Chapter FifteenPrentice-Hall ©2002Slide 1 of 31 920131 1 http:\\asadipour.kmu.ac.ir 48slides.

Chapter FifteenPrentice-Hall ©2002 Slide 2 of 31• http:\\asadipour.kmu.ac.ir 48slides • 2

•Neutralization•Reactions

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• Standard Solutions: strong acids or strong bases because they will react completely– Acids: (HCl), (HClO4), (H2SO4)– Bases: (NaOH), (KOH)

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• Thymol blue (thymolsulphonephthalein) is

a brownish-green or

reddish-brown crystaline

powder that is used as an pH indicator. It is insoluble in water but soluble in

alcohol and dilute alkali solutions. It transitions from red to yellow at pH 1.2–2.8 and

from yellow to blue from at pH 8.0–9.6.

• Bromophenol blue

(3',3",5',5"-tetrabromophenolsulfonphthalein) is an

acid-base indicator

whose useful range as an

indicator lies between pH

3.0 and 4.6. It changes from yellow at pH 3.0 to purple at pH 4.6; this

reaction is reversible.

• Chlorophenol red is

an indicator dye that changes

color from yellow to violet in the pH

range 4.8 to 6.7.

The lamda max is at 572 nm.

• A solution of phenol red is used as a pH indicator: its color exhibits

a gradual transition

from yellow to red over the pH range 6.6 to 8.0. Above

pH 8.1, phenol red turns a bright pink

(fuchsia) color. This observed color change

is because phenol red

loses protons (and changes color) as the pH increases.

• Bromothymol Blue (also

known as dibromothymolsulfonephthalein,

Bromthymol Blue, and BTB) is a chemical

indicator for weak acids and bases

The pKa for bromothym

ol blue is 7.10.

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Acid/Base IndicatorsMany substances display colors that depend on the pH of the solutions in which they are dissolved. An acid/base indicator is a weak organic acid or a weak organic base whose undissociated form differs in color from its conjugate form. e.g., the behavior of an acid-type indicator, HIn, is described by the equilibrium

HIn + H2O In- + H3O+

acid color base colorThe equilibrium for a base-type indicator, In, is

In + H2O InH+ + OH-

base color acid color• 920131


…continued…

The equilibrium-constant expression for the dissociation of an acid-type indicator takes the form

Rearranging leads to

The hydronium ion concentration determines the ratio of the acid to the conjugate base form of the indicator and thus determines the color developed by the solution.

K

H O In

H Ina

3

H O K

H In

Ina3

• + • -

• +• -

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…continued…

The color imparted to a solution by a typical indicator appears to the average observer to change rapidly only within the limited concentration ratio of approximately 10 to 0.1

and its base color when

The color appears to be intermediate for ratios between these two values. These ratios vary considerably from indicator to indicator.

H In

In

1 0

1

H In

In

1

1 0

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…continued…

For the full acid color,[H3O+] = 10Ka

and similarly for the full base color,[H3O+] = 0.1Ka

To obtain the indicator pH range, we take the negative logarithms of the two expression:

pH (acid color) = -log (10Ka) = pKa - 1pH (basic color) = -log (0.1Ka) = pKa + 1indicator pH range = pKa 1

H O K

H In

Ina3

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Variables: 1)temperature,2)ionic strength of medium 3)presence of organic solvents 4)presence of colloidal particles• 920131

Chapter FifteenPrentice-Hall ©2002 Slide 16 of 31• http"\\asadipour.kmu.ac.ir........52 slid

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Neutralization Reactions

• Neutralization is the reaction of an acid and a base.• Titration is a common technique for conducting a

neutralization.• At the equivalence point in a titration, the acid and base

have been brought together in exact stoichiometric proportions.

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

• The indicator endpoint and the equivalence point for a neutralization reaction can be best matched by plotting a titration curve, a graph of pH versus volume of titrant.

• In a typical titration, 50 mL or less of titrant that is 1 M or less is used.

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Drawing titration Curve ForStrong Acid - Strong Base

HCl + NaOH → NaCl +H2O

Calculate the pH at the

some points

and draw the curve.

4 essential points.

1)initial point

2)equivalence point

3)before the equivalence point

4)beyond the equivalence point

Ml تیتران ت

pH محی ط

0151919.519.92020.120.5212540

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4 questions.

1)What are the present compounds?

2)Which of them is effective on pH?

3)How much are the concentrations?

4)What is the relationship between their Conc. And pH?

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Calculate the pH at the following points in the

titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH.

