ANALYTICAL CHEMISTRY - CLUTCH 1E CH.12 - ADVANCED TOPICS IN EQUILIBRIUMlightcat-files.s3.amazonaws.com/packets/admin_analytical... · 2019-11-23 · PRACTICE: SYSTEMATIC APPROACH

! www.clutchprep.com

!

ANALYTICAL CHEMISTRY - CLUTCH 1E

CH.12 - ADVANCED TOPICS IN EQUILIBRIUM

CONCEPT: SYSTEMATIC APPROACH – ACID-BASE SYSTEMS

Strong Acids and Bases are considered _________________ Electrolytes so they ionize completely in water.

• In general, the ___________ the Ka the stronger the acid and the ___________ the concentration of H+.

When calculating the pH of a solution we must take into consideration the concentration of the strong acid or base.

Concentration ≥10−6M

The concentrations of either of H+ and OH– are significant enough to determine pH and pOH directly.

1.5 x 10-3 M 1.5 x 10-3 M 1.5 x 10-3 M 0.00075 M 0.00075 M 0.00075 MHNO3 (aq) H+ (aq) + NO3

– (aq) NaOH (aq) Na+ (aq) + OH– (aq)

H2O H2O

Concentration ≤10−8M

The concentrations either of H+ and OH– are too small to be significant and so pH equals _______.

8.4 x 10-11 M 8.4 x 10-11 M 8.4 x 10-11 M 7.0 x 10-9 M 7.0 x 10-9 M 7.0 x 10-9 MHNO3 (aq) H+ (aq) + NO3

– (aq) NaOH (aq) Na+ (aq) + OH– (aq)

H2O H2O

Between10−6M to10−8M

The concentrations of H+ and OH– must compete with the auto-ionization of water so a systematic approach is used.

0.200 M 0.200 M 0.200 M



Page 2

PRACTICE: SYSTEMATIC APPROACH – ACID-BASE SYSTEMS CALCULATIONS 1 EXAMPLE: Determine the pH of a 3.5 x 10-8 M HBr. PRACTICE: Determine the pH of a 6.7 x 10-8 M NaOH.



Page 3

CONCEPT: FRACTIONAL COMPOSITION Fractional compositions are used as an illustration for the amount of acid or base species at a specific pH.

Monoprotic Systems The dissociation of a weak monoprotic acid creates an equilibrium and expression that can be tied to its mass balance.

HA H+ +A– Ka =[H+ ][A− ][HA] Mass Balance : F = [HA]+[A– ]

The fraction of HA molecules is represented by αHA .

pH

(frac

tion

in e

ach

form

)α

0

0

.25

0.50

0.7

5

1.0

0

Diprotic Systems From the derived equilibrium expressions we can determine the mass balance.

H2A H+ +HA– HA– H+ +A2–

Ka2=[H+ ][A2− ][HA− ]

⇒ [A2− ]=[HA− ][Ka2

][H+ ]

=[H2A]Ka1

Ka2

[H+ ]2Ka1=[H+ ][HA][H2A]

⇒ [HA− ]=[H2A][Ka1

][H+ ]

F = [H2A] [H2A]⋅[H+ ]2 +[H+ ]Ka1

+Ka1Ka2

[H+ ]2⎤

⎦⎥

⎡

⎣⎢⎢

The fractions of the three major diprotic forms can be seen as:

[HA]= [H+ ]F[H+ ]+Ka

Dividing by F αHA =HAF

=[H+ ]

[H+ ]+Ka

The fraction in the conjugate base form, A–, can be represented as αA− .

αA− =

A−

F=

Ka

[H+ ]+Ka

pH

0

0

.25

0.50

0.7

5

1.0

0(fr

actio

n in

eac

h fo

rm)

α

αH2A=H2AF

=[H+ ]2

[H+ ]2 +[H+ ]Ka1+Ka1

Ka2

αHA− =

HA−

F=

Ka1[H+ ]

[H+ ]2 +[H+ ]Ka1+Ka1

Ka2

αA2−

=A2−

F=

Ka1Ka2

[H+ ]2 +[H+ ]Ka1+Ka1

Ka2



Page 4

PRACTICE: FRACTIONAL COMPOSITION CALCULATIONS 1 EXAMPLE 1: A dibasic compound, B, has pKb1 = 5.00 and pKb2 = 8.00. Find the fraction of the acidic form when the pH = 9.00. EXAMPLE 2: What fraction of tyrosine (pKa1 = 2.37, pKa2 = 8.67) exists in all of its forms at pH = 10.00? PRACTICE: Calculate the fraction of the intermediate for sulfurous acid, H2SO3, at pH = 11.00? pKa1 = 1.80, pKa2 = 7.19.



Page 5

CONCEPT: FRACTIONAL COMPOSITION AND CONCENTRATIONS Monoprotic Systems

Recall that the fraction of HA and A– molecules are represented by αHA and αA− .

pH

(frac

tion

in e

ach

form

)α

0

0

.25

0.50

0.7

5

1.0

0

When taking into account the formal concentrations of HA and A– molecules we now restructure the equations:

[HA]= αHAFHA =[H+ ]FHA[H+ ]+Ka

[A− ]= αA−FHA =

KaFHA[H+ ]+Ka

Diprotic Systems From the derived equilibrium expressions we can determine the mass balance.

