Acid-base titration. Titration In an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration

Acid-base titration

Titration• In an acid-base titration, a solution of unknown

concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete– when the reaction is complete we have reached the

endpoint of the titration• An indicator may be added to determine the endpoint

– an indicator is a chemical that changes color when the pH changes

• When the moles of H3O+ = moles of OH−, the titration has reached its equivalence point

2Tro: Chemistry: A Molecular Approach, 2/e

Titration


Titration Curve• A plot of pH vs. amount of added titrant• The inflection point of the curve is the equivalence point of the

titration• Prior to the equivalence point, the known solution in the flask is

in excess, so the pH is closest to its pH• The pH of the equivalence point depends on the pH of the salt

solution– equivalence point of neutral salt, pH = 7 – equivalence point of acidic salt, pH < 7 – equivalence point of basic salt, pH > 7

• Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH


Titration Curve:Unknown Strong Base Added to Strong Acid


Before Equivalence(excess acid)

After Equivalence(excess base)

Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH

Equivalence Point equal moles of HCl and NaOH

pH = 7.00

Because the solutions are equal concentration, and 1:1 stoichiometry, the equivalence point is at equal volumes


HCl NaCl NaOH

mols before 2.50E-3 0 5.0E-4

mols change −5.0E-4 +5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new 0.0667 0.017 0

HCl NaCl NaOH


mols change

mols end

molarity, new

HCl NaCl NaOH


mols change −5.0E-4 +5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new

5.0 x 10−4 mole NaOH added


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

• Initial pH = −log(0.100) = 1.00• Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3

• Before equivalence point added 5.0 mL NaOH


Continued...

Table

HCl NaOH NaCl

mols before 2.50E-3 5.0E-4 0

mols change−5.0E-4 −5.0E-4 +5.0E-4

mols end2.00E-3 0 5.0E-4

molarity, new0.0667M 0 0.017M


• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(aq)

• To reach equivalence, the added moles NaOH = initial moles of HCl = 2.50 x 10−3 moles

• At equivalence, we have 0.00 mol HCl and 0.00 mol NaOH left over

• Because the NaCl is a neutral salt, the pH at equivalence = 7.00


HCl NaCl NaOH


mols change

mols end

molarity, new

HCl NaCl NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 0

molarity, new

HCl NaCl NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 0

molarity, new 0 0.050 0





• At equivalence point added 25.0 mL NaOH


HCl NaCl NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new 0 0.045 0.0091

HCl NaCl NaOH


mols change

mols end

molarity, new

HCl NaCl NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new




• After equivalence point added 30.0 mL NaOH


L of added NaOH0.100 mol NaOH

1 Lmoles added NaOH

added 5.0 mL NaOH0.00200 mol HClpH = 1.18


added 25.0 mL NaOHequivalence pointpH = 7.00added 40.0 mL NaOH0.00150 mol NaOHpH = 12.36

added 50.0 mL NaOH0.00250 mol NaOHpH = 12.52



Adding 0.100 M NaOH to 0.100 M HCl


Practice – Calculate the pH of the solution that results when 10.0 mL of 0.15 M NaOH is added to 50.0 mL of

0.25 M HNO3


HNO3 NaNO3 NaOH


mols change −1.5E-3 +1.5E-3 −1.5E-3

mols end 1.1E-3 1.5E-3 0

molarity, new

HNO3 NaNO3 NaOH


mols change −1.5E-3 +1.5E-3 −1.5E-3

mols end 1.1E-3 1.5E-3 0

molarity, new 0.018 0.025 0


0.25 M HNO3 • HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)• Initial pH = −log(0.250) = 0.60• Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2

• Before equivalence point added 10.0 mL NaOH


Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M

HNO3 to reach equivalence


Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M

HNO3 to reach equivalence

• HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)

• Initial pH = −log(0.250) = 0.60• Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2

• At equivalence point: moles of NaOH = 1.25 x 10−2



0.25 M HNO3


HNO3 NaNO3 NaOH


mols change −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new 0 0.0833 0.017

HNO3 NaNO3 NaOH


mols change −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new


0.25 M HNO3 • HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)• Initial pH = −log(0.250) = 0.60• Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2


