Tro, Chemistry: A Molecular Approach 1 Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCHO 2(aq) + NaOH (aq) NaCHO 2 (aq) + H 2 O (aq) after equivalence.

Tro, Chemistry: A Molecular Approach 2

added 30.0 mL NaOH0.00050 mol NaOH xspH = 11.96


Adding NaOH to HCHO2

added 12.5 mL NaOH0.00125 mol HCHO2

pH = 3.74 = pKa

half-neutralization

initial HCHO2 solution0.00250 mol HCHO2

pH = 2.37


pH = 3.14


pH = 3.56


pH = 3.92


pH = 4.34


added 25.0 mL NaOHequivalence point0.00250 mol CHO2

−

[CHO2−]init = 0.0500 M

[OH−]eq = 1.7 x 10-6

pH = 8.23



Titrating Weak Acid with a Strong Base• the initial pH is that of the weak acid solution– calculate like a weak acid equilibrium problem • e.g., 15.5 and 15.6

• before the equivalence point, the solution becomes a buffer– calculate mol HAinit and mol A−

init using reaction stoichiometry

– calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−

init

• half-neutralization pH = pKa


Titrating Weak Acid with a Strong Base• at the equivalence point, the mole HA = mol Base, so

the resulting solution has only the conjugate base anion in it before equilibrium is established– mol A− = original mole HA

• calculate the volume of added base like Ex 4.8– [A−]init = mol A−/total liters– calculate like a weak base equilibrium problem

• e.g., 15.14• beyond equivalence point, the OH is in excess– [OH−] = mol MOH xs/total liters– [H3O+][OH−]=1 x 10-14


Ex 16.7a – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the volume of KOH at the

equivalence point

Write an equation for the reaction for B with HA.

Use Stoichiometry to determine the volume of added B

HNO2 + KOH NO2 + H2O

KOH L .02000

KOH mol 0.200

KOH L 1

NO mol 1

KOH mol 1

NO L 1

NO mol 0.100NO L .04000

22

22

L 0400.0mL 1

L 0.001mL 0.40

mL 0.20L 0.001

mL 1L 0200.0


Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH

Write an equation for the reaction for B with HA.

Determine the moles of HAbefore & moles of added B

Make a stoichiometry table and determine the moles of HA in excess and moles A made


KOH mol 00100.0L 1

KOH mol 200.0

mL 1

L 0.001mL 00.5

HNO2 NO2- OH−

mols Before 0.00400 0 ≈ 0

mols added - - 0.00100

mols After ≈ 0

22 HNO mol 00400.0

L 1

HNO mol 100.0

mL 1

L 0.001mL 0.40

0.00300 0.00100


Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH.

Write an equation for the reaction of HA with H2O

Determine Ka and pKa for HA

Use the Henderson-Hasselbalch Equation to determine the pH

HNO2 + H2O NO2 + H3O+

HNO2 NO2- OH−



mols After 0.00300 0.00100 ≈ 0

Table 15.5 Ka = 4.6 x 10-4

15.3106.4loglogp 4 aa KK

2

2

HNO

NOlogppH aK

67.20.00300

00100.0log15.3pH


Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at

the half-equivalence pointWrite an equation for the reaction for B with HA.

Determine the moles of HAbefore & moles of added B

Make a stoichiometry table and determine the moles of HA in excess and moles A made


HNO2 NO2- OH−



mols After ≈ 0

22 HNO mol 00400.0

L 1

HNO mol 100.0

mL 1

L 0.001mL 0.40

0.00200 0.00200

at half-equivalence, moles KOH = ½ mole HNO2


Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at

the half-equivalence point.

