applications of aqueous equilibria buffers, acid-base titrations and precipitation reactions

Applications of Applications of Aqueous Aqueous

EquilibriaEquilibriaBuffers, Acid-Base Titrations Buffers, Acid-Base Titrations and Precipitation Reactionsand Precipitation Reactions

BuffersBuffers

BuffersBuffers are solutions of a weak acid are solutions of a weak acid and its conjugate base. and its conjugate base.

Buffers resist changes in pH when Buffers resist changes in pH when small amounts of strong acid or base small amounts of strong acid or base are added. This is because there is both are added. This is because there is both an acid and its conjugate base present an acid and its conjugate base present initially. Buffers typically have the acid initially. Buffers typically have the acid and its conjugate base (or salt) present and its conjugate base (or salt) present in roughly equal concentrations.in roughly equal concentrations.

BuffersBuffers

Consider a buffer of weak acid HA Consider a buffer of weak acid HA and NaA in equimolar concentrations. and NaA in equimolar concentrations. The predominant reaction is:The predominant reaction is:

HA(HA(aqaq) + H) + H22O(O(ll) ↔ H) ↔ H33OO++((aqaq) + A) + A--((aqaq))

If a small amount of strong acid is If a small amount of strong acid is added, there is enough Aadded, there is enough A-- available to available to become protonated. The equilibrium become protonated. The equilibrium shifts to the left, and the [Hshifts to the left, and the [H33OO++] and pH ] and pH do not change much.do not change much.

BuffersBuffers


Likewise, if a small amount of strong Likewise, if a small amount of strong base is added, some of the hydronium will base is added, some of the hydronium will react with it.react with it.

HH33OO++((aqaq) + OH) + OH--((aqaq) ) 2 H 2 H22O(O(ll))

As hydronium ion reacts, more HA will As hydronium ion reacts, more HA will dissociate so as to replenish the [Hdissociate so as to replenish the [H33OO++]. ]. As a result, the pH remains fairly constant.As a result, the pH remains fairly constant.

BuffersBuffersBecause buffers contain both a weak acid and its conjugate base, they can react with strong acids or bases and maintain their pH.

BuffersBuffers

Problem: BuffersProblem: Buffers

Determine the pH of a solution that Determine the pH of a solution that contains 0.50M HClO and 0.60M contains 0.50M HClO and 0.60M NaClO. (KNaClO. (Kaa for HClO = 3.5 x 10 for HClO = 3.5 x 10-8-8))

- Write the major reactions:- Write the major reactions:

NaClO(NaClO(aqaq) ) Na Na++((aqaq) + ClO) + ClO--((aqaq))

The NaClO serves as a source of The NaClO serves as a source of ClOClO-- ions. The sodium ion is a ions. The sodium ion is a spectator.spectator.

Problem: BuffersProblem: Buffers Determine the pH of a solution that contains Determine the pH of a solution that contains

0.50M HClO and 0.60M NaClO. 0.50M HClO and 0.60M NaClO.

Even though both HClO and ClOEven though both HClO and ClO -- are are present initially, it’s important to realize present initially, it’s important to realize that they are that they are in equilibriumin equilibrium with each other, with each other, and go on and go on opposite sides of the equationopposite sides of the equation..

HClO(HClO(aqaq) + H) + H22O(O(ll) ↔ H) ↔ H33OO++((aqaq) + ClO) + ClO--((aqaq))



HClO(HClO(aqaq) + H) + H22O(O(ll) ↔ H) ↔ H33OO++((aqaq) + ) +

ClOClO--((aqaq))

KKaa = [H = [H33OO++][ClO][ClO--] = 3.5 x 10] = 3.5 x 10-8-8

[HClO][HClO]



HClO(HClO(aqaq) + H) + H22O(O(ll) ↔ H) ↔ H33OO++((aqaq) + ) + ClOClO--((aqaq))HClOHClO HH33OO++ ClOClO--

initiainitiall

0.500.50 00 0.600.60

chanchangege -x-x +x+x +x+x

equilequil..

