119 ionic equilibria 離子平衡 : part ii buffers 緩衝液 and titration curves 滴定曲線...

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1919Ionic EquilibriaIonic Equilibria 離子平衡離子平衡 : :

Part IIPart IIBuffers Buffers 緩衝液緩衝液 and and

Titration CurvesTitration Curves 滴定曲線滴定曲線

花青素

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Chapter GoalsChapter Goals1.The Common Ion Effect and Buffer

Solutions ( 共同離子效應及緩衝溶液 )2.Buffering Action ( 緩衝作用 )3.Preparation of Buffer Solutions ( 緩衝溶液的

製備 )4.Acid-Base Indicators ( 酸鹼指示劑 )Titration Curves ( 滴定曲線 )5.Strong Acid/Strong Base Titration Curves6.Weak Acid/Strong Base Titration Curves7.Weak Acid/Weak Base Titration Curves8.Summary of Acid-Base Calculations

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The Common Ion Effect The Common Ion Effect and Buffer Solutionsand Buffer Solutions

• Common ion effect 共同離子效應– When a solution of a weak electrolyte is

altered by adding one of its ions from another source, the ionization of the weak electrolyte is suspressed.(當弱電解質溶液中加入具有相同離子的強電解質時，弱電解質的電離平衡會移動，使弱電解質的電離度下降，這種現象叫共同離子效應 )

• If a solution is made in which the same ion is produced by two different compounds the common ion effect is exhibited.

• Buffer solutions are solutions that resist changes in pH when acids or bases are added to them.– Buffering is due to the common ion effect.

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•Buffer solution 緩衝溶液– Resists changes in pH when strong acids or

strong bases are added.(當加入強酸或強鹼時 ,pH的變化不大 )

– contains a conjugate acid-base pair in reasonable concentrations. It can react with added base or acid ( 具共軛酸鹼對 )

– Buffer solution contain ( 緩衝溶液含有 :)• A weak acid and a soluble ionic salt of the

weak acid ( 弱酸和具此弱酸根的可溶性鹽類 ) – CH3COOH plus NaCH3COO

• A weak base and a soluble ionic salt of the weak base ( 弱鹼和具此弱鹼根的可溶性鹽類 ) – NH3 plus NH4Cl

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1. Solutions made of weak acids plus a soluble ionic salt of the weak acid– Solution that contain a weak acid plus a salt of weak

acid are always less acidic than solutions that contain the same concentration of weak acid alone

– One example of this type of buffer system is:• The weak acid - acetic acid CH3COOH• The soluble ionic salt - sodium acetate NaCH3COO

CH3COOH H+ + CH3COO- 部分解離Na+CH3COO- Na+ + CH3COO- 完全解離 100%

[CH3COO-] 增加反應向左減少 [H+] pH 增加

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Example 19-1: Calculate the concentration of H+and the pH of a solution that is 0.15 M in acetic acid and 0.15 M in sodium acetate.– This is another equilibrium problem with a starting

concentration for both the acid and anion.

CH3COOH H+ + CH3COO-

Na+CH3COO- Na+ + CH3COO- 100%(0.15-x) Mx M x M

0.15 M 0.15 M 0.15 M

Ka=[H+] [CH3COO-]

[CH3COOH]=1.8x10-5=

(x)(0.15+x)(0.15-x)

(0.15+x) 0.15 and (0.15-x) 0.15

= 1.8x10-5(x)(0.15)

(0.15)x =1.8x10-5= [H+]pH=4.74

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• Compare the acidity of a pure acetic acid solution and the buffer described in Example 19-1.

Solution [H+] pH

0.15 M CH3COOH 1.6 x 10-3 2.80

0.15 M CH3COOH & 0.15 M NaCH3COO

buffer1.8 x 10-5 4.74

[H+] is 89 times greater in pure acetic acid than in buffer solution.

