4.9 0.1 0.1/99=1.01 10-3 pH = 3.00

50 5.0 5 - 5 0 0/100 = 0 pH = 7

51 5.1 5.1 [OH] - 5 0 H+ 0.1/101=9.9e-4 pOH=3, pH=11

60 6.0 6 [OH] - 5 1 OH- 1/110= 9.9e-3 pOH=2, pH=12

Titration: Result Summary


Summary of the titration of 0.100 M HCl with 0.100 M NaOH.

Titration: Result Summary


Result: Titration Curve A 0.100M HCl (50.00mL) is to be titrated with 0.100 M NaOH. What is the pH after addition of 0.00, 20.00, 49.00, 50.00, 51.00 and 60.00 mL of base.

Titration curve between strong acid and strong base. Analyte is HCl and titrant is NaOH.

Rxn: NaOH g OH 100% H Cl g H+ 100%

Net H+ + OH- g H2O


Titration Curve In or Out of Class Exercise

Design a problem in which NaOH (analyte) of some concentration and 50-mL volume is titrated with HCl of some concentration. Design the problem in such a way that the the following conditions are met.

HCl Added pH soln

1 0.00 mL 13.500

2 10.00 mL 13.199

3 15.0 mL 12.988

4 20.0 ml 12.655

5 24.6 ml 11.530

6 25.4 ml 2.475

7 30.0 ml 1.403

8 35.0 ml 1.290

9 40.0 ml 0.977

10 50.0 ml 0.801


... remember That a titration problem is nothing more than a Stoichiometry problem

( )Volume L

( )Temperature K

Pressure atm( )

R (.0821 atm • L

mol • K)

Density (g / cc)

Molar Mass (g / mol)

MolesABalance equation

← → # # # # # # # # # Stoic. coefficient .

MolesB

Mass (g)

L# of molecules / atoms

Conc. (mol / L )

Vol (L)

Volume (L )

Temperature (K)

Pressure (atm)

N Av (6.02 •1023)

R (.0821 atm • L

mol • K)

Vol (L)

)Conc. (mol / L

Density (g / cc)

Mass (g)Molar Mass ( g / mol)

Gas phase

Aqueous phase

Solid phase

Liquidphase particle

(atomic) phase

Vol ( )

Gas phase

Aqueous phase

Solid phase

Vol (L)(l)Liquid

phase(atomic) phase

N Av (6.02 •1023) particle

# of molecules / atoms

January 13 1 Weak Acids Weak Bases Titration

16.3 Weak Acids Weak Bases Titration

Titration of Weak Acid with Strong Base Titration of Base Acid with Strong Acid



Weak Acid (or Weak Base) with Strong Base (or strong Acid)

Experimental technique and the concept is similar to that of the titration of a strong acid with a strong base (or vice versa) with equilibrium concept applied.

pH calculation involves 4 different type of calculations.

i) The analyte alone

(equilibrium calculation)

ii) Buffer region

(Henderson-Hasselbach eqn)

iii) Equivalence point

(Hydrolysis)

iv) Excess titrant

(Stoichiometric calculation) Equilb Buffer Hydrolysis Stoic Excess

Click for simulation


Titration (WA-SB): Weak Acid (or Weak Base) with Strong Base (or strong Acid)

Consider the titration problem:

Titration curve for a weak acid (HOCl) and an strong base (KOH). Generate a titration curve upon addition of KOH @ 0%-, 50%-, 95%-, 100%- and 105% of

equivalent point. Analyte : 10.00 ml 0.400M HOCl: Titrant: 0.400 M KOH


Titration (WA-SB): Weak Acid (or Weak Base) with Strong Base (or strong Acid)

Consider the titration problem:

Titration curve for a weak acid (HOCl) and an strong base (KOH). Generate a titration curve upon addition of KOH @ 0%-, 50%-, 95%-, 100%- and 105% of

equivalent point. Analyte : 10.00 ml 0.400M HOCl: Titrant: 0.400 M KOH

HOCl + KOH g H2O + OCl- + K+

HOCl = 0.400M • 10.00ml = 4.0 mmol HOCl

Volume KOH corresponding to 0%-, 50%-, 95%-, 100%- and 105%

0.400 M KOH = 0ml, 5.0ml, 9.5ml, 10.0ml, 10.5 ml

Misc. Info.: HOCl: Ka = 3•10-8 pKa = 7.5


Type i: Weak Acid with Strong Base 0 % Type 1: Calculation EQUILBRIUM 0% KOH added (TYPE 1 Weak acid calc.) 0%, VT = 10.0 mlWeak acid pKa - (Type 1 Calculation)

pH of solution is determined by the dissociation of the weak acid.

