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Chapter 9 Aqueous Solutions and Chemical Equilibria. 1) Non electrolytes : Are substances that dissolve in water but do not produce any ions and do not conduct an electric current.

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chapter 9 aqueous solutions and chemical equilibria
Chapter 9Aqueous Solutions and Chemical Equilibria

1) Non electrolytes:Are substances that dissolve in water but do not produce any ions and do not conduct an electric current.

2) Electrolytes: Form ions when dissolved in water or other solvents and produce solutions that conduct electricity.

1) Strong Electrolytes: Ionize essentially completely. Strong conductor of electricity.

2) Weak Electrolytes: Ionize only partially. Poorer conductor than strong electrolyte.

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slide4
Acids and Bases: An acid is a proton donor and a base is a proton acceptor (Bronsted-Lowry concept).

Conjugate Acids and Bases: A conjugate base is the species formed when an acid loses a proton.

NH3 + H2O NH4+ + OH-

base1 acid2conjugate acid1conjugate base2

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Amphiprotic Solvents: A solvent that can act either as an acid or as a base depending on the solute. Water is the classic example.

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Three general types of solvents :

1) Protic solvent : amphiprotic solvents

- possess both acidic and basic properties

- undergoes self-ionization (=autoprotolysis)

ex. Water, alcohols, acetic acid, ammonia

ethylenediamine

2) Aprotic solvents

- have no appreciable acidic or basic character

- do not undergo autoprotolysis

- ex. Benzene, carbon tetrachloride, pentane

3) Basic solvents

- have basic properties but essentially no acidic

tendencies

- do not undergo autoprotolysis

ex. ethers, esters, pyridine and amines

2H2O  H 3O+ + OH–

2C2H5OH  C2H5OH2+ + C2H5O–

2HOAc  H2OAc+ + OAc–

2NH3 NH4+ + NH2–

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Strong acid: Reacts with water so completely that no undissociated solute molecules remain.

Weak acid: Reacts incompletely with water to give solution that contain significant amounts of both the parent acid and its conjugate base.

Strong base: Completely dissociated in water solution

Weak base: Incomplete dissociation in water solution

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relative strength
Relative Strength

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Strengths of Acids and Bases

In a leveling solvent, several acids are completely dissociated and show the same strength.

In a differentiating solvent, various acids dissociate to different degrees and have different strengths.

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The tendency of a solvent to accept or donate protons determine the strength of a solute acid or base dissolved in it.

H2O + HClO4→ H3O+ + ClO4-

H2O + HCl → H3O+ + Cl-

CH3COOH + HClO4  CH3COOH2+ + ClO4-

CH3COOH + HCl CH3COOH2+ + Cl-

base1 acid2 acid1 base2

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Chemical Equilibrium & Equilibrium Constants

wW + xX yY + zZ

Where, the capital letters represent the formulas of participating chemical species and the lowercase letters are the small whole numbers (# of moles) required to balance the equation. The equilibrium-constant expression is

Where, the bracketed terms are molar concentration if the species is a dissolved solute or partial pressure (atm) if the species is a gas. If one of the species is a pure liquid, a pure solid, or the solvent in excess (dilute soln.), no term for this species appear in the equilibrium-constant expression. Equilibrium constant K is a temperature dependent quantity.

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The Ion-Product Constant for Water

Aqueous solutions contain small concentrations of hydronium and hydroxide ions as a consequence of the dissociation reaction.

2H2O H3O+ + OH-

Equilibrium constants,

The concentration of water in dilute aqueous solution is enormous when compared with the concentration of hydrogen and hydroxide ions. [H2O] can be taken as constant,

K[H2O]2 = Kw = [H3O+][OH-] where the new constant is the ion-product constant for water. At 25oC, Kw 1.00 x 10-14

-logKw = - log[H3O]+ - log[OH-]

pKw = pH + pOH = 14.00

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Temperature dependence of pH

H+ + OH-  H2O +Q

Body temperature = 37 oC

Blood pH = 7.35 ~ 7.45

[HCl] in gastric juice = 0.1 ~ 0.02M

if [H+] = 0.02M pH = 1.7

pOH = 13.6 –1.7 = 11.9 at 37 oC

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Solubility-Products

Ba(IO3)2(s) Ba2+(aq) + 2IO3-(aq)

K[Ba(IO3)2(s)] = Ksp = [Ba2+][IO3-]2

Where, the new constant, Ksp, is called the solubility-product. The position of this equilibrium is independent of the amount of Ba(IO3)2 as long as some solid is present.

