ch7 acid base rev.ppt - cal state la
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• Isoionic Point: The pH of pure, neutral polyprotic acid (the natural zwitterion) – HAwater: H2A+, A-, H+, and OH-
– Most HA and [H2A+][A-] • Isoelectric Point: pH at which the average
charge of the polyprotic acid is zero (amino acids) – Most HA and [H2A+]=[A-]
Acid-Base Titrations
Titration: Increments of reagent solution-the titrant-are added to analyte until their reaction is complete.
Titration type/endpoint determination:
•Volumetric: volume of standard is measured •Gravimetric: mass of standard is measured, e.g., Mohr method (Cl-Ag-CrO4). •Coulometric: time at constant current is measured, i.e., amount of charge. •Spectrophotometric: Beer’s Law, the absorbance
Standard Solution (Standard Titrant): Reagent of known concentration that is used to carry out the analysis.
Titration: Performed slowly by adding standard solution from a buret until the reaction is judged to be complete.
Terms
Indicators: Added to a tritration to give an observable physical change at or near the equivalence point. Usually a color or turbidity change.
Equivalence Point: The point when the amount of added titrant is chemically equivalent to the amount of analyte in the sample.
End Point: The point at which an indicator physically changes representing the estimated equivalence point (color or turbidity change).
Titration error = Vep - Veq
Terms
Back-titration: Add a known excess of one standard titrant to the analyte, and then titrate a second standard titrant to determine the excess reagent. May be required when a reaction is slow or the standard solution lacks stability.
Blank titration: Carry out the same procedure without analyte. To estimate the titration error.
Terms
2
1. Sufficiently stable (only need to determine concentration once).
2. React rapidly with the analyte (less time between additions).
3. React more or less completely with the analyte (satisfactory end points).
4. Undergo a selective reaction with analyte described by a simple chemical equation
Standard Solutions (Titrants) Standard Solutions (Titrants)
Methods for Establishing the Concentration of Standard Solutions
1. Direct Method: A carefully weighed quantity of primary standard is diluted to a known volume. First choice if possible.
2. Standardization: A titrant to be standardized is used to titrate a weighed quantity of a primary, secondary standard or a known volume of another standard solution.
Primary Standard: Highly purified compound that serves as a reference material. • High purity, stability (air, heat, vacuum), • Standard solution (direct method) • To standardize a standard solution (standardization)
Secondary Standard: A reference material (a compound) whose purity has been established by a chemical analysis. Often used in lieu of a primary standard because there are not many primary standards.
Standard Solutions (Titrants) Summary titration titrant analyte indicator equivalence point vs. end point titration error blank titration Standard: Primary standard, second standard, standard solution
Acid-Base Titrations
How does the pH change as titrant is added?
Titration Curve: A graph showing how the concentration of one of the reactants varies as titrant is added. (how the pH changes as titrant is added in acid and base titration)
Titration of Strong Base with Strong Acid
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Titration of Strong Base with Strong Acid
1. Before the equivalence point: excess OH-
2. At the equivalence point: pH = 7.00 3. After the equivalence point: excess H+
0.00, initial point 10.00 ml, 50% titration 19.98 ml, half-drop before equivalence point 20.00 ml, 100% titration (equivalence point) 20.02 ml, half-drop after equivalence point 40.00 ml, 200% titration (far-over)
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
Titration of Strong Acid with Strong Base
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
At 0.00 mL of NaOH added, initial point
[H3O+] = CHCl = 0.1000 M
at 10.00 mL of NaOH added, Va * Ma > Vb * Mb
(Va Ma) - (Vb Mb) [H3O+] = ----------------------------
(Va + Vb)
((20.00mL)*(0.1000M)) - ((10.00mL)*(0.1000M)) = ------------------------------------------------------------
(20.00 + 10.00)mL
= 3.33x10-2M pH =1.48
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
at 19.98 mL of NaOH added, Va * Ma > Vb * Mb
(Va Ma) - (Vb Mb) [H3O+] = ----------------------------
(Va + Vb)
((20.00mL)*(0.1000M)) - ((19.98mL)*(0.1000M)) = ------------------------------------------------------------
(20.00 + 19.98)mL
=5.0x10-5M pH =4.30
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
at 20.00 mL of NaOH added Va * Ma = Vb * Mb , equivalence point at equivalence point of a strong acid - strong base titration pH = 7.00
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
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at 20.02 mL of NaOH added, Vb * Mb > Va * Ma , post equivalence point
(Vb * Mb) - (Va * Ma) [OH-] = ---------------------------
(Va + Vb) ((20.02mL)*(0.1000M)) - ((20.00mL)*(0.1000M))
=5.0x10-5M pOH =4.30 pH=9.70
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
at 40.00 mL of NaOH added, Vb * Mb > Va * Ma , post equivalence point
(Vb * Mb) - (Va * Ma) [OH-] = ---------------------------
(Va + Vb) ((40.00mL)*(0.1000M)) - ((20.00mL)*(0.1000M))
=3.33x10-2M, pOH =1.48 pH=12.52
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
Strong Acid - Strong Base Titration
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Volume of Base Added
(mL)
The Effect of Concentration • The change in pH at the equivalence point is less pronounced at decreased concentrations. • The pH at the equivalence point is still 7.
