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Chapter 16

“AqueousEquilibrium”

The Danger of Antifreeze• Each year, thousands of pets and wildlife die from consuming

antifreeze• Most brands of antifreeze contain ethylene glycol

– sweet taste and initial effect is drunkenness• Metabolized in the liver to glycolic acid

– HOCH2COOH• If present in high enough concentration in the bloodstream, it

overwhelms the buffering ability of HCO3−, causing the blood pH

to drop• When the blood pH is low, it ability to carry O2 is compromised

– acidosis• The treatment is to give the patient ethyl alcohol, which has a

higher affinity for the enzyme that catalyzes the metabolism ofethylene glycol

Buffers

• Buffers are solutions that resist changes inpH when an acid or base is added

• They act by neutralizing the added acid orbase

• But just like everything else, there is a limit towhat they can do, eventually the pH changes

• Many buffers are made by mixing a solutionof a weak acid with a solution of soluble saltcontaining its conjugate base anion

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Making an Acid Buffer

How Acid Buffers WorkHA(aq) + H2O(l) ⇔ A−(aq) + H3O+

(aq)

• Buffers work by applying Le Châtelier’s Principle toweak acid equilibrium

• Buffer solutions contain significant amounts of theweak acid molecules, HA – these molecules reactwith added base to neutralize it– you can also think of the H3O+ combining with the OH− to

make H2O; the H3O+ is then replaced by the shiftingequilibrium

• The buffer solutions also contain significant amountsof the conjugate base anion, A− - these ions combinewith added acid to make more HA and keep the H3O+

constant

Common Ion Effect HA(aq) + H2O(l) ⇔ A−(aq) + H3O+

(aq)

• Adding a salt containing the anion, NaA,that is the conjugate base of the acid(the common ion) shifts the position ofequilibrium to the left

• This causes the pH to be higher thanthe pH of the acid solution– lowering the H3O+ ion concentration

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Common Ion Effect

Practice• What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M

NaC2H3O2?

Practice• What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and

0.071 M KF?

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Henderson-HasselbalchEquation

• Calculating the pH of a buffer solution can besimplified by using an equation derived fromthe Ka expression called the Henderson-Hasselbalch Equation

• The equation calculates the pH of a bufferfrom the Ka and initial concentrations of theweak acid and salt of the conjugate base– as long as the “x is small” approximation is valid

Henderson-HasselbalchEquation

!!"

#$$%

&+=

[HA]

][AlogppH

-

aK

Example• What is the pH of a buffer that is 0.050 M HC7H5O2 and 0.150 M

NaC7H5O2?

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Example• What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and

0.071 M KF?

Using the Henderson-Hasselbalch Equation

• The Henderson-Hasselbalch equation isgenerally good enough when the “x is small”approximation is applicable

• Generally, the “x is small” approximation willwork when both of the following are true:

a)the initial concentrations of acid and salt are not verydilute

b)the Ka is fairly small

• For most problems, this means that the initialacid and salt concentrations should be over1000x larger than the value of Ka

Example• What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol

NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it

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Basic BuffersB:(aq) + H2O(l) ⇔ H:B+

(aq) + OH−(aq)

• buffers can also be made by mixing aweak base, (B:), with a soluble salt of itsconjugate acid, H:B+Cl−

Example• What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and

0.20 M NH4Cl?

Buffering Effectiveness• A good buffer should be able to neutralize

moderate amounts of added acid or base– there is a limit to how much can be added before

the pH changes significantly• The buffering capacity is the amount of acid

or base a buffer can neutralize• The buffering range is the pH range the buffer

can be effective• The effectiveness of a buffer depends on two

factors• (1) the relative amounts of acid and base• (2) the absolute concentrations of acid and base

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Effect of Relative Amounts ofAcid and Conjugate Base

• A buffer will be most effective when the[base]:[acid] = 1– equal concentrations of acid and base

• Effective when 0.1 < [base]:[acid] < 10• A buffer will be most effective when the

[acid] and the [base] are large

Buffering Range• A buffer will be effective when

0.1 < [base]:[acid] < 10• Substituting into the Henderson-Hasselbalch

we can calculate the maximum and minimumpH at which the buffer will be effectiveLowest pH

( )

1ppH

10.0logppH

!=

+=

a

a

K

K

Highest pH( )

1ppH

10logppH

+=

+=

a

a

K

K

therefore, the effective pH range of a buffer is pKa ± 1when choosing an acid to make a buffer, chooseone whose is pKa is closest to the pH of the buffer

Example• Which of the following acids would be the best choice to

combine with its sodium salt to make a buffer with pH 4.25?

