Acid-Base Titrations

Titration of Strong Acid and Strong Base

Strong Acid - Strong Base Titration Curve

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.at 0.00 mL of NaOH added, initial point

[H3O+] = CHCl = 0.1000 M

pH = 1.0000

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.at 15.00 mL of NaOH added,

Va * Ma > Vb * Mb

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.at 15.00 mL of NaOH added,

Va * Ma > Vb * Mb

(Va * Ma) - (Vb * Mb)[H3O+] = ----------------------------

(Va + Vb)

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.at 15.00 mL of NaOH added, Va * Ma > Vb * Mb

(Va * Ma) - (Vb * Mb)[H3O+] = ----------------------------

(Va + Vb)

((35.00mL)*(0.1000M)) - ((15.00mL)*(0.1000M))

= ------------------------------------------------------------(35.00 + 15.00)mL

= 4.000x10-2M pH = 1.3979

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.at 35.00 mL of NaOH added,

Va * Ma = Vb * Mb , equivalence point

at equivalence point of a strong acid - strong base titration

pH = 7.0000

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.at 50.00 mL of NaOH added,

Vb * Mb > Va * Ma , post equvalence point

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.at 50.00 mL of NaOH added,

Vb * Mb > Va * Ma , post equvalence point

(Vb * Mb) - (Va * Ma)[OH-] = ---------------------------

(Va + Vb)

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

at 50.00 mL of NaOH added,

Vb * Mb > Va * Ma , post equvalence point

(Vb * Mb) - (Va * Ma)[OH-] = ---------------------------

(Va + Vb)

((50.00mL)*(0.1000M)) - ((35.00mL)*(0.1000M))

= ------------------------------------------------------------(35.00 + 50.00)mL

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

at 50.00 mL of NaOH added, Vb * Mb > Va * Ma , post equvalence point

(Vb * Mb) - (Va * Ma)[OH-] = --------------------------- (Va + Vb)

((50.00mL)*(0.1000M)) - ((35.00mL)*(0.1000M))

= ------------------------------------------------------------(35.00 + 50.00)mL

= 1.765x10-2M pOH = 1.7533

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.at 50.00 mL of NaOH added, Vb * Mb > Va * Ma , post equvalence point

[OH-] = 1.765x10-2M pOH = 1.7533pH = 14.00 - 1.7533 = 12.25

Strong Acid - Strong Base Titration

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Strong Acid - Strong Base Titration

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Strong Acid - Strong Base Titration

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Titration of Weak Acid with Strong BaseEXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5M

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5Mat 0.00 mL of NaOH added, initial point

[H3O+] = Ka * CHA

= (1.75e-5 M)*(0.1000 M) = 1.32e-3 M

pH = 2.878

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5M

at 15.00 mL of NaOH added,

Va * Ma > Vb * Mb

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5Mat 15.00 mL of NaOH added,

Va * Ma > Vb * Mb

(Va * Ma) - (Vb * Mb)[HC2H3O2]excess = ---------------------------

(Va + Vb)

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5M

at 15.00 mL of NaOH added, Va * Ma > Vb * Mb

(Va * Ma) - (Vb * Mb)[HC2H3O2]excess = ------------------------------

(Va + Vb)

((35.00mL)*(0.1000M)) - ((15.00mL)*(0.1000M))

= ------------------------------------------------------------(35.00 + 15.00)mL

= 4.000x10-2M

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5M

at 15.00 mL of NaOH added, Va * Ma > Vb * Mb

[HC2H3O2]excess = 4.000x10-2M (Vb * Mb)

[C2H3O2-] = -------------

(Va + Vb) (15.00 mL)(0.1000 M)= ---------------------------- = 3.000x10-

2M(35.00 + 15.00)mL

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5Mat 15.00 mL of NaOH added, Va * Ma > Vb * Mb

[HC2H3O2]excess = 4.000x10-2M[C2H3O2-] = 3.000x10-2M

[HC2H3O2]excess[H3O+] = Ka * ----------------------

[C2H3O2-]

4.000x10-2M = 1.75e-5 M * ------------------- = 2.33e-5 M

3.000x10-2M

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5Mat 15.00 mL of NaOH added, Va * Ma > Vb * Mb