HOCl + H2O !Ka

H3O+ + OCl-

i 0.4M Ex 0 0C -x -x +x +xe 0.4M-x Ex +x +x

Ka = 3 ⋅10−8 =x2

.4M −x , 0.4M−x ≈0.4M

x = .4 3•10−8( )H30+[ ]= x = 1.10• 10-4 M

pH = 3.96


Titration (4ii): Weak Acid (or Weak Base) with Strong Base (or strong Acid)

50 % Type 2: Calculation BUFFER, Henderson- Hasselbalch Eqn 50% KOH added (TYPE 2 Buffer) 50%, VT = 15.0 ml

HOCl + KOH ! H2O + OCl- + K+ s 4mmol 2mmol Ex 0 - R -2 -2 +2 +2 - f 2mmol 0 Ex+2 2mmol - c .133M - Ex .133M

Notice that the concentration of the acid and its conjugate are equal. In the mass action expression these two terms cancel. pH = pKa. Buffer situation-(Type-2) A. Long Approach Note that the excess 2mmol of HOCl will dissociate in water to

HOCl + H2O " Ka

H3O+ + OCl-

i .133M Ex 0 0.133M C -x -x +x +x e .133-x Ex x .133+x

The calculation is a simple equilibrium analysis-

Ka = 3⋅10−8 =x 0.133+x( )

0.133-x( ) , 0.133M ±x ≈ 0.133M

x = 3⋅10−8 ,Assumption checks!

H30+[ ]= 3⋅10−8M

pH = 7.5

B Simple ApproachBuffer solution using Henderson Hasselbalchequation and the sRfc table above

pH = pKa + logCbCa

pH = 7.5 + log.133M.133M

pH = 7.5


Titration (4ii): Weak Acid (or Weak Base) with Strong Base (or strong Acid)

95% Type II Calculation BUFFER, Henderson- Hasselbalch Eqn 95% KOH added (TYPE 2 Buffer): 95%, VT = 19.5 ml

HOCl + KOH ! H2O + OCl- + K+ s 4 mmol 3.8 mmol Ex 0 - R -3.8 -3.8 +3.8 +3.8 - f 0.2 mmol 0 Ex 3.8 mmol - c 1.03•10-2M - Ex 0.195M

Noticed that the acid and conjugate are present in the same solution. This is a common ion effect or buffer type problem. Use the HH equation. Buffer situation-(Type-2) A. Long Approach Note that the excess 0.2mmol (or 1.03•10-2M) of HOCl will dissociate

in water HOCl + H2O "

Ka H3O+ + OCl-

i 1.03•10-2M Ex 0 0.195M

C -x -x +x +x e 1.03•10-2M - x Ex x 0.195M+x

Ka = 3⋅10−8 =x .195 +x( )

1.03⋅10-2 -x( )

0.195M +x ≈ 0.195M1.03⋅10-2M − x ≈ 1.03⋅10-2M

1.03⋅10-2 3⋅10−8% & ' (

) *

0.195= x

1.58 ⋅10-9 = x ,Assumption checks!

1.58 ⋅10-9M = H30++ , -

. / 0

8.80 = pH

B Simple ApproachBuffer solution using HendersonHasselbach equation and sRfc table

pH = pKa + logCbCa

pH = 7.5 + log 0.195M1.03⋅10-2 M

pH = 7.5+ 1.28pH = 8.78


Titration (4iii): Weak Acid (or Weak Base) with Strong Base (or strong Acid)

100% Type III Calculation HYDROLYSIS, conjugate back to original

100% KOH added (TYPE 3 Hydrolysis): 100%, VT = 20.0 ml

HOCl + KOH ! H2O + OCl- + K+ s 4mmol 4mmol Ex 0 - R -4 -4 +4 +4 - f 0 0 Ex 4mmol - c - - Ex 0.2M -

How is the pH or pOH calculated since there are no H30+ or OH-?

Actually the excess HOCl does react with water to form H3O+. Equivalence point calculation, Hydrolysis (Type-3) Note that all the HOCl acid is neutralize by the base. HOCl

cannot dissociate in water since there is no excess. But the conjugate base OCl- can react with water in a hydrolysis reaction according to-

OCl- + H2O " Kb

OH- + HOCl

i 0.2M Ex 0 0 C -x -x +x +x e 0.2-x Ex x x

Kb =KwKa

= 1 ⋅10−14

3⋅10−8 =x2

0.2-x( ) 0.2 -x ≈ 0.2M

x2

0.2= 3.33⋅10−7 x = 2.58 ⋅10−4M

OH−% & '

( ) * = 2.58 ⋅10−4M, pOH = 3.59

pH = 10.41


Titration (4iv): Weak Acid (or Weak Base) with Strong Base (or strong Acid)

105% Type IV Calculation Stoichiometry 105% KOH added (TYPE 4 Strong Base)

HOCl + KOH ! H2O + OCl- + K+ s 4mmol 4.2mmol Ex 0 - R -4 -4 +4 +4 - f 0 0.2mmol Ex 4mmol - c - 9.76•10-3M Ex .195M -

Strong Base calculation (Type-4) Since the excess is KOH, a strong acid, then the pH (or pOH in

this case) is determine by the following dissociation of KOH:

KOH ! OH- + K

i 9.76•10-3M 0 0 C -9.76•10-3M +9.76•10-3M +9.76•10-3M e - +9.76•10-3M +9.76•10-3M

[OH-] 1 = 9.76 ⋅ 10−3MpOH = 2.01pH = 11.99

Note that the 0.195 M of OCl- will contribute negligible

amounts of OH- as it back reacts with water in a hydrolysis type reaction.

