8 - 1 titrations an acid-base titration is a volumetric analysis in which a solution of one reactant...

8 - 1

TitrationsTitrations

An acid-base titration is a volumetric analysisin which a solution of one reactant (acid orbase) is gradually added to a solution ofanother reactant (base or acid).

One way to describe acid-base titrations is tocalculate the [H+] or pH as a function of thevolume of titrant added.

When equivalent amounts of the tworeactants have been mixed, na = nb, and this point is known as the equivalence point.

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Neutralization ReactionsNeutralization Reactions

A neutralization reaction is a doublereplacement reaction that occurs between

anacid and a base to form a salt and water.

All ionizable hydrogens from the acid reactwith all of the hydroxides from the base.

H+(aq) + OH-(aq) → H2O(l)

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Equal chemical equivalent amounts ofsulfuric acid and potassium hydroxideproduce salt and water.

H2SO4(aq) + 2KOH(aq) → K2SO4(aq) +

2H2O(l)

2H+(aq) + SO42-(aq) + 2K+(aq) + 2OH-

(aq) → 2K+(aq) + SO42-(aq) +

2H2O(l)

H+(aq) + OH-(aq) → H2O(l)

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Neutralization ProblemNeutralization Problem

25.0 mL of 0.50 M sodium hydroxide isneutralized by 15.0 mL of sulfuric acid.Determine the concentration of the sulfuricacid.

[NaOH] = 0.50 M [H2SO4] = ?Vb = 25.0 mL Va = 15.0 mL

2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

H+(aq) + OH-(aq) → H2O(l)

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The reason for including the molecularequation is to illustrate what is involved

whenyou have a diprotic acid and a monoproticbase.

The same result is obtained by comparing 1 mol of NaOH to one mole

of H2SO4.

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.

nb =0.50 mol NaOH

1 L×25.0 mL ×

1 L103 mL

×1 mol OH-

1 mol NaOH

=0.0125 mol OH-

=0.0125 mol NaOH ×2 mol NaOH

1 mol H2SO4

= 6.25 x 10-3 mol H2SO4

na

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[H2SO4] = n/V

[H2SO4] =6.25 x 10-3 mol H2SO4

15.0 mL ×103 mL

1 L

[H2SO4] = 0.417 M

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Strong Acid-Strong Base TitrationStrong Acid-Strong Base Titration

(a) Calculate the pH when 24.9 mL of 0.100 M HI is added to 25.0 mL of 0.100 M NaOH.

[HI] = 0.100 M [NaOH] = 0.100 M Va = 24.9 mL Vb = 25.0 mL

HI(aq) H+(aq) + I-(aq)

NaOH(aq) Na+(aq) + OH-(aq)

→

→

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[H+] = n/V

na =0.100 mol HI

L× 0.100 mol H+

0.100 mol HI×

24.9 mL ×1 L

103 mL=2.49 x 10-3 mol H+

nb =0.100 mol NaOH

L× 0.100 mol OH-

0.100 mol NaOH×

25.0 mL ×1 L

103 mL=2.50 x 10-3 mol OH-

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nb 0.00249 0.00250(mol)

H+(aq) + OH-(aq) → H2O(l)

na 0 1.00 x 10-5 (mol)

[OH-] = n/V[OH-] = 1.00 x 10-5 mol OH-

49.9 mL×1 L

103 mL

= 2.00 x 10-4 M

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pOH = -log[OH-] = -log(2.00 x 10-4) = 3.700

pH + pOH = 14.00

pH = 14.00 – 3.700 = 10.30

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(b) Calculate the pH when 25.1 mL of 0.100 M

HI is added to 25.0 mL of 0.100 M NaOH.

[H+] = n/Vna =

0.100 mol HI

L× 0.100 mol H+

0.100 mol HI×

25.1 mL ×1 L

103 mL=2.51 x 10-3 mol H+

nb =2.50 x 10-3 mol OH-

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nb 0.00251 0.00250 (mol)

H+(aq) + OH-(aq) → H2O(l)

na 1.00 x 10-5 0 (mol)

[H+] = n/V[H+] =

1.00 x 10-5 mol H+

50.1 mL×1 L

103 mL

= 2.00 x 10-4 M

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pH = -log[H+] = -log(2.00 x 10-4) = 3.70

It does not looklike a pH of 3.70!