Common Ion Affect

The common-ion effect is responsible for the reduction in solubility of an ionic precipitate when a soluble compound combining one of the ions of the precipitate is added to the solution in equilibrium with the precipitate.

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Dissociation Constants for Acids and Bases

When a weak acid or a weak base is dissolved in water, partial dissociation occurs,

HNO2 + H2O H3O+ + NO2- ,

NH3 + H2O NH4+ + OH- ,

Where Ka and Kb are acid dissociation constant for nitrous acid and base dissociation constant for ammonia respectively.

Dissociation Constants for Conjugate Acid/Base Pairs:

NH3 + H2O NH4+ + OH- ,

NH4+ + H2O NH3 + H3O+ ,

Ka. Kb = [H3O+ ][OH- ] = Kw

Kw = Ka. Kb

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Calculation of pH : strong acids and bases

1. Strong acid or base Cacid or base >> 10–6 M

2. Both 10–6 < Cacid or base < 10–8 M

3. Water Cacid or base << 10–8 M

Calculated pH as a function of the concentration of a strong acid or strong base dissolved in water.

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A buffered solution is one that resists change of pH

on adding acids or bases, or on diluting it with solvent.

The buffer consists of mixture of an acid and its conjugated base.

HA  A– + H+

Ka = [A–][H+] / [HA]

[H+] = Ka [HA] / [A–]

– log [H+] = – log Ka [HA] / [A–] = – log Ka – log [HA] / [A–] = – log Ka + log [A–]/[HA]

pH = pKa+ log [A–] / [HA]

pKa is constant [A–]/[HA]  pH

100:1 or 1:100 pKa 2

10:1 or 1:10 pKa 1

1:1 pKa 0

Buffer solutions

Henderson-Hasselbalch

equation

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preparation
preparation
  • 1) HA + A- B + HB+
  • 2) HA + OH-B + H+
  • 3) A- + H+HB+ + OH-

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Ex. Tris buffer

NH2

N+H3

C

C

HOH2C

HOH2C

+ H+

CH2OH

CH2OH

HOH2C

HOH2C

Tris hydrochloride(FW=157.597) : BH+Cl– BH+ + Cl–

BH+ =B + H+

Find the pH of a solution prepared by dissolving 12.43g of tris plus 4.67g of tris hydrochloride in 1.00 L of water. pKa = 8.075

Tris = B : 121.136 g/L = 1.00M Tris hydrochloride = BH+ : 157.597 g/L = 1.00 M

12.43 g/L = x M 4.67 g/L = y M

x = 0.1026 M y = 0.0296 M

pH = pKa+ log [B] / [BH+] = 8.075 + log (0.1026 / 0.0296) = 8.61

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Effect of dilution of a buffer solution

Ex. 1) Calculate the pH of a buffer prepared by mixing 0.250 mol of acetic acid with

0.100 mol of sodium acetate and diluting to 1.00 L.

pKa = – log 1.75×10–5 = 4.75

2) Calculate the pH when 10.0 ml of this buffer is diluted to 250 mL with water.

HOAC H+ + OAC–

Ka = [OAC–][H+] / [HOAC] [H+] = Ka [HOAC] / [OAC–]

pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.100M)/(0.250M) = 4.36

HOAC 0.250 M×10.0 ml = x M×250 ml x = 0.0100 M

NaOAC 0.100 M ×10.0 ml = x M×250 ml x = 0.00400 M

pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.00400M)/(0.0100M) = 4.36

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Effect of addition of acids or bases to buffer.

Buffer HA A-+ H+

NaOH + HA A-

HCl + A- HA

pH of 100 ml of buffer consisting of 0.20M HA and 0.10M NaA

pH = pKa+ log [NaA] / [HA] = pKa+ log (0.10/0.20) = pKa+ 0.30

If 0.01 mole of strong base is added

HA =0.10M.