.50-x.50-x xx .60+x.60+x

The Henderson-Hasselbalch The Henderson-Hasselbalch EquationEquation

For buffers, regardless of the value of For buffers, regardless of the value of KKaa, you can , you can alwaysalways assume that x will be assume that x will be small compared to the concentrations of small compared to the concentrations of the acid or its conjugate base. the acid or its conjugate base.

This is because the reaction will This is because the reaction will proceed only slightly to the right due to proceed only slightly to the right due to the presence of the conjugate base the presence of the conjugate base initially.initially.



As a result, the KAs a result, the Kaa expression expression can be written in logarithmic form.can be written in logarithmic form.

KKaa = [H = [H33OO++][A][A--]] [HA][HA]

[H[H33OO++] = K] = Kaa [HA] [HA] [A[A--]]

pH = pKpH = pKaa + log [A + log [A--]] [HA][HA]


In more general terms, the In more general terms, the equation is written as:equation is written as:

pH = pKpH = pKaa + log + log

This equation is This equation is for buffers onlyfor buffers only, , and is a quick way to calculate pH.and is a quick way to calculate pH.

[base][acid]

Buffers Maintain pHBuffers Maintain pH

If 5.00 mL of 0.010M HNOIf 5.00 mL of 0.010M HNO33 is added is added to 50.0 mL of the HClO/NaClO buffer, the to 50.0 mL of the HClO/NaClO buffer, the pH pH will remain unchangedwill remain unchanged..

The nitric acid will protonate some of The nitric acid will protonate some of the ClOthe ClO-- to form additional HClO. to form additional HClO.

HNOHNO33((aqaq) + H) + H22O(O(ll) ) H H33OO++((aqaq) + NO) + NO33--((aqaq))

HH33OO++((aqaq) + ClO) + ClO--((aqaq)) HClO HClO ((aqaq) + H) + H22O(O(ll))

Problem: Adding Acid to Problem: Adding Acid to a Buffera Buffer

HNOHNO33((aqaq) + H) + H22O(O(ll) ) H H33OO++((aqaq) + NO) + NO33--

((aqaq))

HH33OO++((aqaq) + ClO) + ClO--((aqaq)) HClO HClO ((aqaq) + ) +

HH22O(O(ll))

Because the addition hydronium Because the addition hydronium

ion is neutralized by the ClOion is neutralized by the ClO --, the pH , the pH

remains constant.remains constant.

Compare Buffers to Compare Buffers to WaterWater

If 5.00 mLs of 0.0100M HNOIf 5.00 mLs of 0.0100M HNO33 is added to is added to 50.0 mL of water, the pH changes by 4 50.0 mL of water, the pH changes by 4 pH units.pH units.

[H[H33OO++] = (5.00 mLs) (0.0100M)/55.00mls] = (5.00 mLs) (0.0100M)/55.00mls

[H[H33OO++] = 9.09 x 10] = 9.09 x 10-4-4

pH = 3.04pH = 3.04

Preparation of Buffer Preparation of Buffer SolutionsSolutions

Scientists often need to make a Scientists often need to make a solution that is buffered to a specific solution that is buffered to a specific pH. They may need to mimic pH. They may need to mimic biological conditions, test the biological conditions, test the corrosiveness of metal parts at a corrosiveness of metal parts at a specific pH, etc.specific pH, etc.

Conjugate acid/base pairs have Conjugate acid/base pairs have a pH range at which they can a pH range at which they can effectively serve as buffers.effectively serve as buffers.

Choosing the Acid and Choosing the Acid and SaltSalt

For optimum buffering, choose an For optimum buffering, choose an acid with a Kacid with a Kaa value close to the desired value close to the desired [H[H33OO++] of the buffer, or with a pK] of the buffer, or with a pKa a value value near the desired pH.near the desired pH.