Chlorous Acid, HClO2 pKa = 1.95Nitrous Acid, HNO2 pKa = 3.34Formic Acid, HCHO2 pKa = 3.74Hypochlorous Acid, HClO pKa = 7.54

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Example• What ratio of NaCHO2 : HCHO2 would be required to make a

buffer with pH 4.25?

Buffering Capacity• Buffering capacity is the amount of acid or

base that can be added to a buffer withoutdestroying its effectiveness

• The buffering capacity increases withincreasing absolute concentration of thebuffer components

• As the [base]:[acid] ratio approaches 1, theability of the buffer to neutralize both addedacid and base improves– Buffers that need to work mainly with added acid

generally have [base] > [acid]– Buffers that need to work mainly with added base

generally have [acid] > [base]

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

unknown concentration (titrant) is slowlyadded to a solution of known concentrationfrom 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

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Titration

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 solutionin the burette is in excess, so the pH approaches itspH

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Titration Curve: Strong BaseAdded to Strong Acid

Example

• Show the titration of 25 mL of 0.100 M HClwith 0.100 M NaOH

Titrating Weak Acid with a StrongBase: Overview

• 1. Before any titrant (pH of weak acid)• 2. After a small amount of titrant (~5mL)• 3. At Half titration point (pH = pKa)• 4. At the equivalence point

– (mol acid = mol base)• 5. A drop beyond the endpoint (~0.5mL)

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Titrating Weak Acid with a StrongBase

• 1. The initial pH is that of the weak acidsolution– calculate like a weak acid equilibrium problem

• 2. Before the equivalence point, the solutionbecomes a buffer– calculate mol HAinit and mol A−init using reaction

stoichiometry– calculate pH with Henderson-Hasselbalch using

mol HAinit and mol A−init

• 3. Half-neutralization pH = pKa

Titrating Weak Acid with a StrongBase

• 4. At the equivalence point, the mole HA = molBase, so the resulting solution has only theconjugate base anion in it before equilibrium isestablished– 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

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

Example• A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH.

Calculate all 5 points.

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Titration Curve of a WeakBase with a Strong Acid

Titration of a Polyprotic Acid

titration of 25.0 mL of0.100 M H2SO3 with0.100 M NaOH

Monitoring pH During aTitration

• The general method for monitoring the pH during thecourse of a titration is to measure the conductivity ofthe solution due to the [H3O+]– using a probe that specifically measures just H3O+

• The endpoint of the titration is reached at theequivalence point in the titration – at the inflectionpoint of the titration curve

• If you just need to know the amount of titrant addedto reach the endpoint, we often monitor the titrationwith an indicator

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Monitoring pH During aTitration

Phenolphthalein

Methyl Red

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)2N N N N

H

NaOOC

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)2N N N N

NaOOC

H3O+ OH-

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Monitoring a Titration with anIndicator

• For most titrations, the titration curve shows avery large change in pH for very smalladditions of base near the equivalence point

• An indicator can therefore be used todetermine the endpoint of the titration if itchanges color within the same range as therapid change in pH– pKa of HInd ≈ pH at equivalence point

Indicators

Solubility Equilibria

• All ionic compounds dissolve in water tosome degree– however, many compounds have such low

solubility in water that we classify them asinsoluble

• The concepts of equilibrium can be applied tosalts dissolving, and then one can use theequilibrium constant for the process tomeasure relative solubilities in water