HNO3 NaNO3 NaOH


mols change −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new 0 0.0833 0.017

HNO3 NaNO3 NaOH


mols change −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new


0.25 M HNO3 • HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)• Initial pH = −log(0.250) = 0.60• initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2



Titration of a Strong Base with a Strong Acid

• If the titration is run so that the acid is in the burette and the base is in the flask, the titration curve will be the reflection of the one just shown


Titration of a Weak Acid with a Strong Base• Titrating a weak acid with a strong base results in

differences in the titration curve at the equivalence point and excess acid region

• The initial pH is determined using the Ka of the weak acid• The pH in the excess acid region is determined as you

would determine the pH of a buffer • The pH at the equivalence point is determined using the

Kb of the conjugate base of the weak acid• The pH after equivalence is dominated by the excess

strong base– the basicity from the conjugate base anion is negligible


Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH

• HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq)

• Initial pH

[HCHO2] [CHO2−] [H3O+]

initial 0.100 0.000 ≈ 0

change −x +x +x

equilibrium 0.100 − x x x

Ka = 1.8 x 10−4



• HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq)

• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3

• Before equivalence added 5.0 mL NaOH

HA A− OH−

mols before 2.50E-3 0 0

mols added – – 5.0E-4

mols after 2.00E-3 5.0E-4 ≈ 0


HA A− OH−

mols before 2.50E-3 0 0

mols added – – 2.50E-3

mols after 0 2.50E-3 ≈ 0


added 25.0 mL NaOHCHO2

−(aq) + H2O(l) HCHO2(aq) + OH−

(aq)

[OH−] = 1.7 x 10−6 M

• HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq)


• At equivalence

Kb = 5.6 x 10−11

HA A− NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new

HA A− NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new 0 0.045 0.0091

HA A− NaOH


mols change

mols end

molarity, new


35

added 30.0 mL NaOH


• HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq)


• After equivalence

added 35.0 mL NaOH0.00100 mol NaOH xspH = 12.22

initial HCHO2 solution0.00250 mol HCHO2

pH = 2.37

added 5.0 mL NaOH0.00200 mol HCHO2

pH = 3.14


pH = 3.56

added 25.0 mL NaOHequivalence point0.00250 mol CHO2

−

[CHO2−]init = 0.0500 M

[OH−]eq = 1.7 x 10−6

pH = 8.23



pH = 4.34


pH = 3.92


pH = 3.74 = pKa

half-neutralization

Adding NaOH to HCHO2




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 as you did in Example 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


Example 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.

40

write an equation for the reaction for B with HA

use stoichiometry to determine the volume of added B

HNO2 + KOH NO2 + H2O

Tro: Chemistry: A Molecular Approach, 2/e

Example 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.

41


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 before 0.00400 0 ≈ 0

mols added – – 0.00100

mols after ≈ 00.00300 0.00100


Example 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.

42

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


Example 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.

44


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 ≈ 00.00200 0.00200

at half-equivalence, moles KOH = ½ mole HNO2


Example 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.

45

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


Titration Curve of a Weak Base with a Strong Acid


Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the initial pH of the

NH3 solution.


Practice – Titration of 25.0 mL of 0.10 M NH3 with 0.10 M HCl. Calculate the initial pH of the NH3(aq)

49

• NH3(aq) + HCl(aq) NH4Cl(aq)

• Initial: NH3(aq) + H2O(l) NH4+

(aq) + OH−(aq)

[HCl] [NH4+] [NH3]

initial 0 0 0.10

change +x +x −x

equilibrium x x 0.10−x

pKb = 4.75Kb = 10−4.75 = 1.8 x 10−5


Practice – Titration of 25.0 mL of 0.10 M NH3 with 0.10 M HCl. Calculate the initial pH of the NH3(aq)

50


• Initial: NH3(aq) + H2O(l) NH4+

(aq) + OH−(aq)

[HCl] [NH4+] [NH3]

initial 0 0 0.10

change +x +x −x

equilibrium x x 0.10−x

pKb = 4.75Kb = 10−4.75 = 1.8 x 10−5


Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution after

adding 5.0 mL of HCl.