(M)
Titration of Strong Acid with Strong Base Titration of Weak Acid with Strong Base
Titration of Weak Acid with Strong Base
Titration of Weak Acid with Strong Base
•The pH change at the equivalence point decreases (the end point become less distinct) as the acid becomes weaker or more dilute
•The Effect of Concentration: the pH at the equivalence point is lower for lower initial weak acid concentrations and initial pH is higher.
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Titration of Weak Acid with Strong Base
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5 M
0.00, initial point 15.00 ml, before equivalence point 35.00 ml, 100% titration (equivalence point) 50.00 ml, after equivalence point
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5 M
at 0.00 mL of NaOH added, initial point
[H+] Ka Ca
= (1.75 x 10-5 M) x (0.1000 M) = 1.32 x 10-3 M pH = 2.878
at 15.00 mL of NaOH added, Va Ca > Vb Cb
(Va Ca) - (Vb Cb) [HAc]ex. = ------------------------------
(Va + Vb) [(35.00 mL)x(0.1000 M)) - ((15.00 mL)x(0.1000M)]
= ----------------------------------------------------------------- (35.00 + 15.00) mL
= 4.000x10-2 M
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5 M
at 15.00 mL of NaOH added, Va Ca > Vb Cb
[HAc]ex = 4.000x10-2 M
(35.00 + 15.00) mL
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5 M
at 15.00 mL of NaOH added, Va Ca > Vb Cb
[HAc]ex = 4.000 x 10-2 M [Ac-] = 3.000 x 10-2 M
[HAc]ex [H+] = Ka ------------------
[ Ac-] 4.000 x10-2 M
= 1.75 x 10-5 M ------------------- = 2.33 x 10-5 M 3.000 x10-2 M
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5 M
at 15.00 mL of NaOH added, Va Ca > Vb Cb
[HAc]ex = 4.000 x 10-2 M
[Ac-] = 3.000 x 10-2 M
[H+] = 2.33 x 10-5 M pH = 4.632
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5 M
632.4 10000.4 10000.3
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at 35.00 mL of NaOH added, Va Ca = Vb Cb , equivalence point
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc 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 Ca = Vb Cb , equivalence point
Kw VaCa[OH-] = ----------------- Ka (Va + Vb)
Kb Cb
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5M
1.00 x10-14 M2 * (35.00mL)(0.1000M) [OH-] = -----------------------------------------------
1.75 x10-5 M * (35.00 + 35.00)mL
= 5.34x10-6 M
at 35.00 mL of NaOH added, Va * Ca = Vb * Cb , equivalence point
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc 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 Ca = Vb Cb , equivalence point
[OH-] = 5.34 x 10-6 M
pOH = 5.272 pH = 8.728
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5M
at 50.00 mL of NaOH added, Vb Cb > Va Ca , post equivalence point
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5M
at 50.00 mL of NaOH added, Vb Cb > Va Ca , post equivalence point
(Vb Cb) - (Va Ca)[OH-] = -------------------------- (Va + Vb)
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc 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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at 50.00 mL of NaOH added, Vb Cb > Va Ca
(Vb Cb) - (Va Ca)[OH-] = -------------------------- (Va + Vb)
[(50.00mL)*(0.1000M)) - ((35.00mL)*(0.1000M)] = -------------------------------------------------------------
(35.00 + 50.00) mL = 1.765 x 10-2 M
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5M
at 50.00 mL of NaOH added, Vb Cb > Va Ca
[OH-] = 1.765 x 10-2M pOH = 1.7533 pH = 14.00 - 1.7533 = 12.25
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5M
(a) Which of the two curves represents the titration of a strong acid, and which a weak acid? (b) What is the approximate pH at the equivalence point for each of the acids? (c) What is the approximate pKa of the weak acid?
