Acid/Base TitrationsAcid/Base Titrations

Titrations Titration Curve – always calculate

equivalent point first Strong Acid/Strong Base

Regions that require “different” calculations B/F any base is added Half-way point region At the equivalence point After the equivalence point

Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr

First -find Volume at equivalence M1V1 = M2V2

(0.050 L)(0.02000M) = 0.1000 V V = 10.0 mL

Strong Acid/Strong Base 50.00 mL of 0.02000 M KOH Titrated with 0.1000 M HBr

Second – find initial pH pH = - logAH ~ -log [H+]

pOH = -logAOH ~ -log [OH-]

pH = 12.30

Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr

Third– find pH at mid-way volume KOH (aq) + HBr (aq) -> H2O (l) +

KBr(aq) BeforeAfter

0.001000 mol 0.0006000 mol0.000400 mol 0 mol

Limiting Reactant

0.0006000 mol0.0006000 mol

pH = 11.8

(~6 ml)

Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr

Fourth – find pH at equivalence point KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq)

BeforeAfter

0.001000 mol 0.0010000 mol0 mol 0 mol 0.0010000 mol0.0010000 mol

pH = 7.0

Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr

Finally – find pH after equivalence point KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq)

BeforeAfter

0.001000 mol 0.001200 mol0 mol 0.0002000 mol 0.0010000 mol

pH = 2.5

12 ml

Limiting Reactant

Typical pH titration

0

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14

0 5 10 15 20

mL of HBr

pH

Titration of WEAK acid with a strong base

Titration of a weak acid solution with a strong base.

25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH

First, calculate the volume at the equivalence-point M1V1 = M2V2

(0.0250 L) 0.1000 M = 0.1000 M (V2) V2 = 0.0250 L or 25.0 mL

Titration of a weak acid solution with a strong base.

25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH

Second, Calculate the initial pH of the acetic acid solution

Titration of a weak acid solution with a strong base.

25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH

Third, Calculate the pH at some intermediate volume

Titration of a weak acid solution with a strong base.

25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH

Fourth, Calculate the pH at equivalence

Titration of a weak acid solution with a strong base.

25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH

Finally calculate the pH after the addition 26.0 mL of NaOH

0

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0 10 20 30 40 50

Volume of NaOH, mL

pH

Initial pH

aa CKH ][

Buffer Region

][

][log

acid

basepKpH a

Equivalence pointM1V1=M2V2

pH @ equivalence?

pH after equivalence?

0

2

4

6

8

10

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14

0 10 20 30 40 50

Volume of NaOH, mL

pH

Initial pH

aa CKH ][

Buffer Region

][

][log

acid

basepKpH a

Equivalence pointM1V1=M2V2

pH @ equivalence

pH after equivalence?

bb CKOH ][

0

2

4

6

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10

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14

0 10 20 30 40 50

Volume of NaOH, mL

pH

Initial pH

aa CKH ][

Buffer Region

][

][log

acid

basepKpH a

Equivalence pointM1V1=M2V2

pH @ equivalence

pH after equivalenceDominated by remaining[OH-]

bb CKOH ][

Weak Base titrated with strong acid Consider a 100 ml of a 0.0100 M

base with 0.0500 M HCl Kb = 1 x 10-5

Initial pH

bb CKOH ][

Buffer Region

][

][log

acid

basepKpH a

pH @ equivalence

aa CKH ][

pH after equivalenceDominated by remaining[H+]