52


• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3

• Before equivalence: after adding 5.0 mL of HCl

NH3 NH4Cl HCl


mols change −5.0E-4 −5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new 0.0667 0.017 0

NH4+

(aq) + H2O(l) NH4+

(aq) + H2O(l) pKb = 4.75pKa = 14.00 − 4.75 = 9.25



adding 5.0 mL of HCl.• NH3(aq) + HCl(aq) NH4Cl(aq)


• Before equivalence: after adding 5.0 mL of HCl

NH3 NH4Cl HCl


mols change −5.0E-4 −5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new 0.0667 0.017 0


Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution

at equivalence.


Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution at

equivalence.

55



• At equivalence mol NH3 = mol HCl = 2.50 x 10−3

added 25.0 mL HClNH3 NH4Cl HCl


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 0

molarity, new 0 0.050 0


Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution at

equivalence.

56

NH3(aq) + HCl(aq) NH4Cl(aq) at equivalence [NH4Cl] = 0.050 M

[NH3] [NH4+] [H3O+]

initial 0 0.050 ≈ 0

change +x −x +x

equilibrium x 0.050−x x

NH4+

(aq) + H2O(l) NH3(aq) + H3O+(aq)


Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution

after adding 30.0 mL of HCl.




58



• After equivalence: after adding 30.0 mL HClNH3 NH4Cl HCl


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new 0 0.045 0.0091

when you mix a strong acid, HCl, with a weak acid, NH4

+, you only need to consider the 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


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 titrant 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


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

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


Example 16.8: Calculate the molar solubility of PbCl2 in pure water at 25 C

71

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


Example 16.8: Calculate the molar solubility of PbCl2 in pure water at 25 C

72

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


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



74



[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



75

substitute into the Ksp expression

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)


Ksp and Relative Solubility

• Molar solubility is related to Ksp

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

• 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 left77


Example 16.10: Calculate the molar solubility of CaF2 in 0.100 M NaF at 25 C

78



[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


Example 16.10: Calculate the molar solubility of CaF2 in 0.100 M NaF at 25 C

79

substitute into the Ksp expression,assume S is small


[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


Practice – Determine the concentration of Ag+ ions in seawater that has a [Cl−] of 0.55 M

Ksp of AgCl = 1.77 x 10−10





[Ag+] [Cl−]

initial 0 0.55

change +S +S

equilibrium S 0.55 + S

AgCl(s) Ag+(aq) + Cl−(aq)

Ksp = [Ag+][Cl−]



83

substitute into the Ksp expression,assume S is small


[Ag+] [Cl−]

Initial 0 0.55

Change +S +S

Equilibrium S 0.55 + S

Ksp = [Ag+][Cl−]

Ksp = (S)(0.55 + S)

Ksp = (S)(0.55)


[Ag+] [Cl−]

Initial 0 0.55

Change +S +S

Equilibrium S 0.55 + S

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 solubilityM2(CO3)n(s) 2 Mn+(aq) + nCO3

2−(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


Example 16.12: Will a precipitate form when we mix Pb(NO3)2(aq) with NaBr(aq) if the concentrations after mixing

are 0.0150 M and 0.0350 M respectively?write the equation for the reaction

determine the ion concentrations of the original salts

determine the Ksp for any “insoluble” product

write the dissociation reaction for the insoluble product

calculate Q, using the ion concentrations

compare Q to Ksp. If Q > Ksp, precipitation

Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq)

Ksp of PbBr2 = 4.67 x 10–6

PbBr2(s) Pb2+(aq) + 2 Br−

(aq)

Pb(NO3)2 = 0.0150 M Pb2+ = 0.0150 M,

NO3− = 2(0.0150 M)

NaBr = 0.0350 M Na+ = 0.0350 M,

Br− = 0.0350 M

Q < Ksp, so no precipitation89


Practice – Will a precipitate form when we mix Ca(NO3)2(aq) with NaOH(aq) if the concentrations after

mixing are both 0.0175 M? Ksp of Ca(OH)2 = 4.68 x 10−6


Practice – Will a precipitate form when we mix Ca(NO3)2(aq) with NaOH(aq) if the concentrations after

mixing are both 0.0175 M? write the equation for the reaction

determine the ion concentrations of the original salts

determine the Ksp for any “insoluble” product

write the dissociation reaction for the insoluble product

calculate Q, using the ion concentrations

compare Q to Ksp. If Q > Ksp, precipitation

Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)