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Solubility Product• The equilibrium constant for the dissociation of a

solid salt into its aqueous ions is called the solubilityproduct, Ksp

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

• The solubility product would beKsp = [Mm+]n[Xn−]m

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

• Its equilibrium constant isKsp = [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 molesof 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

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Example• Calculate the molar solubility of PbCl2 in pure water at 25°C

Example• Determine the Ksp of PbBr2 if its molar solubility in water at 25°C

is 1.05 x 10-2 M

Ksp and Relative Solubility

• Molar solubility is related to Ksp

• But you cannot always comparesolubilities of compounds by comparingtheir Ksps

• In order to compare Ksps, thecompounds must have the samedissociation stoichiometry

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The Effect of Common Ion onSolubility

• Addition of a soluble salt that contains one ofthe ions of the “insoluble” salt, decreases thesolubility of the “insoluble” salt

• For example, addition of NaCl to the solubilityequilibrium of solid PbCl2 decreases thesolubility of PbCl2

PbCl2(s) ⇔ Pb2+(aq) + 2 Cl−(aq)

addition of Cl− shifts the equilibrium to the left

Example

• Calculate the molar solubility of CaF2 in 0.100 M NaF at 25°C

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

pH, the lower the solubility of the ionichydroxide– 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, thehigher the solubility

M2(CO3)n(s) ⇔ 2 Mn+(aq) + nCO32−(aq)

H3O+(aq) + CO32− (aq) ⇔ HCO3

− (aq) + H2O(l)

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Precipitation• Precipitation will occur when the

concentrations of the ions exceed thesolubility of the ionic compound

• If we compare the reaction quotient, Q, for thecurrent solution concentrations to the value ofKsp, we can determine if precipitation willoccur

• 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 notprecipitate unless disturbed – these are calledsupersaturated solutions

precipitation occursif Q > Ksp

a supersaturated solution willprecipitate if a seed crystal is added

Selective Precipitation

• A solution containing several different cationscan often be separated by addition of areagent that will form an insoluble salt withone of the ions, but not the others

• A successful reagent can precipitate withmore than one of the cations, as long as theirKsp values are significantly different

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Example• What is the minimum [OH−] necessary to just begin to precipitate Mg2+

(with [0.059]) from seawater?

Qualitative Analysis• An analytical scheme that utilizes selective

precipitation to identify the ions present in asolution is called a qualitative analysisscheme– wet chemistry

• A sample containing several ions is subjectedto the addition of several precipitating agents

• Addition of each reagent causes one of theions present to precipitate out

Qualitative Analysis

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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

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

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Group 3• Group three cations are Fe2+, Co2+, Zn2+,

Mn2+, Ni2+ precipitated as sulfides; as well asCr3+, Fe3+, and Al3+ precipitated ashydroxides

• 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 withPO4

3− that are insoluble in water at highpH

• Precipitated by the addition of(NH4)2HPO4

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

+

• All these cations form compounds thatare soluble in water – they do notprecipitate

• Identified by the color of their flame

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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 astronger bond than the current ligand, it willreplace the current ligand

Ag(H2O)2+

(aq) + 2 NH3(aq) ⇔ Ag(NH3)2+

(aq) + 2 H2O(l)

– Generally H2O is not included, since its complexion is always present in aqueous solution

Ag+(aq) + 2 NH3(aq) ⇔ Ag(NH3)2

+(aq)

Complex Ion Formation

• The reaction between an ion andligands to form a complex ion is called acomplex ion formation reaction

Ag+(aq) + 2 NH3(aq) ⇔ Ag(NH3)2

+(aq)

• The equilibrium constant for theformation reaction is called theformation constant, Kf

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Example• 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?

The Effect of Complex IonFormation on Solubility

• The solubility of an ionic compound thatcontains a metal cation that forms a complexion increases in the presence of aqueousligands

AgCl(s) ⇔ Ag+(aq) + Cl−(aq) Ksp = 1.77 x 10-10

Ag+(aq) + 2 NH3(aq) ⇔ Ag(NH3)2

+(aq) Kf = 1.7 x 107

• Adding NH3 to a solution in equilibrium withAgCl(s) increases the solubility of Ag+

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