You can then use the Henderson-You can then use the Henderson-Hasselbalch equation (or Ka expression) Hasselbalch equation (or Ka expression) to determine the relative concentration of to determine the relative concentration of base to acid needed to prepare the buffer.base to acid needed to prepare the buffer.

Preparing a BufferPreparing a Buffer

Choose an appropriate acid and base Choose an appropriate acid and base to make a buffer with a pH of 6.50. to make a buffer with a pH of 6.50. Calculate the relative concentration Calculate the relative concentration of acid and base needed.of acid and base needed.

Acid-Base TitrationsAcid-Base Titrations

TitrationTitration is a laboratory is a laboratory technique in which the amount and technique in which the amount and concentration of one reactant is concentration of one reactant is known, and the concentration of the known, and the concentration of the other reactant is determined.other reactant is determined.

Titrations can be applied to redox Titrations can be applied to redox reactions, and precipitation reactions, reactions, and precipitation reactions, but they are most commonly used for but they are most commonly used for analyzing solutions of acids or bases.analyzing solutions of acids or bases.

Acid-Base TitrationsAcid-Base Titrations

Acid-base titrations are based Acid-base titrations are based on on neutralizationneutralization reactions, in which reactions, in which an acid is completely neutralized by an acid is completely neutralized by a base. The progress of the reaction a base. The progress of the reaction may be monitored using a pH meter, may be monitored using a pH meter, or pH indicators.or pH indicators.

TitrationsTitrations

In either method, base is added In either method, base is added to the acid (or vice versa) until the to the acid (or vice versa) until the equivalence pointequivalence point is reached. is reached.

The equivalence point has been The equivalence point has been reached when the moles of acid reached when the moles of acid exactly equals the moles of base. It exactly equals the moles of base. It is sometime called the is sometime called the stoichiometric pointstoichiometric point..

TitrationsTitrations

Not all titrations reach the Not all titrations reach the equivalence point at a pH of 7. The equivalence point at a pH of 7. The nature of the acid and base (weak or nature of the acid and base (weak or strong) will influence the pH at the strong) will influence the pH at the equivalence point. As a result, equivalence point. As a result, indicators much be carefully chosen indicators much be carefully chosen to change color at the desired pH.to change color at the desired pH.

Strong Acid & Strong Strong Acid & Strong BaseBase

When a strong acid is titrated When a strong acid is titrated with a strong base, a “neutral” salt with a strong base, a “neutral” salt and water result. Since the and water result. Since the conjugate base of a strong acid has conjugate base of a strong acid has no tendency to accept protons, and no tendency to accept protons, and the conjugate acid of a strong base the conjugate acid of a strong base has no tendency to donate protons, has no tendency to donate protons, the pH at the equivalence point will the pH at the equivalence point will be 7.be 7.


If the pH is If the pH is monitored during monitored during the titration using the titration using a pH meter, a a pH meter, a graphical graphical presentation of presentation of the pH versus the pH versus volume of titrant volume of titrant can be obtained.can be obtained.


The The graph has graph has certain certain features features characteristicharacteristic of a strong c of a strong acid-strong acid-strong base base titration.titration.


Note the Note the equivalence equivalence point of 7, point of 7, and the and the nearly nearly vertical rise vertical rise in pH from in pH from 4-10.4-10.


Because the rise Because the rise in pH is vertical in pH is vertical over such a over such a broad range, broad range, there are several there are several indicators that indicators that may be used with may be used with accurate results.accurate results.


Phenolphthalein, Phenolphthalein, which changes which changes color at a pH of color at a pH of around 9, and around 9, and methyl red, which methyl red, which changes color at a changes color at a pH of pH of approximately 5 approximately 5 will both give will both give highly accurate highly accurate results.results.


Phenolphthalein is Phenolphthalein is usually used usually used because it goes because it goes from colorless to from colorless to pink, and our eyes pink, and our eyes are better at are better at detecting the detecting the presence of color presence of color than a change in than a change in color.color.