Ksp of Ca(OH)2 = 4.68 x 10–6

Ca(OH)2(s) Ca2+(aq) + 2 OH−

(aq)

Ca(NO3)2 = 0.0175 M Ca2+ = 0.0175 M,

NO3− = 2(0.0175 M)

NaOH = 0.0175 M Na+ = 0.0175 M, OH− = 0.0175 M

Q > Ksp, so precipitation


92

Example 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

Mg(OH)2(s) Mg2+(aq) + 2 OH−

(aq)


Practice – What is the minimum concentration of Ca(NO3)2(aq) that will precipitate Ca(OH)2 from 0.0175 M

NaOH(aq)? Ksp of Ca(OH)2 = 4.68 x 10−6


94

Practice – What is the minimum concentration of Ca(NO3)2(aq) that will precipitate Ca(OH)2 from 0.0175 M

NaOH(aq)?


[Ca(NO3)2] = [Ca2+] = 0.0153 M

Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)


(aq)


95

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

seawater?



(aq)


96

Example 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 precipitating Ca2+ begins when [OH−] = 2.06 x 10−2 M

when Ca2+ just begins to precipitate out, the [Mg2+] has dropped from 0.059 M to 4.8 x 10−10 M

Mg(OH)2(s) Mg2+(aq) + 2 OH−

(aq)


Practice – A solution is made by mixing Pb(NO3)2(aq) with AgNO3(aq) so both compounds have a concentration of 0.0010

M. NaCl(s) is added to precipitate out both AgCl(s) and PbCl2(aq). What is the [Ag+] concentration when the Pb2+ just begins to

precipitate?


98

Practice – What is the [Ag+] concentration when the Pb2+(0.0010 M) just begins to precipitate?


AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)



99

Practice – What is the [Ag+] concentration when the Pb2+(0.0010 M) just begins to precipitate?


Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)

PbCl2(s) Pb2+(aq) + 2 Cl−(aq)


100

Practice – What is the [Ag+] concentration when the Pb2+(0.0010 M) just begins to precipitate

precipitating Ag+ begins when [Cl−] = 1.77 x 10−7 M precipitating Pb2+ begins when [Cl−] = 1.08 x 10−1 M

when Pb2+ just begins to precipitate out, the [Ag+] has dropped from 0.0010 M to 1.6 x 10−9 M



101

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


102

Qualitative Analysis



104

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


105

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


106

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


107

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


108

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, because 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


Formation Constants


Example 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)


Example 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?

Create an ICE table. Because 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

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


Example 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?

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

and solve for x

confirm the “x is small” approximation

[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

2.7 x 10−13 << 6.7 x 10−4, so the approximation is valid


Practice – What is [HgI42−] when 125 mL of 0.0010 M KI

is reacted with 75 mL of 0.0010 M HgCl2?4 KI(aq) + HgCl2(aq) 2 KCl(aq) + K2HgI4(aq)


Practice – What is [HgI42−] when 125 mL of 0.0010 M KI is

reacted with 75 mL of 0.0010 M HgCl2?4 KI(aq) + HgCl2(aq) 2 KCl(aq) + K2HgI4(aq)

Write the formation reaction and Kf expression.Look up Kf value

determine the concentration of ions in the diluted solutions

Hg2+(aq) + 4 I−(aq) HgI42−(aq)




Create an ICE table. Because Kf is large, assume all the lim. rgt. is converted into complex ion, then the system returns to equilibrium.

[Hg2+] [I−] [HgI42−]

initial 3.75E-4 6.25E-4 0

change ≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)

equilibrium 2.19E-4 x 1.56E-4


I− is the limiting reagent




substitute in and solve for x

confirm the “x is small” approximation

2 x 10−8 << 1.6 x 10−4, so the approximation is valid


[HgI42−] = 1.6 x 10−4

[Hg2+] [I−] [HgI42−]

initial 3.75E-4 6.25E-4 0

change ≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)

equilibrium 2.19E-4 x 1.56E-4


122

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+



124

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+


125

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)



Documents

Acid-base titration. Titration In an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration