Download - 2. titrasi netralisasi
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Acid-Base Titrations
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Titration of Strong Acid and Strong Base
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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.
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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
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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
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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)
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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
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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
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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
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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)
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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
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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
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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
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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
pH Series1
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Strong Acid - Strong Base Titration
0
2
4
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8
10
12
14
0 10 20 30 40 50 60
Volume of Base Added
pH
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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
pH
1
0.1
0.01
0.001
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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Weak Acid - Strong Base Titration
0
2
4
6
8
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14
0 10 20 30 40 50 60
Volume of Base Added
pH
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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
pH
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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
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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
pH
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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
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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
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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)
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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
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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
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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
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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
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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)
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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
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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)
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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
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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
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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)
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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
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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)
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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
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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
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Finding the End Point with a pH Electrode
Titration curve for a weak acid - strong base titration.
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Finding the End Point with a pH Electrode
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
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Indicators
Indicator Behavior
HIn + H2O H3O+ + In-
acid basecolor color
[H3O+][In-]Ka = -------------- (1)
[HIn]
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Indicators
Indicator Behavior
In + H2O InH+ + OH-
base acid color color
[InH+][OH-]Kb = --------------- (2)
[In]
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Indicators
Indicator Behavior
rearranging Equation (1) gives
[In-] Ka------ = ----------[HIn] [H3O+]
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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]
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Indicators
Indicator Behavior
acid color shows when
pH + 1 = pKa
and base color shows when
pH - 1 = pKa
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Indicators
Indicator Behavior
Color change range is
pKa = pH + 1 or pH = pKa + 1
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Choosing the Proper Indicator color change range should be in area where
titration 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
pH
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
pH
9.6
8.0phenolphthalein
3.8
5.4bromocresol green