H2A H+ +HA– HA– H+ +A2–

Ka2=[H+ ][A2− ][HA− ]

⇒ [A2− ]=[HA− ][Ka2

][H+ ]

=[H2A]Ka1

Ka2

[H+ ]2Ka1=[H+ ][HA][H2A]

⇒ [HA− ]=[H2A][Ka1

][H+ ]

F = [H2A] [H2A]⋅[H+ ]2 +[H+ ]Ka1

+Ka1Ka2

[H+ ]2⎤

⎦⎥

⎡

⎣⎢⎢

The formal concentrations of the three major diprotic forms can be restructured as:

[HA]= [H+ ]F[H+ ]+Ka

Dividing by F αHA =HAF

=[H+ ]

[H+ ]+Ka

αA− =

A−

F=

Ka

[H+ ]+Ka

pH

0

0

.25

0.50

0.7

5

1.0

0(fr

actio

n in

eac

h fo

rm)

α

αH2A=H2AF

=[H+ ]2

[H+ ]2 +[H+ ]Ka1+Ka1

Ka2

αHA− =

HA−

F=

Ka1[H+ ]

[H+ ]2 +[H+ ]Ka1+Ka1

Ka2

αA2−

=A2−

F=

Ka1Ka2

[H+ ]2 +[H+ ]Ka1+Ka1

Ka2

[H2A]= αH2AFH2A =

[H+ ]2 FH2A[H+ ]2 +[H+ ]Ka1

+Ka1Ka2

[HA− ]= αHA−FH2A =

Ka1[H+ ]FH2A

[H+ ]2 +[H+ ]Ka1+Ka1

Ka2

[A2− ]= αA2−FH2A =

Ka1Ka2FH2A

[H+ ]2 +[H+ ]Ka1+Ka1

Ka2



Page 6

PRACTICE: FRACTIONAL COMPOSITION AND CONCENTRATIONS CALCULATIONS 1 EXAMPLE 1: A dibasic compound, B, has pKb1 = 4.00 and pKb2 = 6.00. Find the concentration of the intermediate form when FH2A = 0.150 M and the pH = 8.00.

EXAMPLE 2: Calculate the concentration of the acidic form for 0.230 M tartaric acid, H2C4H4O6, at pH = 6.00? pKa1 = 3.00, pKa2 = 4.34.



Page 7

CONCEPT: DAVIES EQUATION We learned that the activity coefficient and ionic strength of a solution could be closely and accurately related by using the extended Debye-Huckel equation:

log γ = − 0.51z2 µ

1+ α µ305

⎞

⎠⎟⎟

⎛

⎝⎜⎜

When the size parameter of the ion is unknown we can instead use the Davies Equation. • Because of the lack of a size parameter this formula is most useful for monovalent ions.

From the Davies Equation, all ions with the same magnitude in charge will have the same activity coefficient. EXAMPLE: Calculate the activity coefficient of Ca2+ in 0.025 M Ca3(PO4)2.

log γ = − 0.51z2 µ

1+ µ− 0.3µ

⎞

⎠⎟

⎛

⎝⎜⎜ Ionic Strength ±1 ± 2 ±3

0.001 0.97 0.87 0.730.005 0.93 0.74 0.510.010 0.90 0.66 0.400.050 0.82 0.45 0.160.100 0.78 0.36 0.100.200 0.73 0.28 0.060.500 0.69 0.23 0.040.700 0.69 0.23 0.04

Ionic Charge (z)



Page 8

CONCEPT: DEPENDENCE OF SOLUBILITY ON PH

Recall that ionic compounds are composed of an anion and cation, either of which can create an acidic, basic or neutral solution.

Cations +

Transition Metals___ or higher charge will be acidic, less than ___ will be neutral

MnI5

Main Group Metals___ or higher charge will be acidic, less than ___ will be neutral

AlF3

Positive Amines

CH3NH3Cl

Positively charged amines are acidic

Cations can create solutions that are either acidic or neutral.

• ______________ the pH increases the solubility of sparingly acidic salts.

Anions –Add an H+ to the anion and if you create a weak acid then your negative ion is basic.

NaNO2H+

Add an H+ to the anion and if you create a strong acid then your negative ion is neutral.

KClH+

Anions can create solutions that are either basic or neutral.

• ______________ the pH increases the solubility of sparingly basic salts.

AmphotericAcidic BasicHSO4

– HSO3– H2PO4

– HCO3– HS – HPO4

2-



Page 9

PRACTICE: DEPENDENCE OF SOLUBILITY ON PH CALCULATIONS 1

EXAMPLE 1: BaCO3 is the slightly soluble ionic salt that results from the reaction between Ba(OH)2 and H2CO3. Identify the effect of increasing acidity on the solubility of the given compound.

EXAMPLE 2: Which salts will be more soluble in an acidic solution than in pure water?

a. CuBr

b. Ag2SO4

c. BaSO3

d. Sn(OH)2

e. KClO4



Page 10

Documents

ANALYTICAL CHEMISTRY - CLUTCH 1E CH.12 - ADVANCED TOPICS IN EQUILIBRIUMlightcat-files.s3.amazonaws.com/packets/admin_analytical... · 2019-11-23 · PRACTICE: SYSTEMATIC APPROACH