Write an equation for the reaction of HA with H2O

Determine Ka and pKa for HA

Use the Henderson-Hasselbalch Equation to determine the pH

HNO2 + H2O NO2 + H3O+

HNO2 NO2- OH−



mols After 0.00200 0.00200 ≈ 0

Table 15.5 Ka = 4.6 x 10-4

15.3106.4loglogp 4 aa KK

2

2

HNO

NOlogppH aK

15.30.00200

00200.0log15.3pH


Titration Curve of a Weak Base with a Strong Acid


Titration of a Polyprotic Acid• if Ka1 >> Ka2, there will be two equivalence

points in the titration– the closer the Ka’s are to each other, the less

distinguishable the equivalence points are

titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH


Monitoring pH During a Titration• the general method for monitoring the pH during the

course of a titration is to measure the conductivity of the solution due to the [H3O+]– using a probe that specifically measures just H3O+

• the endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve

• if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator


Monitoring pH During a Titration


Indicators• many dyes change color depending on the pH of the solution• these dyes are weak acids, establishing an equilibrium with the

H2O and H3O+ in the solutionHInd(aq) + H2O(l) Ind

(aq) + H3O+(aq)

• the color of the solution depends on the relative concentrations of Ind:HInd– when Ind:HInd ≈ 1, the color will be mix of the colors of Ind and

HInd – when Ind:HInd > 10, the color will be mix of the colors of Ind

– when Ind:HInd < 0.1, the color will be mix of the colors of HInd

18

Phenolphthalein


Methyl Red

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)2N N N NH

NaOOC

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)2N N N N

NaOOC

H3O+ OH-


Monitoring a Titration with an Indicator

• for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point

• an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH– pKa of HInd ≈ pH at equivalence point

21

Acid-Base Indicators


Solubility Equilibria

• all ionic compounds dissolve in water to some degree – however, many compounds have such low

solubility in water that we classify them as insoluble

• we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water


Solubility Product• the equilibrium constant for the dissociation of a solid

salt into its aqueous ions is called the solubility product, Ksp

• for an ionic solid MnXm, the dissociation reaction is:MnXm(s) nMm+(aq) + mXn−(aq)

• the solubility product would be Ksp = [Mm+]n[Xn−]m

• for example, the dissociation reaction for PbCl2 isPbCl2(s) Pb2+(aq) + 2 Cl−(aq)

• and its equilibrium constant is Ksp = [Pb2+][Cl−]2



Molar Solubility• solubility is the amount of solute that will dissolve in a

given amount of solution– at a particular temperature

• the molar solubility is the number of moles of solute that will dissolve in a liter of solution– the molarity of the dissolved solute in a saturated solution

• for the general reaction MnXm(s) nMm+(aq) + mXn−(aq)

mnmn

sp

mn

K

solubilitymolar


Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C

Write the dissociation reaction and Ksp expression

Create an ICE table defining the change in terms of the solubility of the solid

[Pb2+] [Cl−]

Initial 0 0

Change +S +2S

Equilibrium S 2S

PbCl2(s) Pb2+(aq) + 2 Cl−(aq)

Ksp = [Pb2+][Cl−]2


Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C

Substitute into the Ksp expression

Find the value of Ksp from Table 16.2, plug into the equation and solve for S

[Pb2+] [Cl−]

Initial 0 0

Change +S +2S

Equilibrium S 2S

Ksp = [Pb2+][Cl−]2

Ksp = (S)(2S)2

M 1043.14

1017.1

4

4

2

35

3

3

S

SK

SK

sp

sp


Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M





[Pb2+] [Br−]

Initial 0 0

Change +(1.05 x 10-2) +2(1.05 x 10-2)

Equilibrium (1.05 x 10-2) (2.10 x 10-2)

PbBr2(s) Pb2+(aq) + 2 Br−(aq)

Ksp = [Pb2+][Br−]2




plug into the equation and solve

Ksp = [Pb2+][Br−]2

Ksp = (1.05 x 10-2)(2.10 x 10-2)2

[Pb2+] [Br−]

Initial 0 0

Change +(1.05 x 10-2) +2(1.05 x 10-2)

Equilibrium (1.05 x 10-2) (2.10 x 10-2)

6

222

1063.4

1010.21005.1

sp

sp

K

K


Ksp and Relative Solubility

• molar solubility is related to Ksp

• but you cannot always compare solubilities of compounds by comparing their Ksps

• in order to compare Ksps, the compounds must have the same dissociation stoichiometry


The Effect of Common Ion on Solubility• addition of a soluble salt that contains one of

the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt

• for example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2

PbCl2(s) Pb2+(aq) + 2 Cl−(aq)

addition of Cl− shifts the equilibrium to the left


Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C



[Ca2+] [F−]

Initial 0 0.100

Change +S +2S

Equilibrium S 0.100 + 2S

CaF2(s) Ca2+(aq) + 2 F−(aq)

Ksp = [Ca2+][F−]2


Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C


assume S is small

Find the value of Ksp from Table 16.2, plug into the equation and solve for S

[Ca2+] [F−]

Initial 0 0.100

Change +S +2S

Equilibrium S 0.100 + 2S

Ksp = [Ca2+][F−]2

Ksp = (S)(0.100 + 2S)2

Ksp = (S)(0.100)2

M 1046.1

100.0

1046.1

100.0

8

2

10

2

S

S

SKsp


The Effect of pH on Solubility• for insoluble ionic hydroxides, the higher the pH, the

lower the solubility of the ionic hydroxide– and the lower the pH, the higher the solubility– higher pH = increased [OH−]

M(OH)n(s) Mn+(aq) + nOH−(aq)

• for insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility

M2(CO3)n(s) 2 Mn+(aq) + nCO32−(aq)

H3O+(aq) + CO32− (aq) HCO3

− (aq) + H2O(l)


Precipitation• precipitation will occur when the concentrations of the

ions exceed the solubility of the ionic compound• if we compare the reaction quotient, Q, for the current

solution concentrations to the value of Ksp, we can determine if precipitation will occur– Q = Ksp, the solution is saturated, no precipitation– Q < Ksp, the solution is unsaturated, no precipitation– Q > Ksp, the solution would be above saturation, the salt

above saturation will precipitate

• some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions


precipitation occurs if Q > Ksp

a supersaturated solution will precipitate if a seed crystal is added


Selective Precipitation

• a solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others

• a successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different


Ex 16.13 What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059]) from seawater?

precipitating may just occur when Q = Ksp

22 ]OH][Mg[ Q

6

13

132

109.1059.0

1006.2]OH[

1006.2]OH][059.0[

spKQ


Ex 16.14 What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from

seawater?

precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M 22 ]OH][Ca[ Q

2

6

62

1060.2011.0

1068.4]OH[

1068.4]OH][011.0[

spKQ


Ex 16.14 What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from

seawater?

precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M

M 108.4

1060.2

1006.2]Mg[

1006.2]1060.2][Mg[

when

1022

132

13222

spKQ

precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M 22 ]OH][Mg[ Q when Ca2+ just begins to

precipitate out, the [Mg2+] has dropped from 0.059 M to 4.8 x 10-10 M


Qualitative Analysis• an analytical scheme that utilizes selective

precipitation to identify the ions present in a solution is called a qualitative analysis scheme– wet chemistry

• a sample containing several ions is subjected to the addition of several precipitating agents

• addition of each reagent causes one of the ions present to precipitate out


Qualitative Analysis

44


Group 1

• group one cations are Ag+, Pb2+, and Hg22+

• all these cations form compounds with Cl− that are insoluble in water– as long as the concentration is large enough– PbCl2 may be borderline• molar solubility of PbCl2 = 1.43 x 10-2 M

• precipitated by the addition of HCl

46

Group 2• group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+,

Pb2+, Sb3+, and Hg2+

• all these cations form compounds with HS− and S2− that are insoluble in water at low pH

• precipitated by the addition of H2S in HCl

47

Group 3

• group three cations are Fe2+, Co2+, Zn2+, Mn2+, Ni2+ precipitated as sulfides; as well as Cr3+, Fe3+, and Al3+ precipitated as hydroxides

• all these cations form compounds with S2− that are insoluble in water at high pH

• precipitated by the addition of H2S in NaOH


Group 4

• group four cations are Mg2+, Ca2+, Ba2+ • all these cations form compounds with PO4