[HC2H3O2]excess = 4.000x10-2M[C2H3O2-] = 3.000x10-2M

[H3O+] = 2.33e-5 M

pH = 4.632

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5M

at 35.00 mL of NaOH added,

Va * Ma = Vb * Mb , equivalence point

35 35

0.1x 35

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5Mat 35.00 mL of NaOH added,

Va * Ma = Vb * Mb , equivalence point

pH = ½ (pKw + pKa + log [G])

pH = ½ (14 –log 1.75x10-5 + log( ))

pH = ½ (14 + 4.76 – 1.301)

pH = 8.728

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5Mat 50.00 mL of NaOH added,

Vb * Mb > Va * Ma , post equvalence point

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5Mat 50.00 mL of NaOH added,

Vb * Mb > Va * Ma , post equvalence point

(Vb * Mb) - (Va * Ma)[OH-] = --------------------------

(Va + Vb)

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5Mat 50.00 mL of NaOH added, Vb * Mb > Va * Ma

(Vb * Mb) - (Va * Ma)[OH-] = --------------------------

(Va + Vb)

((50.00mL)*(0.1000M)) - ((35.00mL)*(0.1000M))

= ------------------------------------------------------------(35.00 + 50.00)mL

= 1.765x10-2M

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HC2H3O2 with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10-5Mat 50.00 mL of NaOH added, Vb * Mb > Va * Ma

[OH-] = 1.765x10-2MpOH = 1.7533

pH = 14.00 - 1.7533 = 12.25

Weak Acid - Strong Base Titration

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Weak Acid - Strong Base Titration

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Titration of Weak Base with Strong AcidMuch the same shape, except the titration

would start at high pH and decrease as acid is added

Weak Base - Strong Acid Titration

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Triprotic Weak Acid - Strong Base Titration Curve

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.

Ka1 = 7.11 X 10-3 M

Ka2 = 6.34 X 10-8 M

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 Mat 0.00 mL of NaOH added, initial point

[H3O+] = Ka1 * CH2A

= (7.11e-3 M)*(0.1000 M) = 2.67e-2 MpH = 1.574

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 M

at 15.00 mL of NaOH added,

Va * Ma > Vb * Mb, 1st buffer region

(Va * Ma) - (Vb * Mb)[H3PO4]excess = ----------------------------

(Va + Vb)

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 Mat 15.00 mL of NaOH added,

Va * Ma > Vb * Mb, 1st buffer region

((35.00mL)*(0.1000M) - (15.00mL)*(0.1000M))

[H3PO4]excess = -------------------------------------------------------- (35.00 + 15.00)mL

= 4.000x10-2M

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 Mat 15.00 mL of NaOH added,

Va * Ma > Vb * Mb, 1st buffer region[H3PO4]excess = 4.000x10-2M

(Vb * Mb) (15.00 mL)(0.1000 M)[H2PO4

-] = ------------ = -------------------------- = 3.000x10-2M (Va + Vb) (35.00 + 15.00)mL

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 M

at 15.00 mL of NaOH added,

Va * Ma > Vb * Mb, 1st buffer region[H3PO4]excess = 4.000x10-2M

[H2PO4-] = 3.000x10-2M

[H3PO4]excess 4.000x10-2M

[H3O+] = Ka1 * ---------------- = 7.11e-3 M * ------------------ [H2PO4

-] 3.000x10-2M

= 9.48e-3 M

pH = 2.023

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 M at 35.00 mL of NaOH added,

Va * Ma = Vb * Mb , 1st equivalence point

[H3O+] = Ka1 * Ka2

= (7.11x10-3 M)(6.34x10-8 M) = 2.12e-5 M

pH = 4.673

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 M

at 50.00 mL of NaOH added,

2 * Va * Ma > Vb * Mb, 2nd buffer region

2*(Va * Ma) - (Vb * Mb)[H2PO4

-]excess = ------------------------------- (Va + Vb)

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 M

at 50.00 mL of NaOH added,

2 * Va * Ma > Vb * Mb, 2nd buffer region (2(35.00mL)*(0.1000M)) -

((50.00mL)*(0.1000M))[H2PO4

-]excess = --------------------------------------------------------------- (35.00 + 50.00)mL