Weak Acid & Strong BaseWeak Acid & Strong Base

When a weak acid is titrated When a weak acid is titrated with a strong base, the salt that is with a strong base, the salt that is produced will contain the conjugate produced will contain the conjugate base of the weak acid. As a result, base of the weak acid. As a result, the pH will be greater than 7 at the the pH will be greater than 7 at the equivalence point.equivalence point.


When a weak When a weak acid is titrated with acid is titrated with a strong base, the a strong base, the salt that is salt that is produced will produced will contain the contain the conjugate base of conjugate base of the weak acid. As the weak acid. As a result, the pH a result, the pH will be greater than will be greater than 7 at the 7 at the equivalence point.equivalence point.


The titration The titration curve for acetic curve for acetic acid with NaOH acid with NaOH shows an shows an equivalence point equivalence point at a pH of 9. The at a pH of 9. The lack of a steep lack of a steep vertical rise in pH vertical rise in pH means that means that selection of the selection of the proper indicator is proper indicator is crucial.crucial.


Another Another feature of the feature of the curve is the curve is the flattening out or flattening out or leveling off of pH leveling off of pH near the half-way near the half-way point. This is not point. This is not observed in strong observed in strong acid-strong base acid-strong base titrations.titrations.


The pH levels The pH levels off due to the off due to the formation of a formation of a buffer mid-way buffer mid-way through the through the titration. Half of titration. Half of the acid has been the acid has been de-protonated to de-protonated to form its conjugate form its conjugate base.base.

Titration of Benzoic AcidTitration of Benzoic Acid

25.0 mL of 0.10M benzoic acid (K25.0 mL of 0.10M benzoic acid (Kaa = = 6.4 x 106.4 x 10-5-5) is titrated with 0.10M ) is titrated with 0.10M NaOH. Calculate the pHNaOH. Calculate the pH

at the equivalence point, and choose at the equivalence point, and choose an appropriate indicator for the an appropriate indicator for the titration.titration.


25.0 mL of 0.10M benzoic acid (K25.0 mL of 0.10M benzoic acid (Kaa = 6.4 = 6.4 x 10x 10-5-5) is titrated with 0.10M NaOH. ) is titrated with 0.10M NaOH.

We will need to know how much We will need to know how much NaOH is needed for complete NaOH is needed for complete neutralization. In this case, since the neutralization. In this case, since the concentrations of acid and base are the concentrations of acid and base are the same, 25.0 mL of 0.10M acid will same, 25.0 mL of 0.10M acid will require an equal volume of 0.10M base.require an equal volume of 0.10M base.

Calculating Volume for Calculating Volume for NeutralizationNeutralization

In less obvious cases, the In less obvious cases, the following relationship can be used. following relationship can be used. We are assuming complete We are assuming complete neutralization of a monoprotic acid neutralization of a monoprotic acid by a monobasic base.by a monobasic base.

moles acid = moles basemoles acid = moles base

MMacidacidVVacidacid = M = MbasebaseVVbasebase


Calculate the pH at the equivalence Calculate the pH at the equivalence point.point.

At the equivalence point, enough At the equivalence point, enough NaOH has been added (25.0 mL) to NaOH has been added (25.0 mL) to completely neutralize the benzoic acid. completely neutralize the benzoic acid.

There are two reactions to consider. There are two reactions to consider.

1. The neutralization reaction.1. The neutralization reaction.

2. The subsequent reaction of the 2. The subsequent reaction of the benzoate ion with water.benzoate ion with water.


At the equivalence point, benzoate At the equivalence point, benzoate ion will react with water, since it is ion will react with water, since it is the conjugate base of a weak acid.the conjugate base of a weak acid.