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Strong Acid-Strong Base TCStrong Acid-Strong Base TC

A graph of pH as a function of the volume ofadded titrant (a solution of knownconcentration which is added to another) iscalled a titration curve.

The simplest titration is that of a strong acidwith a strong base.

The following titration curve shows a 50.0 mL sample of 0.10 M HCl being

titrated with 0.10 M NaOH.

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In the case of a strong acid-strong base

titration, the reactant species will be the

H+ and OH-.

Remember, the anion of a strong acid and the cation of a strong base are spectator ions.

Spectator ions do not undergo any hydrolysis (reaction with H2O).

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pH vs Strong Acid vs Volume of Strong Base

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0.0 20.0 40.0 60.0 80.0 100.0 120.0

Volume of NaOH (mL)

pH

unionized acid

excess OH-

pH = 7.00

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Strong Acid-Strong Base TCStrong Acid-Strong Base TC

Characteristics of titration curve:

The initial pH is low, ≈ 1.00, which is indicative of a strong acid.

There is a small pH change as the base is initially added.

An equivalent amount of base neutralizes an equivalent amount of

acid.

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The neutralization of an acid and base produces water which decreases the [H+] and increases the pH.

As you approach the equivalence point the pH changes very rapidly.

The midpoint of the vertical section of the curve is the equivalence point indicating there are equivalents of

acid and base.

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For a strong acid-base titration, neutralization occurs at a pH = 7 because the [H+] and the [OH-] are

equal and form water molecules.

H+(aq) + OH-(aq) → H2O(l)

Na+ and Cl- do not hydrolyze leaving the

solution at a pH = 7.

After the equivalence point, the pH depends on the excess OH- in solution.

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Weak Acid-Strong Base TitrationWeak Acid-Strong Base Titration

Calculate the pH when 50.0 mL of 1.00 MLiOH is added to 50.0 mL of 1.00 MHC2H3O2.

[HC2H3O2] = 1.00 M [LiOH] = 1.00 MVa = 50.0 mL Vb = 50.0 mLKb = 1.8 x 10-5

LiOH(aq) Li+(aq) + OH-(aq)→

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[OH-] = n/V

nb =1.00 mol LiOH

L× 1 mol OH-

1 mol LiOH×

50.0 mL × 1 L103 mL

=5.00 x 10-2 mol OH-

na =1.00 mol Ha

L× 50.0 mL ×

1 L103 mL

na = 5.00 x 10-2 mol HC2H3O2

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nb 5.00 x 10-2 5.00 x 10-2 0 (mol)

HC2H3O2(aq) + OH-(aq) → C2H3O2-(aq) +

H2O(l)

na 0 0 5.00 x 10-2 (mol)[C2H3O2

-] =5.00 x 10-2 mol C2H3O2

-

100.0 mL × 1 L103 mL

[C2H3O2-] =0.500 M

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C2H3O2-(aq) + H2O(l) HC2H3O2

(aq) + OH-(aq)

[ ]i 0.500 0 0[ ]c -x +x +x[ ]e 0.500-x x x

Kb =[OH-] [HC2H3O2]

[C2H3O2-]

5.6 x 10-10[OH-]= ×x2

0.500 - x≈1.7 x 10-5 M

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pOH = -log[OH-] = -log(1.75 x 10-5) = 4.76

pH + pOH = 14.00pH = 14.00 – 4.76 = 9.24

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Weak Acid-Strong Base TCWeak Acid-Strong Base TC

A weak acid-strong base titration is morecomplicated to analyze due to the bufferingeffect.

The following titration curve shows a 50.0 mL sample of 1.00 M HC2H3O2

being titrated with 1.00 M NaOH.

In the case of a weak acid-strong base titration, the reacting species will be HC2H3O2 and OH-.

