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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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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.
at 0.00 mL of NaOH added, initial point
[H3O+] = C
HCl = 0.1000 M
pH = 1.0000
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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.
at 15.00 mL of NaOH added,Va * Ma > V b * Mb
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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.
at 15.00 mL of NaOH added,Va * Ma > V b * Mb
(Va * Ma) - (V b * Mb)[H3O+] = ----------------------------(Va + V b)
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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.
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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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.
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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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.
at 50.00 mL of NaOH added,
Vb * Mb > V a * Ma , post equvalence point
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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.
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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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.
at 50.00 mL of NaOH added,Vb * 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
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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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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.
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
0
2
4
6
8
10
12
14
0 10 20 30 40 50 60
Volume of Base
p H
Series1
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Strong Acid - Strong Base Titration
0
2
4
6
8
10
12
14
0 10 20 30 40 50 60
Volume of Base Added
p H
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Strong Acid - Strong Base Titration
0
2
4
6
8
10
12
14
0 10 20 30 40 50 60
Volume of Base Added
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
at 15.00 mL of NaOH added,Va * Ma > V b * Mb
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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
at 15.00 mL of NaOH added,Va * Ma > V b * Mb
(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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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
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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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 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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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
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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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
at 35.00 mL of NaOH added,Va * Ma = V b * Mb , equivalence point
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3535
0.1x35
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
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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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
at 50.00 mL of NaOH added,Vb * Mb > V a * Ma , post equvalence point
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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
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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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 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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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
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
0
2
4
6
8
10
12
14
0 10 20 30 40 50 60
Volume of Base Added
p H
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Weak Acid - Strong Base Titration
0
2
4
6
8
10
12
14
0 10 20 30 40 50 60
Volume of Base Added
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
0
2
4
6
8
10
12
0 10 20 30 40 50 60
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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Strong Acid - Strong Base Titration
0
2
4
6
8
10
12
14
0 10 20 30 40 50 60
Volume of Base Added
p H
phenolphthalein
bromocresol green3.8
5.4
8.0
9.6
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Weak Acid - Strong Base Titration
0
2
4
6
8
10
12
14
0 10 20 30 40 50 60
Volume of Base Added
p H
9.6
8.0 phenolphthalein
3.8
5.4 bromocresol green