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• The general expression for the ionization of a weak monoprotic acid is:

• The generalized ionization constant expression for a weak acid is:

HA H+ + A-

Ka=[H+] [A-]

[HA]

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• If we solve the expression for [H+], this relationship results:

• By making the assumption that the concentrations of the weak acid and the salt are reasonable, the expression reduces to:

[H+] = Ka x [A-][HA] acid

salt

[H+] = Ka x[salt][acid]

• The relationship developed in the previous slide is valid for buffers containing a weak monoprotic acid and a soluble, ionic salt.

• If the salt’s cation is not univalent the relationship changes to:

[H+] = Ka xn[salt][acid]

Where n=charge on cation

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• Simple rearrangement of this equation and application of algebra yields the Henderson-Hasselbach equation

The Henderson-Hasselbach equation is one method to calculate the pH of a buffer given the concentrations of

the salt and acid.

log [H+] =log Ka + log[salt][acid]

Multiply by -1

-log [H+] =-log Ka + log[acid][salt]

pH= pKa + log[acid][salt]

Example 19-1-1: Use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution in Example 19-1


pH= pKa + log[acid][salt]

pH= 4.74 + log0.15M0.15M

Example 19-1: Calculate the concentration of H+and the pH of a solution that is 0.15 M in acetic acid and 0.15 M in sodium acetate.

pKa =-logKa =-log(1.8x10-5) = 4.74

pH= 4.74 + log1 = 4.74 + 0 = 4.74

[H+] = Ka x[salt][acid]

[H+] = 1.8x10-5 x(0.15)(0.15)

[H+] = 1.8x10-5

pH=4.74

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Weak Bases plus Salts of Weak Bases plus Salts of

Weak BasesWeak Bases2.Buffers that contain a weak base plus

the salt of a weak base• One example of this buffer system is

ammonia plus ammonium nitrate.NH3 + H2O NH4

+ + OH-

NH4NO3 NH4+ + NO3

- 100%

Kb=[NH4

+] [OH-][NH3]

=1.8x10-5

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Weak BasesWeak BasesExample 19-2: Calculate the concentration of OH-

and the pH of the solution that is 0.15 M in aqueous ammonia, NH3, and 0.30 M in ammonium nitrate, NH4NO3.NH3 + H2O NH4

+ + OH-

NH4NO3 NH4+ + NO3

- 100%

Kb=[NH4

+] [OH-][NH3]

(0.15-x) M x M x M

0.30 M 0.30 M 0.30 M

=1.8x10-5=(x)(0.30+x)

(0.15-x)

(0.30+x) 0.30 and (0.15-x) 0.15

= 1.8x10-5(x)(0.30)(0.15)

x =9.0x10-6= [OH-]pOH=5.05, pH=8.95

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Weak BasesWeak Bases• A comparison of the aqueous ammonia

concentration to that of the buffer described above shows the buffering effect.

Solution [OH-] pH

0.15 M NH31.6 x 10-3

M11.20

0.15 M NH3 &0.15 M NH4NO3

buffer

9.0 x 10-6 M 8.95The [OH-] in aqueous ammonia is 180 times greater

than in the buffer.

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Weak BasesWeak Bases• We can derive a general relationship for

buffer solutions that contain a weak base plus a salt of a weak base similar to the acid buffer relationship.– The general ionization equation for weak

bases is:B + H2O BH+ + OH-

Where B represents a weak base • The general form of the ionization

expression is:

• Solve for the [OH-]

Kb=[BH+] [OH-]

[B]

[OH-] = Kb x[BH+][B] base

salt

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Weak BasesWeak Bases• For salts that have univalent ions:

• For salts that have divalent or trivalent ions:

[OH-] = Kb x[salt][base]

[OH-] = Kb xn[salt][base]

Where n= charge on anion

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Weak BasesWeak Bases• Simple rearrangement of this equation

and application of algebra yields theHenderson-Hasselbach equation

log [OH-] =log Kb + log[salt][base]

Multiply by -1

-log [OH-] =-log Kb + log[base][salt]

pOH= pKb + log[base][salt]

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Example 19-3: If 0.020 mole of gaseous HCl is added to 1.00 liter of a buffer solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the HCl.