The following plot shows two titration curves, each representing the titration of 50.0 mL of 0.100 M acid with 0.100 M NaOH:
Vb (mL)
Acid - Strong Base Titration Curve
Much 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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Volume of Acid Added
(mL)
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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
Diprotic Systems: Titration of Weak Acid with Strong Base
0.00, Initial point 15.00 ml, Before the 1st equivalence point 35.00 ml, At the 1st equivalence point 50.00 ml, After the 1st equivalence point 70.00 ml, At the 2nd equivalence point 90.00 ml, After the 2nd equivalence point
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EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 0.00 mL of NaOH added, initial point
pH = 1.632
Ka
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 Ca > Vb Cb, the 1st buffer region
(Va Ca) - (Vb Cb) [H2A]ex = ----------------------------
(Va + Vb)
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca > Vb · Cb, the 1st buffer region
[(35.00 mL)·(0.1000 M) - (15.00 mL)·(0.1000 M)] [H2A]ex = ----------------------------------------------------------------
(35.00 + 15.00) mL = 4.000x10-2M
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca > Vb · Cb, the 1st buffer region
[H2A]ex = 4.000x10-2 M
(Va + Vb) (35.00 + 15.00)mL
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca > Vb · Cb, the 1st buffer region
[H2A]ex = 4.000 x 10-2 M; [HA-] = 3.000 x 10-2 M [H2A]ex 4.000 x 10-2 M
[H+] = Ka1 x -------------- = 7.11x10-3 M x ------------------ [HA-] 3.000 x 10-2 M
= 9.48 x 10-3 M pH = 2.023
023.2 10000.4 10000.3
M M
AH HA
pKpH a
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca = Vb · Cb, the 1st equivalence point
[H+] Ka1 · Ka2
= (7.11 x 10-3 M)(6.34 x 10-8 M) = 2.12 x 10-5 M pH = 4.673
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EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca) > Vb · Cb, the 2nd buffer region
2 (Va · Ca) - (Vb · Cb)
[HA-]ex = ------------------------------- (Va + Vb)
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca) > Vb · Cb, the 2nd buffer region
[2(35.00mL)*(0.1000M)) - ((50.00mL)*(0.1000M)] [HA-]ex = ---------------------------------------------------------------
= 2.353x10-2M
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca) > Vb · Cb, the 2nd buffer region [HA-]ex = 2.353x10-2M
(Vb · Cb) - (Va · Ca) [A2-] = ---------------------------
(Va + Vb)
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca) > Vb · Cb, the 2nd buffer region [HA-]ex = 2.353x10-2M
(50.00 mL)(0.1000 M) - (35.00 mL)(0.1000 M) [A2-] = ---------------------------------------------------------
(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 H2A 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 · Ca) > Vb · Cb, the 2nd buffer region [HA-]ex = 2.353x10-2 M [A2-] = 1.764x10-2 M
Ka2 [HA-]ex 6.34 x 10-8 M x 2.353x10-2 M [H+] = ------------------ = ------------------------------------
[A2-] 1.764x10-2 M = 8.46x 10-8 M
pH = 7.073
pKpH
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca) = Vb · Cb, the 2nd equivalence point
Kw · Va · Ca [OH-] = ---------------------
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EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca) = Vb · Cb, the 2nd equivalence point
1.00 x 10-14 M2 x (35.00 mL)x(0.1000 M) [OH-] = ---------------------------------------------------- = 7.25x10-5 M
6.34 x 10-8 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 H2A 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 90.00 mL of NaOH added, Vb · Cb > 2(Va · Ca), after 2nd equivalence point
Vb · Cb - 2(Va · Ca) [OH-] = -----------------------------------
(Va + Vb)
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 90.00 mL of NaOH added, Vb · Cb > 2(Va · Ca), after 2nd equivalence point
(90.00 mL)(0.1000 M) - 2 x (35.00 mL)(0.1000 M) [OH-] = ---------------------------------------------------------------
(35.00 + 90.00) mL
Diprotic Weak Acid - Strong Base Titration
0
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0 10 20 30 40 50 60 70 80 90 100
Volume of Base Added
pH Diprotic Systems: Titration of Weak Acid Base with Strong Base
(mL)
Standard Solutions for Acid/Base Titrations
Standard acid or base reagents are always strong acids or bases • HCl, HClO4, H2SO4, NaOH, and KOH • They yield sharper end points. • Weak reagents react incompletely. • Nitric acid and hot sulfuric and perchloric yield undesirable
ox/red side reactions.