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Remember, the anion of a weak acid is a

relatively strong base and the cation of a

strong base is a spectator ion.

A spectator ion does not undergo hydrolysis but the anion of a weak

base does.

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pH vs Volume Of NaOH

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0.00 10.00 20.00 30.00 40.00 50.00 60.00Volume Of NaOH (mL)

pH

pH of Weak Acid vs Volume of Strong Base

Volume of NaOH (mL)

pH

indicative of a weak acid

pH = 8.10

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Weak Acid-Strong Base TCWeak Acid-Strong Base TC

Characteristics of titration curve:

The initial pH is somewhat higher, > 2.00,

which is indicative of a weak acid.

This results from a weak acid not ionizing completely (≈ 5%) which

supplies fewer H+ to the solution.

The anion of the weak acid, C2H3O2-,

undergoes hydrolysis .

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C2H3O2-(aq) + H2O(l)

HC2H3O2(aq) + OH-

(aq)

This is the cause of the pH rising more rapidly in the early stages when compared to a strong acid-strong

base titration.

The combination of the weak acid and the conjugate base of that acid produces a buffering effect.

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HC2H3O2(aq) + OH-(aq) → C2H3O2

-(aq) + H2O(l)

C2H3O2-(aq) + H2O(l)

HC2H3O2(aq) + OH-(aq)

Remember this buffering effect resists changes in pH.

Compared to the strong-acid strong-base titration, there are smaller changes in pH found near the equivalence point.

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The acid-base reaction produces both water and salt.

The weaker the acid, the stronger the conjugate base which in turn

increases the pH to above 7.00.

After reaching the equivalence point, the

titration curve for both the strong acid- strong base and the weak acid-strong base will be identical because the pH depends on the excess OH- in both titrations.

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Weak Base-Strong Acid TitrationWeak Base-Strong Acid Titration

Calculate the pH when 10.0 mL of 0.100 MHBr is added to 20.0 mL of 0.100 M NH3.

[HBr] = 0.100 M [NH3] = 0.100 MVa = 10.0 mL Vb = 20.0 mL

HBr(aq) H+(aq) + Br-(aq)→

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[H+] = n/V

na =0.100 mol HBr

L× 0.100 mol H+

0.100 mol HBr×

10.0 mL ×1 L

103 mL=1.00 x 10-3 mol H+

nb =0.100 mol NH3

L× 20.0 mL ×

1 L103 mL

nb =2.00 x 10-3 mol NH3

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nb 1.00 x 10-3 2.00 x 10-3 0 (mol)

H+(aq) + NH3(aq) → NH4+(aq)

na 0 mol 1.00 x 10-3 1.00 x 10-3 (mol)[NH3] =

1.00 x 10-3 mol NH3

30.0 mL × 1 L103 mL

1.00 x 10-3 mol NH31.00 x 10-3 mol NH3

[NH3] = 3.33 x 10-2 M

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NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

[ ]i 0.0333 0.0333 0[ ]c -x +x +x[ ]e 0.0333 - x 0.0333 + x x

Kb =[NH4

+] [OH-][NH3]

1.8 x 10-5[H+]= ×0.0333 - x

≈1.8 x 10-5 M[H+]

0.0333 + x

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Weak Base-Strong Acid TCWeak Base-Strong Acid TC

A weak base-strong acid titration is morecomplicated to analyze due to the bufferingeffect.

The following titration curve shows a 50.0 mL sample of 0.10 M NH3 being

titrated with 0.10 M HCl.

In the case of a weak base-strong acid titration, the reacting species will be NH3 and H+.

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Remember, the conjugate acid of a weak

base is a relatively strong acid and the cation of a strong base is a spectator ion.

A spectator ion does not undergo hydrolysis but the conjugate acid of a weak base does.

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C

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0.0 20.0 40.0 60.0 80.0 100.0

T

C

pH of Weak Base vs Volume of Strong Acid pH

Volume of HCl (mL)

indicative of a weak base

pH = 5.28

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Weak Base-Strong Acid TCWeak Base-Strong Acid TC

Characteristics of titration curve:

The initial pH is high, ≈ 11.00, which is indicative of a weak base.