1. Calculate the pH of the original buffer solution. [OH-] = Kb x[NH4Cl][NH3] [OH-] = 1.8x10-5 x

0.20M0.10M

[OH-] = 9.0x10-6 pOH=5.05 pH=8.952. Next, calculate the concentration of all species

after the addition of the gaseous HCl.–The HCl will react with some of the ammonia and change the concentrations of the species.–This is another limiting reactant problem.

HCl + NH3 → NH4ClInitial 0.02 mol0.1 mol 0.2 molchange-0.02 mol-0.02 mol+0.02 mol

After rxn 0 mol 0.08 mol0.22 mol

MNH3= 1.0L0.08mol

=0.08M

MNH4Cl= =0.22M

[OH-] = Kb x[NH4Cl][NH3] = 1.8x10-5 x0.22M

0.08M = 6.5x10-6

pOH=5.19 pH=8.81pH=pHnew-pHoriginal

=8.81-8.95=-0.14

1.0L0.22mol

?mol NH3=0.1M x 1L=0.1mol?mol NH4Cl=0.2M x 1L=0.2mol

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Buffering ActionBuffering ActionExample 19-4: If 0.020 mole of NaOH is added to

1.00 liter of solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the solid NaOH.•pH of the original buffer solution is 8.95, from above.

1. First, calculate the concentration of all species after the addition of NaoH.– NaOH will react with some of the ammonium

chloride.– The limiting reactant is the NaOH.NH4Cl + NaOH → NH3 + H2O + NaCl

Initial 0.2 mol 0.02 mol0.1 molchange-0.02 mol-0.02 mol+0.02 mol

After rxn 0.18 mol0 mol 0.12 mol

MNH3= =0.12M

MNH4Cl= 1.0L0.18mol=0.18M

[OH-] = Kb x[NH4Cl][NH3] = 1.8x10-5 x

0.18M0.12M = 1.2x10-5

pOH=4.92 pH=9.08 pH=pHnew-pHoriginal

=9.08-8.95=0.13

1.0L0.12mol

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Buffering ActionBuffering Action• This table is a summary of examples 19-3 and 19-4.

• Notice that the pH changes only slightly in each case.

Original Solution Original pH

Acid or base

addedNew pH pH

1.00 L of solution

containing 0.100 M NH3 and 0.200 M NH4Cl

8.95

0.020 mol

NaOH9.08 +0.1

3

0.020 mol HCl 8.81 -0.14

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Preparation of Buffer SolutionsPreparation of Buffer SolutionsExample 19-5: Calculate the concentration of H+

and the pH of the solution prepared by mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium hydroxide solutions.

Determine the amounts of acetic acid and sodium hydroxide prior to the acid-base reaction.

?mmol CH3COOH =200ml x 0.15M = 30.0 m mol

?mmol NaOH =100ml x 0.1M = 10.0 m mol

NaOH + CH3COOH → NaCH3COO + H2O Initial 10.0 mmol30.0 mmolchange-10.0 mmol-10.0 mmol+10.0 mmol

After rxn 0.0 mmol 20.0 mmol 10.0 mmolAfter the two solutions are mixed, the total volume of

the solution is 300 mL (100 mL of NaOH + 200 mL of acetic acid).– The concentrations of the acid and base are:

MCH3COOH= =0.0667M MNaCH3COO= =0.0333M300mL20 mmol

300mL10 mmol

Ka=[H+] [CH3COO-]

[CH3COOH] =1.8x10-5=[H+] [0.0333]

[0.0667]

[H+]=3.6x10-5M pH =4.44

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For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH.

Example 19-6: Calculate the number of moles of solid ammonium chloride, NH4Cl, that must be used to prepare 1.00 L of a buffer solution that is 0.10 M in aqueous ammonia, and that has a pH of 9.15.

Because pH = 9.15Then pOH = 14.00 - 9.15 = 4.85[OH-] = 10-4.85

= 1.4x10-5MNH3 + H2O NH4+ + OH-

NH4Cl → NH4+ + Cl-

(0.1-1.4x10-5) M(1.4x10-5) M(1.4x10-5) M

x M x Mx M

Kb=[NH4

+] [OH-][NH3]

= 1.8x10-5

Kb=(1.4x10-5+x)(1.4x10-5)

(0.1-1.4x10-5)=(x)(1.4x10-5)

(0.1)= 1.8x10-5

x= 0.13 M = [NH4Cl]original

0.13Mx1L =0.13 mol

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Acid-Base IndicatorsAcid-Base Indicators

• The point in a titration at which chemically equivalent amounts of acid and base have reacted is called the equivalence point 當量點 .