Major reactions:Major reactions:

HA(HA(aqaq) + OH) + OH--(aq) (aq) H H22O(O(ll) + A) + A--((aqaq))

AA--((aqaq) + H) + H22O(O(aqaq) ↔ OH) ↔ OH--((aqaq) + HA() + HA(aqaq))


- At the equivalence point, the - At the equivalence point, the solution will be basic, due to the solution will be basic, due to the high concentration of benzoate ion. high concentration of benzoate ion.

Major ReactionMajor Reaction::AA--((aqaq) + H) + H22O(O(aqaq) ↔ OH) ↔ OH--((aqaq) + HA() + HA(aqaq))

The [AThe [A--] equals the number of moles ] equals the number of moles of HA initially present divided by the of HA initially present divided by the total volume of solution at the total volume of solution at the equivalence point. equivalence point.



The [AThe [A--] equals the number of moles ] equals the number of moles of HA initially present divided by the total of HA initially present divided by the total volume of solution at the equivalence volume of solution at the equivalence point. point.

[A[A--] = 25.0 mL(0.10M)/50.0 mL = 0.050M] = 25.0 mL(0.10M)/50.0 mL = 0.050M


Major ReactionMajor Reaction::AA--((aqaq)) + H + H22O(O(aqaq) ↔ ) ↔ OHOH--((aqaq) + ) + HAHA((aqaq))

init. 0.050init. 0.050 00 00chg. -xchg. -x +x +x +x +xequil .050-xequil .050-x xx xx

Since the reaction involves a Since the reaction involves a conjugate base reacting with water, conjugate base reacting with water, we need to write the Kwe need to write the Kbb expression, and expression, and calculate the value of Kcalculate the value of Kbb..



Since the reaction involves a conjugate Since the reaction involves a conjugate base reacting with water, base reacting with water, we need to write we need to write the Kthe Kbb expression, and calculate the value expression, and calculate the value of Kof Kbb..

KKbb = [OH = [OH--][HA] = K][HA] = Kww

[A[A--] ] KKaa

KKbb = 1.0x 10 = 1.0x 10-14-14/ 6.4 x 10/ 6.4 x 10-5 -5 = 1.6 x 10= 1.6 x 10-10-10


Major ReactionMajor Reaction::AA--((aqaq)) + H + H22O(O(aqaq) ↔ ) ↔ OHOH--((aqaq) + ) + HAHA((aqaq))init. 0.050init. 0.050 00 00chg. -xchg. -x +x +x +x +xequil .050-xequil .050-x xx xx

1.6 x 101.6 x 10-10-10 = (x)(x)/(.050-x) = (x)(x)/(.050-x)


The pH at the The pH at the equivalence point equivalence point will be 8.45. will be 8.45.

Selection of a Selection of a indicator must be indicator must be made carefully, as made carefully, as there isn’t a steep there isn’t a steep vertical rise in pH vertical rise in pH near the end point near the end point when titrating a when titrating a weak acid.weak acid.

Titration of Weak AcidsTitration of Weak Acids

As the acid As the acid being titrated being titrated gets weaker, the gets weaker, the titration curve titration curve flattens out, and flattens out, and the end point the end point becomes less becomes less obvious.obvious.

Titration of Weak AcidsTitration of Weak Acids

Very Very weak acids weak acids cannot be cannot be analyzed analyzed using using titration.titration.

Titration of a Weak BaseTitration of a Weak Base

NHNH33((aqaq) + HCl() + HCl(aqaq) ) NH NH44++((aqaq) + ) +

ClCl--((aqaq))

The titration of ammonia with a The titration of ammonia with a strong acid will produce a solution of strong acid will produce a solution of ammonium ion. Since ammonium ion ammonium ion. Since ammonium ion is the conjugate acid of a weak base, is the conjugate acid of a weak base, the pH at the equivalence point will the pH at the equivalence point will be less than 7.be less than 7.

Titration of a Weak BaseTitration of a Weak Base

Again, Again, selection of selection of the proper the proper indicator is indicator is crucial. crucial.

applications of aqueous equilibria buffers, acid-base titrations and precipitation reactions

Documents

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