2. Titrasi Netralisasi_v060312

8/12/2019 2. Titrasi Netralisasi_v060312

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Acid-Base Titrations


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Titration ofStrong Acid and Strong Base


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Strong Acid - Strong BaseTitration Curve

EXAMPLE: Derive the titration curvefor the titration of 35.00 mL of 0.1000M HCl with 0.00, 15.00, 35.00, and50.00 mL of 0.1000 M NaOH.


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at 0.00 mL of NaOH added, initial point

[H3O+] = C

HCl = 0.1000 M

pH = 1.0000


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


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


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at 15.00 mL of NaOH added, Va * Ma > V b * Mb (Va * Ma) - (V b * Mb)[H3O +] = ----------------------------(Va + V b)

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

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

= 4.000x10 -2M pH = 1.3979


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at 35.00 mL of NaOH added,Va * Ma = V b * Mb , equivalence point

at equivalence point of a strong acid - strongbase titrationpH = 7.0000


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at 50.00 mL of NaOH added,

Vb * Mb > V a * Ma , post equvalence point


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

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


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(Vb * Mb) - (V a * Ma )[OH -] = ---------------------------

(Va + V

b)

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

(35.00 + 50.00)mL


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EXAMPLE: Derive the titration curvefor the titration of 35.00 mL of 0.1000

M HCl with 0.00, 15.00, 35.00, and50.00 mL of 0.1000 M NaOH.

at 50.00 mL of NaOH added, V b * Mb > V a * Ma , post equvalence point

(Vb * Mb) - (V a * Ma )[OH -] = ---------------------------

(Va + V b)((50.00mL)*(0.1000M)) - ((35.00mL)*(0.1000M))

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

= 1.765x10 -2M pOH = 1.7533


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at 50.00 mL of NaOH added, V b * Mb > V a * Ma , postequvalence point

[OH-

] = 1.765x10-2

M pOH = 1.7533pH = 14.00 - 1.7533 = 12.25


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Strong Acid - Strong Base Titration

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Volume of Base

p H

Series1


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Volume of Base Added

p H


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p H

1

0.1

0.01

0.001


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Titration ofWeak Acid with Strong BaseEXAMPLE: Derive the titration curve for

the titration of 35.00 mL of 0.1000 MHC 2H3O 2 with 0.00, 15.00, 35.00, and 50.00

mL of 0.1000 M NaOH.Ka = 1.75 x 10 -5M


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EXAMPLE: Derive the titration curve forthe titration of 35.00 mL of 0.1000 M

HC 2H3O 2 with 0.00, 15.00, 35.00, and 50.00mL of 0.1000 M NaOH.

Ka = 1.75 x 10 -5M

at 0.00 mL of NaOH added, initial point

[H3O+] = Ka * C HA

=

(1.75e-5 M)*(0.1000 M) = 1.32e-3 MpH = 2.878


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EXAMPLE: Derive the titration curvefor the titration of 35.00 mL of 0.1000 M

HC 2H3O 2 with 0.00, 15.00, 35.00, and50.00 mL of 0.1000 M NaOH.

Ka = 1.75 x 10 -5M



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Ka = 1.75 x 10 -5M


(Va * Ma) - (V b * Mb)[HC 2H3O 2]excess = ---------------------------

(Va + V b)


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EXAMPLE: Derive the titration curvefor the titration of 35.00 mL of 0.1000

M HC 2H3O 2 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 > V b * Mb (Va * Ma) - (V b * Mb)

[HC 2H3O 2]excess = ------------------------------(Va + V b)

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

= 4.000x10 -2M


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Ka = 1.75 x 10 -5M

at 15.00 mL of NaOH added, Va * Ma > V b * Mb [HC 2H3O 2]excess = 4.000x10 -2M

(Vb * Mb)[C2H3O 2-] = -------------

(Va + V b)(15.00 mL)(0.1000 M)

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


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HC 2H3O 2 with 0.00, 15.00, 35.00, and 50.00mL of 0.1000 M NaOH.Ka = 1.75 x 10 -5M

at 15.00 mL of NaOH added, Va * Ma > V b * Mb [HC2H3O 2]excess = 4.000x10 -2M[C2H3O2-] = 3.000x10 -2M