• The point in a titration at which a chemical indicator changes color is called the end point 終點 .

Three common indicators in solutions that cover the pH range 3 to 11.

a. Methyl red 甲基紅•Red at pH4 and below•Yellow at pH 7 and above•Red → orange → yellow

b. Bromthymol blue 溴瑞香草酚藍•yellow at pH6 and below•blue at pH 8 and above•yellow → green → blue

c. Phenolphthalein 酚酞 (most common use)•Colorless below pH8•Bright pink above pH10

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Acid-Base IndicatorsAcid-Base Indicators• Many acid-base indicators are weak organic

acid, HIn, where “In” represents various complex organic gups

• A symbolic representation of the indicator’s color change at the end point is:

HIn H+ + In-

color1 color2

• The equilibrium constant expression for an indicator would be expressed as:

Ka=[H+] [In-]

[HIn]

HIn represents nonionized acid molecules

In- represents the anion (conjugate base) of HIn

[In-][HIn]

=Ka

[H+]

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Acid-Base IndicatorsAcid-Base IndicatorsColor change ranges of some acid-base indicators

Indicator

Color in acidic range pH range

Color in basic range

Methyl violet 甲基紫

Yellow 0 - 2 Purple

Methyl orange Pink 3.1 – 4.4 Yellow

Litmus 石蕊 Red 4.7 – 8.2 Blue

Phenolphthalein

Colorless 8.3 – 10.0

Red

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Titration CurvesTitration CurvesStrong Acid/Strong Base Titration Curves

– These graphs are a plot of pH vs. volume of acid or base added in a titration.

– As an example, consider the titration of 100.0 mL of 0.100 M perchloric acid with 0.100 M potassium hydroxide.• In this case, we plot pH of the mixture vs. mL of KOH added.•Note that the reaction is a 1:1 mole ratio.

HClO4 + KOH → KClO4 + H2O

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Strong Acid/Strong Base Strong Acid/Strong Base Titration CurvesTitration Curves

• Before any KOH is added the pH of the HClO4 solution is .– Remember perchloric acid is a strong acid

that ionizes essentially 100%.

HClO4 H+ + ClO4- 100%

0.10M0.10M 0.10M

[H+] =0.10M

pH =-log(0.10)=1.0

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• After a total of 20.0 mL 0.100 M KOH has been added the pH of the reaction mixture is ___?

HClO4 + KOH → KClO4 + H2OStart 10.0 mmol 2.0 mmolchange -2.0 mmol-2.0 mmol+2.0 mmol

After rxn 8.0 mmol 0.0 mmol 2.0 mmol

? mmol KOH = 20ml x (0.1M) =2.0 mmol? mmol HClO4 = 100ml x (0.1M) =10.0 mmol

MHClO4= =0.067M120mL

8.0 mmol[H+]=0.067MpH = 1.17

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• After a total of 50.0 mL of 0.100 M KOH has been added the pH of the reaction mixture is ___?



MHClO4= =0.033M150mL

5.0 mmol[H+]=0.033MpH = 1.48

? mmol KOH = 50ml x (0.1M) =5.0 mmol

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• After a total of 90.0 mL of 0.100 M KOH has been added the pH of the reaction mixture is ____?



MHClO4= =0.0053M190mL

1.0 mmol[H+]=0.0053MpH = 2.28


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• After a total of 100.0 mL of 0.100 M KOH has been added the pH of the reaction mixture is ___?

HClO4 + KOH → KClO4 + H2OStart 10.0 mmol 10.0 mmolchange-10.0 mmol-10.0 mmol+10.0 mmol

After rxn 0.0 mmol 0.0 mmol 10.0 mmolNo acid or baseNeutralpH=7.0


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Strong acid-strong base

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•We have calculated only a few points on the titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.