= 2.353x10-2M

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 Mat 50.00 mL of NaOH added,

2 * Va * Ma > Vb * Mb, 2nd buffer region

[H2PO4-]excess = 2.353x10-2M

(Vb * Mb) - (Va * Ma)[HPO4

-2] = ---------------------------- (Va + Vb)

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 Mat 50.00 mL of NaOH added,

2 * Va * Ma > Vb * Mb, 2nd buffer region[H2PO4

-]excess = 2.353x10-2M

(50.00 mL)(0.1000 M) - (35.00 mL)(0.1000 M)

[H2PO4-] = ---------------------------------------------------------

(35.00 + 50.00)mL

= 1.764x10-2 M

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 M

at 50.00 mL of NaOH added,

2 * Va * Ma > Vb * Mb, 2nd buffer region

[H2PO4-]excess = 2.353x10-2M

[HPO4-2] = 1.764x10-2 M

[H2PO4-]excess

2.353x10-2M[H3O+] = Ka2 * ------------------ = 6.34e-8 M * ------------------

[HPO4-2] 1.764x10-2M

= 8.46x10-8 M pH = 7.073

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 M

at 70.00 mL of NaOH added,

2 * Va * Ma = Vb * Mb , 2nd eq. pt.

Kw * VaMa[OH-] = ------------------

Ka2 (Va + Vb)

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 Mat 70.00 mL of NaOH added,

2 * Va * Ma = Vb * Mb , 2nd eq. pt.

1.00e-14 M2 * (35.00 mL)(0.1000 M)[OH-] = -------------------------------------------- = 7.25x10-5 M

6.34e-8 M * (35.00 + 70.00)mL

pOH = 4.139 pH = 9.860

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 Mat 90.00 mL of NaOH added,

Vb * Mb > 2 * Va * Ma , post 2nd eq. pt.

(Vb * Mb) - 2 * (Va * Ma)[OH-] = -----------------------------------

(Va + Vb)

EXAMPLE:Derive the titration curve for the titration of 35.00 mL of 0.1000 M H3PO4 with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH.Ka1 = 7.11 X 10-3 M Ka2 = 6.34 X 10-8 Mat 90.00 mL of NaOH added,

Vb * Mb > 2 * Va * Ma , post 2nd eq. pt.

(90.00 mL)(0.1000 M) - 2 * (35.00 mL)(0.1000 M)

[OH-] = --------------------------------------------------------------- (35.00 + 90.00)mL

= 1.600 X 10-2 M

pOH = 1.795 pH = 12.204

Diprotic Weak Acid - Strong Base Titration

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Finding the End Point with a pH Electrode

Titration curve for a weak acid - strong base titration.

Finding the End Point with a pH Electrode

First derivative of a weak acid- strong base titration

Finding the End Point with a pH Electrode

Second derivative of a weak acid - strong base titration

Indicators

Indicator Behavior

HIn + H2O H3O+ + In-

acid basecolor color

[H3O+][In-]Ka = -------------- (1)

[HIn]

Indicators

Indicator Behavior

In + H2O InH+ + OH-

base acid color color

[InH+][OH-]Kb = --------------- (2)

[In]

Indicators

Indicator Behavior

rearranging Equation (1) gives

[In-] Ka------ = ----------[HIn] [H3O+]

IndicatorsIndicator Behavior

acid color shows when

[In-] 1 [H3O+][In-] 1 ------ ---- ------------- = ---*[H3O+] = Ka[HIn] 10 [HIn] 10

base color shows when

[In-] 1 [H3O+][In-] ------ ---- ------------- = 10*[H3O+] = Ka[HIn] 10 [HIn]

Indicators

Indicator Behavior

acid color shows when

pH + 1 = pKa

and base color shows when

pH - 1 = pKa

Indicators

Indicator Behavior

Color change range is

pKa = pH + 1 or pH = pKa + 1

Choosing the Proper Indicator color change range should be in area where

titration curve is most vertical

Strong Acid - Strong Base Titration

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phenolphthalein

bromocresol green3.8

5.4

8.0

9.6

Weak Acid - Strong Base Titration

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9.6

8.0phenolphthalein

3.8

5.4bromocresol green