A simple check shows that [OH-]2 is negligible. OCl- + H2O !

Kb OH- + HOCl

i 0.195 M Ex 0 0

C -x -x +x +x e 0.195 M-x Ex x x

Kb =KwKa

= 1 ⋅ 10−14

3 ⋅ 10−8=

x2

0.195- x( ) , 0.195 - x ≈ 0.195 M

x2

0.195= 3.33 ⋅ 10−7 x = [OH-]2 = 2.54 ⋅ 10−4M

Therefore you see that -

[OH−]Total

= [OH-]1 + [OH-]2 = 2.54 ⋅10−4M + 9.76 ⋅ 10−3

[OH−]Total

= 1.00 ⋅10−2 ≈ 9.76 ⋅ 10−3M

pOH = 1.99 ≈ 2.000 pH = 12.000


Titration Curve: Strong Acid / Strong Base

Titration of 0.500M HCl with 0.500M NaOH

A strong acid-strong base titration curve, showing how the pH increases as 0.5000M NaOH is added to 20.00mL of 0.5000M HCl. The equivalence point pH is 7.00. The steep portion of the curve includes the transition intervals of bromothymol blue, phenolphthalein and bromophenol blue .


Titration Curve Features: WA - SB

Titration of 40.00 mL 0.1000M HCH3CH2OOO with 0.1000 M NaOH Weak acid - Strong base Titration curve for a weak acid by a strong base: 40.00mL of 0.1000M CH3CH2OOOH by 0.1000M NaOH. When exactly one-half the acid is neutralized, [CH3CH2COOH] = [CH3CH2COO-] and pH = pKa = 8.80 The equivalent point is above 7.00 because the solution contains the weak base CH3CH2COO-. Phenolphthalein is a suitable indicator for this titration but Methyl red is not because its color changes over a large volume range.


Titration Curve Features: Monoprotic Acid

Titration curve for a series of acids (A - F) being titrated with a strong base

F

A

B

C

D

E

Acid F is the strongest acid, Acid E is the next strongest acid followed by acid D, acid C, Acid B and acid A. Acid A is the weakest among the weak acid. The Ka’s of each acid is determined by reading the pH half way to the equivalent volume for each acid.


Titration Curve Features: WB -SA

Titration of 0.1000 M NH3 with 0.1000 M HCl Weak base - Strong acid A weak base-strong acid titration curve, showing how the pH decreases as 0.1000M HCl is added to 40.00mL of 0.1000M NH3. When exactly one-half the base is neutralized, [NH3] = [NH4

+] and pOH = pKb of NH3 (4.76) or the pH = 14 - pKb of NH3 (9.24). Note that this pH value (9.24) is actually the pka of NH4

+, the conjugate of NH3.

Note that the equivalent point is below 7.00 because the solution contains the weak acid NH4

+ . Methyl red is a suitable indicator for this titration but phenolphthalein is not because its color changes over a large volume range.


Titration Curve: Polyprotic

Titration Curve of 0.100M H2SO3 with 0.100 M NaOH

Curve for the titration of a weak polyprotic acid. Titrating 40.00mL of 0.1000M H2SO3 with 0.1000M NaOH leads to a curve with two buffer regions and two equivalence points. Because the Ka values are separated by several orders of magnitude, in effect the titration curve looks like two weak acid-strong base curves attached. The pH of the first equivalence point is below 7 because the solution contains HSO3

-, which is a stronger acid than it is a base. Ka of HSO3

- = 6.5•10-8; Kb of HSO3

- = 7.1•10-13


Titration Curve: Poly-Basic

Titration Curve of 0.10 M Na2CO3 with 0.10 M HCl. 1. CO3

2- + H+ g HCO3- + H2O

2. HCO3

- + H+ g H2CO3 + H2O


Various Titration Curve

Example of Various: Titration Curves

Strong Weak Poly Analytes

Acids

Base


Summary There are two main type of titration problems.

The strategy to solve them are:

1) SA-SB: Strong acid being titrated with a strong base (or vice versa). This is a stoichiometry type of problem. The pH at the equivalent point = 7.0.

2a) Weak acid being titrated with a strong base. In this type of problem, the there are four sub-problems that must be solved.

a) At 0% titrant, the problem is an equilibrium type. The Ka of the weak acid will determine the extent of ionization and therefore the pH.

b) At 1-99% titrant, the problem is a buffer type. Use the Henderson-Hasselbach Equation to solve for the pH

c) At equivalence point, this is now a hydrolysis problem. Recall that the conjugate base of the weak acid now reacts with water to produce OH-. The solution will be basic at the equivalent point.

d) Pass the equivalence point, this is now a stoichiometry problem. The excess titrant base will dictate the pH of the solution.

2b) Weak base being titrated with a strong acid. This is the same type of problem as (2a) above except in this case a weak base is being neutralized by the strong acid titrant.

15.1 Acids - Bases in Water

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Transcript of 15.1 Acids - Bases in Water