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Weak Acid/Strong Base Weak Acid/Strong Base Titration CurvesTitration Curves

• As an example, consider the titration of 100.0 mL of 0.100 M acetic acid, CH3COOH, (a weak acid) with 0.100 M KOH (a strong base).– The acid and base react in a 1:1 mole ratio.

CH3COOH + KOH → K+CH3COO- + H2O1mol 1mol 1mol

1mmol 1mmol 1mmol• Before the equivalence point is reached,

both CH3COOH and KCH3COO are present in solution forming a buffer.– The KOH reacts with CH3COOH to form

KCH3COO.– A weak acid plus the salt of a weak acid

form a buffer.• Hypothesize how the buffer production will

effect the titration curve.

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• Before the equivalence point is reached, both CH3COOH and KCH3COO are present in solution forming a buffer.– The KOH reacts with CH3COOH to form KCH3COO.

• A weak acid plus the salt of a weak acid form a buffer.• Hypothesize how the buffer production will effect the titration curve.

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1. Determine the pH of the acetic acid solution before the titration is begun.

CH3COOH CH3COO- +H+

(0.1-x) M x M x MKa=

[H+] [CH3COO-][CH3COOH]

= 1.8x10-5=(x)(x)(0.1-x)

x2= 1.8x10-6x= 1.3x10-3=[H+] pH= 2.89• After a total of 20.0 mL of KOH solution has

been added, the pH is:

KOH + CH3COOH → K+CH3COO- + H2OStart 2.0 mmol 10.0 mmolchange -2.0 mmol -2.0 mmol +2.0 mmol


MCH3COOH= =0.067M120mL8.0 mmol

? mmol KOH = 20ml x (0.1M) =2.0 mmol? mmol CH3COOH = 100ml x (0.1M) =10.0 mmol

MCH3COO-= =0.017M120mL2.0 mmol

= (1.8x10-5) x [H+] = Ka x[CH3COOH][CH3COO-]

0.0670.017= 7.1x10-5

pH= 4.15

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• At the equivalence point, the solution is 0.500 M in KCH3COO, the salt of a strong base and a weak acid which hydrolyzes to give a basic solution.– This is a solvolysis process as

discussed in Chapter 18.– Both processes make the solution

basic.• The solution cannot have a pH=7.00 at

equivalence point.• Let us calculate the pH at the

equivalence point.

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• Set up the equilibrium reaction:KOH + CH3COOH → K+CH3COO- + H2O

Start 10.0 mmol10.0 mmolchange -10.0 mmol-10.0 mmol+10.0 mmol


MKH3COOH= =0.05M200mL10.0 mmol

0.05M CH3COO-

CH3COO- +H2O CH3COOH +OH-

(0.05-x) M x M x M

Kb=[OH-] [CH3COOH]

[CH3COO-]=

(x)(x)(0.05-x)

=5.6x10-10

x2=2.8x10-11 x=5.27x10-6= [OH-]pOH=5.28pH=8.72 the equivalence point

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After the equivalence point is reached, the pH is determined by the excess KOH just as in the strong acid/strong base example.

KOH + CH3COOH → K+CH3COO- + H2OStart 11.0 mmol10.0 mmolchange -10.0 mmol-10.0 mmol+10.0 mmol


MKOH= =4.8x10-3 M210mL1.0 mmol

[OH-]=4.8x10-3 M

pOH=2.32pH=11.68

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Weak acid-strong base

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Strong Acid/Weak BaseStrong Acid/Weak BaseTitration CurvesTitration Curves

• Titration curves for Strong Acid/Weak Base

Titration Curves look similar to Strong

Base/Weak Acid Titration Curves but they are

inverted.

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Weak Acid/Weak BaseWeak Acid/Weak BaseTitration CurvesTitration Curves

•Weak Acid/Weak Base Titration curves have very short vertical sections.•The solution is buffered both before and after the equivalence point.•Visual indicators cannot be used.

Table 19-7a, p. 763

Table 19-7b, p. 763

119 ionic equilibria 離子平衡 : part ii buffers 緩衝液 and titration curves 滴定曲線...

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