© 2014 pearson education, inc. 1 chapter 16 aqueous ionic equilibrium

© 2014 Pearson Education, Inc. 1

Chapter 16

Aqueous Ionic Equilibrium


Buffer Solutions

• Resist changes in pH when an acid or base is added.

• Act by neutralizing acid or base that is added to the buffered solution.

• Contain either– Significant amounts of a weak acid and its conjugate base– Significant amounts of a weak base and its conjugate acid

– Blood has a mixture of H2CO3 and HCO3−


Making an Acidic Buffer Solution

It must contain significant amounts of both a weak acid and its conjugate base.


An Acidic Buffer Solution• If a strong base is added, it is neutralized by the weak

acid (HC2H3O2) in the buffer.

NaOH(aq) + HC2H3O2(aq) H2O(l) + NaC2H3O2(aq)

• If the amount of NaOH added is less than the amount of acetic acid present, the pH change is small.

• If a strong acid is added, it is neutralized by the conjugate base (NaC2H3O2) in the buffer.

HCl(aq) + NaC2H3O2 HC2H3O2(aq) + NaCl(aq)

• If the amount of HCl is less than the amount of NaC2H3O2

present, the pH change is small.


How Acid Buffers Work: Addition of Base HA(aq) + H2O(l) A⇌ −

(aq) + H3O+(aq)

• Buffers work by applying Le Châtelier’s principle to weak acid equilibrium.

• Buffer solutions contain significant amounts of the weak acid molecules, HA .

• These molecules react with added base to neutralize it.

HA(aq) + OH−(aq) → A−(aq) + H2O(l)

– You can also think of the H3O+ combining with the OH− to make H2O; the H3O+ is then replaced by the shifting equilibrium.


How Acid Buffers Work: Addition of AcidHA(aq) + H2O(l) A⇌ −

(aq) + H3O+(aq)

• The buffer solution also contains significant amounts of the conjugate base anion, A−.

• These ions combine with added acid to make more HA.

H+(aq) + A−(aq) → HA(aq)• After the equilibrium shifts, the concentration of

H3O+ is kept constant.


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

(aq)

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

• This causes the pH to be higher than the pH of the acid solution, lowering the H3O+ ion concentration.


Henderson–Hasselbalch Equation

• An equation derived from the Ka expression that allows us to calculate the pH of a buffer solution.

• The equation calculates the pH of a buffer from the pKa and initial concentrations of the weak acid and salt of the conjugate base, as long as the “x is small” approximation is valid.


Deriving the Henderson–Hasselbalch Equation

Consider the equilibrium expression for a generic acidic buffer:

Solving for H3O+ we have the following:



Taking the logarithm of both sides and expanding since log (AB) = log A + log B



HA(aq) + H2O(l) H⇌ 3O+(aq) + A–(aq)

Multiplying both sides by –1 and rearranging we get the following:

–log[H3O+] = –logKa + log([A–]/[HA) (see page 758)



HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)

Since pH = –log[H3O+] and pKa = logKa

pH = pKa + log([A–]/[HA) (see page 758)

We can then generalize as follows:


Do I Use the Full Equilibrium Analysis or the Henderson–Hasselbalch Equation?

• The Henderson–Hasselbalch equation is generally good enough when the “x is small” approximation is applicable.

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

a) The initial concentrations of acid and salt are not very dilute.

b) The Ka is fairly small.

• For most problems, this means that the initial acid and salt concentrations should be over 100 to 1000 times larger than the value of Ka.


How Much Does the pH of a Buffer Change When an Acid or Base Is Added?

• Calculating the new pH after adding acid or base requires breaking the problem into two parts:

1. A stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other. Added acid reacts with the A− to make more HA Added base reacts with the HA to make more A−

2. An equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]


Action of a Buffer

Figure 16.3 pg 762


Basic BuffersB:(aq) + H2O(l) H:B⇌ +

(aq) + OH−(aq)

• Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl−

H2O(l) + NH3 (aq) ⇌ NH4+

(aq) + OH−(aq)


• The Henderson–Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product.

• The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product.

B: + H2O ⇌ H:B+ + OH−

• To apply the Henderson–Hasselbalch equation, the chemical equation of the basic buffer must be looked at like an acid reaction.H:B+ + H2O B: + H⇌ 3O+

This does not affect the concentrations, just the way we are looking at the reaction.

Henderson–Hasselbalch Equation for Basic Buffers


Relationship between pKa and pKb

• Just as there is a relationship between the Ka of a weak acid and Kb of its conjugate base, there is also a relationship between the pKa of a weak acid and the pKb of its conjugate base.

Ka Kb = Kw = 1.0 x 10−14

−log(Ka Kb) = −log(Kw) = 14

−log(Ka) + −log(Kb) = 14

pKa + pKb = 14


• The Henderson–Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product.

• The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product.

B: + H2O ⇌ H:B+ + OH−

• We can rewrite the Henderson–Hasselbalch equation for the chemical equation of the basic buffer in terms of pOH.