3− that are insoluble in water at high pH

• precipitated by the addition of (NH4)2HPO4

49

Group 5• group five cations are Na+, K+, NH4

+

• all these cations form compounds that are soluble in water – they do not precipitate

• identified by the color of their flame


Complex Ion Formation

• transition metals tend to be good Lewis acids• they often bond to one or more H2O molecules to form

a hydrated ion– H2O is the Lewis base, donating electron pairs to form

coordinate covalent bondsAg+(aq) + 2 H2O(l) Ag(H2O)2

+(aq)• ions that form by combining a cation with several

anions or neutral molecules are called complex ions– e.g., Ag(H2O)2

+

• the attached ions or molecules are called ligands– e.g., H2O


Complex Ion Equilibria

• if a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand

Ag(H2O)2+

(aq) + 2 NH3(aq) Ag(NH3)2+

(aq) + 2 H2O(l) – generally H2O is not included, since its complex ion is

always present in aqueous solution

Ag+(aq) + 2 NH3(aq) Ag(NH3)2

+(aq)


Formation Constant

• the reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction


+(aq)

• the equilibrium constant for the formation reaction is called the formation constant, Kf

23

23

]NH][[Ag

])[Ag(NH

fK


Formation Constants


Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the

[Cu2+] at equilibrium?

Write the formation reaction and Kf expression.

Look up Kf value

Determine the concentration of ions in the diluted solutions

Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)

134

32

243 107.1

]NH][Cu[

])Cu(NH[

fK

M 107.6L 0.250 L 200.0

L 1mol 101.5

L 200.0]Cu[ 4

-3

2

M 101.1L 0.250 L 200.0

L 1mol 100.2

L 250.0]NH[ 1

-1

3

55



Create an ICE table. Since Kf is large, assume all the Cu2+ is converted into complex ion, then the system returns to equilibrium

[Cu2+] [NH3] [Cu(NH3)22+]

Initial 6.7E-4 0.11 0

Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4

Equilibrium x 0.11 6.7E-4


132

43

43

2

107.1])Cu(NH[

]NH][Cu[

fK





134

32

243 107.1

]NH][Cu[

])Cu(NH[

fK

Substitute in and solve for x

confirm the “x is small” approximation [Cu2+] [NH3] [Cu(NH3)2

2+]

Initial 6.7E-4 0.11 0

Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4

Equilibrium x 0.11 6.7E-4

13413

4

4

413

107.211.0107.1

107.6

11.0

107.6107.1

x

x

since 2.7 x 10-13 << 6.7 x 10-4, the approximation is valid


The Effect of Complex Ion Formation on Solubility

• the solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands

AgCl(s) Ag+(aq) + Cl−

(aq) Ksp = 1.77 x 10-10


+(aq) Kf = 1.7 x 107

• adding NH3 to a solution in equilibrium with AgCl(s) increases the solubility of Ag+

58


Solubility of Amphoteric Metal Hydroxides

• many metal hydroxides are insoluble• all metal hydroxides become more soluble in acidic

solution– shifting the equilibrium to the right by removing OH−

• some metal hydroxides also become more soluble in basic solution– acting as a Lewis base forming a complex ion

• substances that behave as both an acid and base are said to be amphoteric

• some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+


Al3+

• Al3+ is hydrated in water to form an acidic solutionAl(H2O)6

3+(aq) + H2O(l) Al(H2O)5(OH)2+

(aq) + H3O+(aq)

• addition of OH− drives the equilibrium to the right and continues to remove H from the molecules

Al(H2O)5(OH)2+(aq) + OH−

(aq) Al(H2O)4(OH)2+

(aq) + H2O (l)

Al(H2O)4(OH)2+

(aq) + OH−(aq) Al(H2O)3(OH)3(s) + H2O

(l)


Tro, Chemistry: A Molecular Approach 1 Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCHO 2(aq) + NaOH (aq) NaCHO 2 (aq) + H 2 O (aq) after equivalence.

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