This results from a weak base not ionizing completely (≈ 5%) which supplies fewer OH- to the solution.

The added acid neutralizes an equivalent amount of base.

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H+(aq) + NH3(aq) → NH4+(aq)

The conjugate acid of the weak base, NH4

+, undergoes hydrolysis

NH4+(aq) + H2O(l) H3O+(aq) +

NH3(aq)

The combination of the strong acid and the conjugate acid of that base produces a buffering effect.

Remember this buffering effect resists changes in pH.

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Compared to the strong acid - strong base titration, there are smaller changes in pH found near the equivalence point.

The midpoint of the vertical section of the curve is the equivalence point indicating there are equivalents of base and acid.

The weaker the base, the stronger the conjugate acid which in turn decreases the pH to below 7.00.

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After reaching the equivalence point, the

titration curve for both the strong acid- strong base and the weak base-strong acid will be identical because the pH depends on the excess H+ in both titrations.

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Acid-Base Color IndicatorsAcid-Base Color Indicators

An appropriate indicator is a weak acidwhose Ka is close to the concentration of [H+]or the pH at the equivalence point.

The equivalence point is the point at whichthe added solute reacts completely with thesolute present in the solution.

H+(aq) + OH-(aq) → H2O(l)

1.00 mol of a monoprotic acid will completelyreact with 1.00 mol of a monoprotic base.

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Do not confuse the equivalence point with theendpoint which is the point at which theindicator changes color.

Knowing the equivalence point helps todetermine the most appropriate indicator.

For the HC2H3O2-NaOH titration the pH ≈ 9.24.

A good choice for indicator would bephenolphthalein (endpoint ≈ 8-10) or thymolblue (endpoint ≈ 8-10).

For the HCl-NaOH titration the pH ≈ 7.00.A good choice for indicator would bebromthymol blue (endpoint ≈ 6-8).

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Diprotic Acid TitrationDiprotic Acid Titration

In the case of a weak diprotic acid (H2SO3)titrated with a strong base (OH-), the reactingspecies are H+, OH-, H2SO3, HSO3

-, and SO32-.

For a typical diprotic acid such assulfurous acid, H2SO3, you expect to findtwo equivalence points.

For 25.0 mL of an equimolar diprotic acid andstrong base, you would expect twoequivalent points at 25.0 and 50.0 mL.

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However, the Ka values must be considered.

For H2SO3, the Ka’s are:

Ka1 = 1.50 x 10-2

Ka2 = 1.0 x 10-7

If the Ka2 value is extremely small as in thecase of H2S (≈ 10-19), only one equivalencepoint will be displayed.

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E

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0.00 10.00 20.00 30.00 40.00 50.00 60.00 70.00

E

E pH

Volume of NaOH (mL)

pH of Diprotic Acid (H2SO3) vs Volume of Strong Base

E.P. 1

E.P. 2

8 - 49

Triprotic Acid TitrationTriprotic Acid Titration

In the case of a weak triprotic acid (H3PO4)titrated with a strong base (OH-), the reactingspecies are H+, OH-, H3PO4, H2PO4

-, HPO42-,

and PO43-.

For a typical triprotic acid such asphosphoric acid, H3PO4, you expect to findthree equivalence points.

For 25.0 mL of an equimolar triprotic acid andstrong base, you would expect threeequivalent points at 25.0, 50.0, and 75.0 mL.

8 - 50

However, the Ka values must be considered.

For H3PO4, the Ka’s are:

Ka1 = 7.50 x 10-3

Ka2 = 6.20 x 10-8

Ka3 = 4.20 x 10-13

Because of the extremely small Ka3 value,phosphoric acid displays only twoequivalence points at 25.0 and 50.0 mL.

It is common for polyprotic acids to show onlyone or two equivalence points depending ontheir Ka values.

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pH

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0

pH

pH

Volume of NaOH (mL)

pH

pH of Triprotic Acid (H3PO4) vs Volume of Strong Base

E.P. 1

E.P. 2

No E.P. 3 becauseKa3 is too small.