[HC 2H3O 2]excess [H3O+] = K a * ----------------------

[C2H3O 2-]4.000x10 -2M

= 1.75e-5 M * ------------------- = 2.33e-5 M3.000x10 -2M


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Ka = 1.75 x 10 -5M

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

[HC 2H3O 2]excess = 4.000x10 -2M[C2H3O 2-] = 3.000x10 -2M

[H3O+] = 2.33e-5 MpH = 4.632


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Ka = 1.75 x 10 -5M

at 35.00 mL of NaOH added,Va * Ma = V b * Mb , equivalence point


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3535

0.1x35



Ka = 1.75 x 10 -5M

at 35.00 mL of NaOH added,Va * Ma = V b * Mb , equivalence pointpH = (pKw + pKa + log [G])

pH = (14

log 1.75x10-5

+ log( ))pH = (14 + 4.76 1.301)pH = 8.728


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Ka = 1.75 x 10 -5M



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Ka = 1.75 x 10 -5M


(Vb * Mb) - (V a * Ma)

[OH-

] = --------------------------(Va + V b)


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HC 2H3O 2 with 0.00, 15.00, 35.00, and 50.00mL of 0.1000 M NaOH.

Ka = 1.75 x 10 -5M

at 50.00 mL of NaOH added, Vb * Mb > V a * Ma (Vb * Mb) - (V a * Ma)[OH -] = --------------------------

(Va + V b)((50.00mL)*(0.1000M)) - ((35.00mL)*(0.1000M))

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

= 1.765x10 -2M


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Ka = 1.75 x 10 -5M

at 50.00 mL of NaOH added, Vb * Mb > V a * Ma

[OH -] = 1.765x10 -2M

pOH = 1.7533pH = 14.00 - 1.7533 = 12.25


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Weak Acid - Strong Base Titration

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p H


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p H


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Titration ofWeak Base with Strong Acid Much the same shape, except the titration

would start at high pH and decrease as acidis added


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Weak Base - Strong Acid Titration

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Volume of Acid Added

p H


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Features of the Titration of aPolyprotic Acid with a Strong Base

1. The loss of each mole of H + shows up asseparate equivalence point (but only if thetwo pK as are separated by more than 3 pKunits).

2. The pH at the midpoint of the buffer region

is equal to the pK a of that acid species.3. The same volume of added base is required

to remove each mole of H +.


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Finding the End Point with a pHElectrode

Titration curve for a weak acid - strong base titration.


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Finding the End Point with a pH Electrode

p H / V versu s V

First derivative of a weak acid- strong base titration


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Finding the End Point with a pH Electrode

Second derivative of a weak acid - strong base titration

(

pH /

V) /

V versus V


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Indicators

Indicator BehaviorHIn + H 2O H3O+ + In - acid base

color color

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

[HIn]


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Indicators

Indicator BehaviorIn + H 2O InH + + OH -

base acid

color color

[InH+][OH -]K

b = --------------- (2)

[In]


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Indicators

Indicator Behaviorrearranging Equation (1) gives

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


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Indicators Indicator Behavioracid color shows when [In-] 1 [H 3O+][In-] 1------ ---- ------------- = ---*[H 3O+] = K a [HIn] 10 [HIn] 10

base color shows when [In-] 10 [H 3O+][In-]------ ---- ------------- = 10*[H 3O+] = K a [HIn] 1 [HIn]


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Indicators

Indicator Behavioracid color shows when pH + 1 = pK a

and base color shows when pH - 1 = pK

a


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Indicators

Indicator BehaviorColor change range is

pK a = pH + 1 or pH = pK a + 1


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Choosing the Proper Indicator

color change range should be in area wheretitration curve is most vertical


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0

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p H

phenolphthalein

bromocresol green3.8

5.4

8.0

9.6


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0

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p H

9.6

8.0 phenolphthalein

3.8

5.4 bromocresol green

2. Titrasi Netralisasi_v060312

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