(a) initial pH. (Before the addition of any NaOH) .

Answer Q1. There are:HCl & H2O

Answer Q2. HCl

Answer Q3. [HCl]

Answer Q4. pH=-log[H+]

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b)equivalence point.

Answer Q1. There are:NaCl & H2O

Answer Q2. H2O

Answer Q3.

Answer Q4. pH=7

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c)before the equivalence point.

Answer Q1. There are:HCl,NaCl & H2O

Answer Q2. HCl

Answer Q3.

Answer Q4. [H+]=N pH=-log[H+]21

2211

VV

VNVNN HCl

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d)after the equivalence point.

Answer Q1. There are:NaOH,NaCl & H2O

Answer Q2. NaOH

Answer Q3.

Answer Q4. [OH-]=N pOH=-log[OH-] pH=14-pOH

21

1122

VV

VNVNN OH

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Titration Curve ForStrong Acid - Strong Base

• pH is low at the beginning.• pH changes slowly until just before equivalence point.• pH changes sharply around equivalence point.• pH = 7.0 at equivalence

point.• Further beyond equivalence point, pH changes slowly.• Any indicator whose color changes in pH range of 4 – 10 can be used in titration.

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Drawing titration Curve Forweak acid- Strong Base

CH3COOH + NaOH → CH3COO- + Na+ +H2O

Calculate the pH at the some points and draw the curve.

Ka=1×10-5

5 essential points.

1)initial point

2)equivalence point

3)beyond the initial point

4)before the equivalence point

5)beyond the equivalence point

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4 questions.

1)What are the present compounds?

2)Which of them is effective on pH?

3)How much are the concentrations?

4)What is the relationship between their Conc. And pH?

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titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH.

(a) initial pH. (Before the addition of any NaOH) .

Answer Q1. There are: CH3COOH & H2O

Answer Q2. CH3OOH

Answer Q3. CH3OOH

Answer Q4. pH=-log[H+]CKH a ][

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b)equivalence point.

Answer Q1. There are: CH3COO- , Na+ & H2O

Answer Q2. CH3COO-

Answer Q3.

Answer Q4. pOH=-log[OH-] Ka×Kb=Kw

21

2211

VV

VNVNN

CKOH b ][• 920118


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c)beyond the initial point.

Answer Q1. There are: CH3COOH, CH3COO- ,Na+ & H2O

Answer Q2. CH3COOH, CH3COO-

Answer Q3.

Answer Q4.

21

2211

VV

VNVNN a

21

22

VV

VNN b

][

][log

A

BpKpH a

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d)before the equivalence point.

Answer Q1. There are: CH3COOH, CH3COO- ,Na+ & H2O

Answer Q2. CH3COOH, CH3COO-

Answer Q3.

Answer Q4.

21

2211

VV

VNVNN a

21

22

VV

VNN b

][

][log

A

BpKpH a

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e)after the equivalence point.

Answer Q1. There are:NaOH, CH3COO- , Na+ & H2O

Answer Q2. NaOH

Answer Q3.

Answer Q4. [OH-]=N pOH=-log[OH-] pH=14-pOH

21

1122

VV

VNVNN OH

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Titration Curve ForWeak Acid - Strong Base

• The initial pH is higher because weak acid is partially ionized.• At the half-neutralization point, pH = pKa.• pH >7 at equivalence point because the anion of the weak acid hydrolyzes.• The steep portion of titration curve around equivalence point has a smaller pH range.• The choice of indicators for the titration is more limited.

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• Titration curves for HCl with NaOH.

A. 50.0 ml of 0.0500 M HCl

• with 0.1000M NaOH.

B. 50.00 ml of 0.000500 M HCl

• with 0.00100 M NaOH.• 920131


• Titration curves for the titration of acetic acid with NaOH.

• A. 0.1000 M acetic acid


• B. 0.001000 M acetic acid

• with 0.00100 M NaOH.• 920131


• Titration curves for the titration of acetic acid with NaOH.

• A. 0.1000 M acetic acid


• B. 0.001000 M acetic acid

• with 0.00100 M NaOH.

• Titration curves for HCl with NaOH.

A. 50.0 ml of 0.0500 M HCl


B. 50.00 ml of 0.000500 M HCl

• with 0.00100 M NaOH.• 920131


• General Shapes of Titration Curves

• Effect of pKa

• Effect of initial • concentration

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• The effect of acid strength (dissociation constant) on titration curves. Each curve represents the titration of 50.00 ml of 0.1000 M acid with 0.1000 M base.