Henderson–Hasselbalch Equation for Basic Buffers


Buffering Effectiveness

• A good buffer should be able to neutralize moderate amounts of added acid or base.

• However, 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, and (2) the absolute concentrations of acid and base.


Effectiveness of Buffers

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

• A buffer will be effective when

0.1 < [base]:[acid] < 10.

• A buffer will be most effective when the [acid] and the [base] are large.


Buffering Range

• We have said that a buffer will be effective when 0.1 < [base]:[acid] < 10.

• Substituting into the Henderson–Hasselbalch equation we can calculate the maximum and minimum pH at which the buffer will be effective.

Lowest pH Highest pH

Therefore, the effective pH range of a buffer is pKa ± 1.When choosing an acid to make a buffer, choose one whose pKa is closest to the pH of the buffer.


Buffering Capacity

A concentrated buffer can neutralize more added acid or base than a dilute buffer.

• Buffering capacity is the amount of acid or base that can be added to a buffer without causing a large change in pH.

• The buffering capacity increases with increasing absolute concentration of the buffer components.


Buffering Capacity

• As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid 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 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.


Titration


Titration Curve

• It is a plot of pH versus the 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: Strong Base Added to Strong Acid


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.


Titrating Weak Acid with a Strong Base

• The initial pH is that of the weak acid solution.– Calculate like a weak acid equilibrium problem

• For example, 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 HA init

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

• For example, 15.14• Beyond equivalence point, the OH is in excess.

– [OH−] = mol MOH xs/total liters– [H3O+][OH−] = 1 × 10−14


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

Figure 16.10 pg 780


Titration of 25.0 mL of a 0.1 M weak acid with a 0.1M strong base

0 5 10 15 20 25 30 35 40 45 50

2.0

4.0

6.0

8.0

10.0

12.0

14.0

pH

Vol Titrant (mL)

a

eqtit

pKpH

VV

2

1

][log

][][

0

HAKpH

HAKH

V

a

a

tit

][log

][][

AKpOH

AKOH

VV

b

b

eqtit

excess

eqtit

OHpOH

VV

][log

][

][log

""

0

HA

ApKapH

regionbuffer

VV equivtit

If strong acid with strong base:

00.7

][log

pH

VV

HpH

VV

eqtit

excess

eqtit


0 5 10 15 20 25 30 35 40 45 50

2.0

4.0

6.0

8.0

10.0

12.0

14.0

pH

Vol Titrant (mL)

a

eqtit

pKpH

VV

2

1

][log

][][

BHKpH

BHKH

VV

a

a

eqtit

][log

][][

0

BKpOH

BKOH

V

b

b

tit

excess

eqtit

HpH

VV

][log

][

][log

][

][log

""

0

._

B

BHpKpOH

BH

BpKpH

regionbuffer

VV

b

acidconja

equivtit

Titration of 25.0 mL of a 0.1 M weak base with a 0.1M strong acid

If strong base with strong acid:

00.7

][log

pH

VV

OHpOH

VV

eqtit

excess

eqtit


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 solution.

HInd(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 a mix of the colors

of Ind and HInd .– When Ind:HInd > 10, the color will be a mix of the

colors of Ind.– When Ind:HInd < 0.1, the color will be a mix of the

colors of Hind.


Phenolphthalein


Methyl Red


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

Table 16.1 pg 783


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 isMnXm(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)


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) Pb⇌ 2+(aq) + 2 Cl−(aq)

Addition of Cl− shifts the equilibrium to the left.


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) M⇌ n+(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 M⇌ n+(aq) + nCO32−(aq)

H3O+(aq) + CO32− (aq) HCO⇌ 3

− (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.

Image top right of page 789

Precipitation


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.


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


Figure 16.16 pg 118

A General Qualitative Analysis Scheme


Group 1

• Group one cations are Ag+, Pb2+, and Hg2

2+.

• 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 × 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.


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.


Group 5

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

• All these cations form compounds that are soluble in water—they do not precipitate.

• They are 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 bonds.

Ag+(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.– For example, Ag(H2O)2

+

• The attached ions or molecules are called ligands.– For example, 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.

Ag+(aq) + 2 NH3(aq) ⇌ Ag(NH3)2

+(aq)

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


Formation Constants


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 ×

10−10

Ag+(aq) + 2 NH3(aq) Ag(NH⇌ 3)2

+(aq) Kf = 1.7 × 107

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


Figure 16.18


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 solution.

Al(H2O)63+

(aq) + H2O(l) Al(H⇌ 2O)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(H⇌ 2O)4(OH)2+

(aq) + H2O (l)

Al(H2O)4(OH)2+

(aq) + OH−(aq) Al(H⇌ 2O)3(OH)3(s) + H2O

(l)


Solubility of an Amphoteric Hydroxide

© 2014 pearson education, inc. 1 chapter 16 aqueous ionic equilibrium

Documents

h aq

concentration of h

addition of acid ha

mixture of h

addition of base ha

added acid

o aq buffers work

o combining