2. titrasi netralisasi

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

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60

Volume of Base

pH Series1

Strong Acid - Strong Base Titration

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60

Volume of Base Added

pH

Strong Acid - Strong Base Titration

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60

Volume of Base Added

pH

1

0.1

0.01

0.001

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

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60

Volume of Base Added

pH

Weak Acid - Strong Base Titration

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60

Volume of Base Added

pH

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

0

2

4

6

8

10

12

0 10 20 30 40 50 60

Volume of Acid Added

pH

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

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60 70 80 90 100

Volume of Base Added

pH

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

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60

Volume of Base Added

pH

phenolphthalein

bromocresol green3.8

5.4

8.0

9.6

Weak Acid - Strong Base Titration

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60

Volume of Base Added

pH

9.6

8.0phenolphthalein

3.8

5.4bromocresol green