• Effect of Ka

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• The effect of base strength (dissociation constant) on titration curves. Each curve represents the titration of 50.00 ml of 0.1000 M base with 0.1000 M HCl.

• Effect of Kb

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non-aqueous acid base titration

1)Solubility 2) acid or base strength• Acid and base strengths that are not distinguished in aqueous solution

may be distinguishable in non-aqueous solvents.

HClO4 > HCl in acetic acid solvent,

• neither acid is completely dissociated.• HClO4 + CH3COOH ClO4

– + CH3COOH2+ K = 1.3×10–5

• strong acid strong base weak base weak acid

• HCl + CH3COOH Cl– + CH3COOH2+ K = 5.8×10–8

• Differentiate acidity or basicity of different acids or bases

• differentiating solvent for acids …… acetic acid

• differentiating sovent for bases …… ammonia• 920131


Reaction between weak acid and weak base

–Both the weak acid and weak base remain largely undissociated and neutralization involves proton transfer from the weak acid to the weak base. Consider acetic acid and ammonia:

•CH3COOH + NH3 CH3COO- + NH4+ is composed of

•CH3COOH + H2O CH3COO- + H3O+ K1 = Ka = 1.8 x 10-5

•NH3 + H2O NH4+ + OH- K2 = Kb = 1.8 x 10-5

•H3O+ + OH- 2 H2O K3 = 1/ Kw = 1 x 1014

Kn = Koverall = K1 x K2 x K3 = Kb Ka / Kw = 3.2 x 104

•Therefore, the Reaction Still Shifts significantly to the right -- ionizing much of the component present in the smaller amount

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Titration problems1. What volume of 0.10 mol/L NaOH is needed

to neutralize 25.0 mL of 0.15 mol/L H3PO4?

2. 25.0 mL of HCl(aq) was neutralized by 40.0 mL of 0.10 mol/L Ca(OH)2 solution. What was the concentration of HCl?

3. A truck carrying sulfuric acid is in an accident. A laboratory analyzes a sample of the spilled acid and finds that 20 mL of acid is neutral-ized by 60 mL of 4.0 mol/L NaOH solution. What is the concentration of the acid?

4. What volume of 1.50 mol/L H2S will neutral-ize a solution containing 32.0 g NaOH?• 920131


Titration problems1. (3)(0.15 M)(0.0250 L) = (1)(0.10 M)(VB)

VB= (3)(0.15 M)(0.0250 L) / (1)(0.10 M) = 0.11 L

2. (1)(MA)(0.0250 L) = (2)(0.10 M)(0.040 L)

MA= (2)(0.10 M)(0.040 L) / (1)(0.0250 L) = 0.32 M

3. Sulfuric acid = H2SO4

(2)(MA)(0.020 L) = (1)(4.0 mol/L)(0.060 L)

MA = (1)(4.0 M)(0.060 L) / (2)(0.020 L) = 6.0 M

4. mol NaOH = 32.0 g x 1 mol/40.00 g = 0.800 (2)(1.50 mol/L)(VA) = (1)(0.800 mol)

VA= (1)(0.800 mol) / (2)(1.50 mol/L) = 0.267 L• 920131


Species concentrations of weak diprotic acids• Evaluate concentrations of species in a 0.10 M H2S solution.

• Solution:H2S = H+ + HS– Ka1 = 1.02e-7(0.10–x) x+y x-y Assume x = [HS–]

• HS– = H+ + S2– Ka2 = 1.0e-13

x–y x+y y Assume y = [S2–]

• (x+y) (x-y) (x+y) y————— = 1.02e-7 ———— = 1.0e-13(0.10-x) (x-y)

• [H2S] = 0.10 – x = 0.10 M[HS–] = [H+] = x y = 1.0e–4 M;

[S2–] = y = 1.0e-13 M• 0.1>> x >> y:

x+ y = x-y = xx = 0.1*1.02e-7 = 1.00e-4y = 1e-13

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Alpha Values• Def.: the relative equilibrium concentration of the weak

acid/base and its conjugate base/acid • (titrating HOAc with NaOH):

α0 + α1 = 1

Ct

][OAc

Ct

[HOAc]

-

10

K ][H3O

K ][H3O

]H3O [

a0

a0

aK

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• Plots of relative amounts of acetic acid and acetate ion during a titration.

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Chapter FifteenPrentice-Hall ©2002Slide 1 of 31 920131 1 http:\\asadipour.kmu.ac.ir 48slides.

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