Titrations

Strong Acid with Strong Base

• Starting pHpH = -log[FAcid]

• Just before the Equivalence Point[H+] = (Vacid·Facid-Vbase·Fbase)/(Vsol+Vbase)

• Very Sharp Equivalence PointpH = 7.00

• Excess basepH = 14 – (-log[OH-])[OH-] = Fbase·{(Vbase-V*)/(Vbase+Vsol)]}

25.00 mL of 0.10 M HCl titrated with 0.10 M NaOH

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Vol NaOH (mL)

pH

25.00 mL of ? M HCl titrated with 0.10 M NaOH

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Vol NaOH (mL)

pH

V* = 12.50 mL of NaOH

[HCl] = (12.50 mL)(0.1000 M)/(25.00 mL)

= 0.0500 M

Strong Base with Strong Acid

• Starting pHpH = 14 – {-log[Base]

• Very Sharp Equivalence PointpH = 7.00

• Excess acidpH = -log[H+])

[H+] = Facid·{(Vacid-V*)/(Vacid+Vsol)]}

25.00 mL of 0.10 M NaOH titrated with 0.10 M HCl

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Vol HCl (mL)

pH

25.00 mL of ? NaOH titrated with 0.20 M HCl

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Vol HCl (mL)

pH

V* = 12.50 mL of HCl

[NaOH] = (12.50 mL)(0.2000 M)/(25.00 mL)

= 0.1000 M

Titrating Monoprotic Weak Acids with Strong Base

• Starting pH: Use weak acid dissociationKa = [H+]2/(FAcid – [H+])pH = -log[H+]

HA + OH- → A- + H2O

• Before the Equivalence Point use H-H eqpH = pKa + log [(FNaOH·VNaOH)/(FHA·VHA- FNaOH·VNaOH)]

½ way to Equivalence point, pH = pKa

• At Equivalence point: use weak base dissociation

Kb = [OH-]2/(Fbase – [OH-]),

where Fbase = FHA·VHA/(Vsol+Veq)

• Excess base

[OH-] = Fbase·{(Vbase-V*)/(Vbase+Vsol)]}pH = 14 – (-log[OH-])

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alpha fraction

pH

HA

A-

Ka 1.0·10-5

Weak acid with 0.10 M NaOH

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V NaOH (mL)

pH

Example for Ka = 10-5

• Starting pH

[H+] = ((1∙10-5∙0.1)1/2 = 0.0010 M

pH = 3.00

• ½ way – pH = pKa = 5.00

• At Equiv. pt

[OH-] = ((1∙10-9∙0.05)1/2 = 7.1·10-6 M

pH = 14.00 – [-log(7.1·10-6)] = 8.85

Titrating Monoprotic Weak Bases with Strong Acid

• Starting pH: Use weak base dissociationKb = [OH-]2/(Fbase – [OH-])pH = 14.00 – (-log[OH-])

A- + H+ → HA

• Before the Equivalence Point use H-H eqpH = pKa + log [(FA-·VA-- FHCl·VHCl)/ (FHCl·VHCl)]

½ way to Equivalence point, pH = pKa

• At Equivalence point: use weak acid dissociation

Kb = [H+]2/(Facid – [H+]),

where FACID = FA-·VA-/(Vsol+Veq)

• Excess HCl

[H+] = Facid·{(Vacid-V*)/(Vacid+Vsol)]}pH = log[H+])

Titrate with 0.10 M HCl

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Vol NaOH (mL)

pH

Example for Ka = 10-7

• Starting pH

[OH-] = ((1∙10-7∙0.1)1/2 = 0.00010 M

pH = 10.00

• ½ way – pH = pKa = 7.00

• At Equiv. pt

[H-] = ((1∙10-7∙0.05)1/2 = 7.1·10-5 M

pH = 4.15

Titration of H2A with strong base

• Starting pH – weak acid equilibrium

• ½ way to first eq.: pH = pKa1

• At first equivalence pt:

pH = (pKa1 + pKa2)/2

• ½ way to 2nd eq.: pH = pKa2

• At 2nd equiv. pt.: weak base equilibrium

alpha fraction plot

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pH

alp

ha

frac

tio

n

H2A

HA-

A2-

Titrating weak acid w ith strong base

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Volume base

pH

Example

• Starting pH – [H+] = (10-3*0.1)1/2 = 0.0100 M

• ½ way to first eq.: pH = 3.00

• At first equivalence pt:

pH = (pKa1 + pKa2)/2 = 6.00

• ½ way to 2nd eq.: pH = 9.00

• At 2nd equiv. pt.: [OH-] = (10-5*0.05)1/2 = 7.07∙10-4 M, pH = 10.85

Amino acids

• Amphoteric– Can act as an acid and a base

• Glycine (GH2+, GH, G-)

pKa1 = 2.350, pKa2 = 9.778

Alpha Fraction Plot for Glycine

alpha fraction plot

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pH

alp

ha

frac

tio

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H2A

HA-

A2-

Titration of Glycine

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Vol

pH

Triprotic system• Example: H3PO4

alpha fraction plot

0.00E+00

2.00E-01

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6.00E-01

8.00E-01

1.00E+00

1.20E+00

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pH

alp

ha

frac

tio

n

H3A

H2A-

HA2-

a3-

Titration of a triprotic acid with strong base

• Starting pH – weak acid equilibrium

• ½ way to first eq.: pH = pKa1

• At first equivalence pt:

pH = (pKa1 + pKa2)/2

• ½ way to 2nd eq.: pH = pKa2

• At 2nd equiv. pt.:

pH = (pKa2 + pKa3)/2

• ½ way to third equiv. pt.: pH = pKa3

• At third equiv pt: weak base equilibrium

Titrating weak acid w ith strong base

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Volume base

pH

pKa’s are 4, 7, and 10

H3PO4 (pKa’s 2.2, 7.2, 12.2)

Titrating weak acid w ith strong base

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Volume base

pH