A- =0.2M

pH = pKa+ log [NaA] / [HA] = pKa+ log (0.20/0.10) = pKa– 0.30

pH = 0.60

Not very large

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Effect of addition of acids or bases to buffer.

Ex. 1) Calculate the pH of a buffer prepared by mixing 0.250 mol of acetic acid with

0.100 mol of sodium acetate and diluting to 1.00 L.

pKa = – log 1.75×10–5 = 4.75

2) Calculate the pH when 10.0 ml of 0.1M NaOH is added.

HOAC H+ + OAC–

Ka = [OAC–][H+] / [HOAC] [H+] = Ka [HOAC] / [OAC–]

pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.100M)/(0.250M) = 4.36

HOAC N1V1-N2V2= 1000 ×0.250 -10 ×0.1=0.240 mole

NaOAC N1V1+N2V2= 1000 ×0.100 +10 ×0.1=0.110 mole

pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.110)/(0.240) = 4.41

Not very different

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Buffer capacity

The relative ability of a buffer solution to resist pH change upon addition of an acid or a base.

 = dCb / dpH = – dCa / dpH

A buffer is most effective in resisting changes in pH when pH= pKa

(i.e., when [HA] = [A–]).

Choose a buffer for an experiment whose pKa is as close as possible to the desired pH.

The useful pH range of a buffer is usually considered to be pKa 1 pH unit.

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buffer capacity
Buffer Capacity
  • [Buffer]=[Acid]+[Base]
  • [Acid]↑ & [Base]↑Capacity↑
  • In equimolar buffersis is important
  • Capacity↑

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Buffer capacity as a function of the logarithm of the ratio CNaA/CHA. The maximum buffer capacity occurs when the concentration of acid and conjugated base are equal; that is, when 0= 1 = 0.5.

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Buffer capacity

Ex. ) Calculate the pH and capacity of a buffer prepared by mixing 0.250 mol of acetic acid with 0.100 mol of sodium acetate and diluting to 1.00 L.

pKa = – log 1.75×10–5 = 4.75

pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.100M)/(0.250M) = 4.36

[OAC–] is less than [HOAC]and reduce sooner,by adding H+ ,

so it is determinant of capacity.

4.36 -1= 3.36= 4.75 + log ( 0.100M) - X/(0.250M) +X

-1.39 = log ( 0.100M) - X/(0.250M) +X

0.041=0.1-X/0.250+X X=0.064M

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Cb versus pH for a solution containing 0.100F HF with pKa =5.00.

  • Buffer capacity versus pH for the same system reaches a maximum when pH = pKa. The lower curve is the derivative of the upper curve.

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The composition of buffer solutions as a function of pH:  value

CT = CHOAC + CNaOAC

0 : fraction of the total concentration of acid that is undissociated

0= [HOAC] / CT

1 : the fraction dissociated

1= [OAC–] / CT

Ka = [OAC–][H3O+] / [HOAC] [OAC–] = Ka [HOAC] /[H3O+]

CT = CHOAC + CNaOAC = [HOAC] + [OAC–] = [HOAC] + Ka [HOAC] /[H3O+]

={[HOAC][H3O+]+[HOAC] Ka)]} / [H3O+]=[HOAC](Ka + [H3O+])/ [H3O+]

0= [HOAC] / CT = [H3O+] / ([H3O+]+ Ka)

1= [OAC–] / CT = Ka / ([H3O+]+ Ka)

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Variation in  of HOAC with pH

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The fraction of dissociation of weak electrolyte increases as the electrolyte is diluted. Stronger acid is more dissociated than the weaker acid at all concentrations.

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1 value : Fraction of dissociation of a weak acid

Ex. 0.0500 M benzoic acid

Ka = [A–][H+] / [HA] = (x)(x) / (F –x)

 x = [H+] = [A–] = 1.77 ×10–3

  = x / F = 1.77 ×10–3 / 0.05

= 0.0354

= 3.54 %

The fraction of dissociation of weak acids increases as the acid is diluted.

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1 value: Fraction of association of a weak base

Ex. 0.0372 M cocaine solution

Kb = 2.6×10–6x = [BH+] =[OH–] = KbF = 3.10 ×10–4

 1 = x / F = 3.10 ×10–4 /0.0372

= 0.0083

= 0.83 %

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slide36

The End!!!!!!!

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