• Tris (THAM) – tris(hydroxymethyl)aminomethane, (HOCH2)3CNH2
• Sodium tetraborate decahydrate, Na2B4O7.10H2O
1. Indicators 2. Derivatives of titration curve 3. Gran Plot
Acid/Base Indicators • Weak acids or bases themselves • The conjugate forms differs in color
HIn In- + H+
In + H2O ⇔ InH+ + OH-
base acid Color1 Color2
base acid Color1 Color2
[InH+][OH-] Kb = --------------- (2)
[In-] Ka------ = ---------- [HIn] [H+]
Indicators Indicator Behavior acid color shows when [In-] 1 [H+][In-] 1 ------ <= ---- ------------- = ---*[H+] = Ka[HIn] 10 [HIn] 10
base color shows when [In-] 10 [H+][In-] ------ >= ----- ------------- = 10*[H+] = Ka[HIn] 1 [HIn]
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Indicators
and base color shows when pH - 1 = pKa
Color change range is pH = pKa + 1 (Transition range)
Choosing the Proper Indicator
• Color change range should be in area where titration curve is most vertical.
• Indicator error: the difference between the observed end point (color change) and the true equivalence point.
Common Indicators
Using Derivatives of Titration Curve to Determine the End Point
Titration curve for a polyprotic acid - strong base titration.
First derivative of a weak acid- strong base titration
Second derivative of a weak acid - strong base titration
Using Derivatives of Titration Curve to Determine the End Point
Using Derivatives of Titration Curve to Determine the End Point
13
)(10
bea
A
)( 1
10
)(10
ae
BH
B
a
• Isoionic Point: The pH of pure, neutral polyprotic acid (the natural zwitterion) – HAwater: H2A+, A-, H+, and OH-
– Most HA and [H2A+][A-] • Isoelectric Point: pH at which the average
charge of the polyprotic acid is zero (amino acids) – Most HA and [H2A+]=[A-]
Acid-Base Titrations
Titration: Increments of reagent solution-the titrant-are added to analyte until their reaction is complete.
Titration type/endpoint determination:
•Volumetric: volume of standard is measured •Gravimetric: mass of standard is measured, e.g., Mohr method (Cl-Ag-CrO4). •Coulometric: time at constant current is measured, i.e., amount of charge. •Spectrophotometric: Beer’s Law, the absorbance
Standard Solution (Standard Titrant): Reagent of known concentration that is used to carry out the analysis.
Titration: Performed slowly by adding standard solution from a buret until the reaction is judged to be complete.
Terms
Indicators: Added to a tritration to give an observable physical change at or near the equivalence point. Usually a color or turbidity change.
Equivalence Point: The point when the amount of added titrant is chemically equivalent to the amount of analyte in the sample.
End Point: The point at which an indicator physically changes representing the estimated equivalence point (color or turbidity change).
Titration error = Vep - Veq
Terms
Back-titration: Add a known excess of one standard titrant to the analyte, and then titrate a second standard titrant to determine the excess reagent. May be required when a reaction is slow or the standard solution lacks stability.
Blank titration: Carry out the same procedure without analyte. To estimate the titration error.
Terms
2
1. Sufficiently stable (only need to determine concentration once).
2. React rapidly with the analyte (less time between additions).
3. React more or less completely with the analyte (satisfactory end points).
4. Undergo a selective reaction with analyte described by a simple chemical equation
Standard Solutions (Titrants) Standard Solutions (Titrants)
Methods for Establishing the Concentration of Standard Solutions
1. Direct Method: A carefully weighed quantity of primary standard is diluted to a known volume. First choice if possible.
2. Standardization: A titrant to be standardized is used to titrate a weighed quantity of a primary, secondary standard or a known volume of another standard solution.
Primary Standard: Highly purified compound that serves as a reference material. • High purity, stability (air, heat, vacuum), • Standard solution (direct method) • To standardize a standard solution (standardization)
Secondary Standard: A reference material (a compound) whose purity has been established by a chemical analysis. Often used in lieu of a primary standard because there are not many primary standards.
Standard Solutions (Titrants) Summary titration titrant analyte indicator equivalence point vs. end point titration error blank titration Standard: Primary standard, second standard, standard solution
Acid-Base Titrations
How does the pH change as titrant is added?
Titration Curve: A graph showing how the concentration of one of the reactants varies as titrant is added. (how the pH changes as titrant is added in acid and base titration)
Titration of Strong Base with Strong Acid
3
Titration of Strong Base with Strong Acid
1. Before the equivalence point: excess OH-
2. At the equivalence point: pH = 7.00 3. After the equivalence point: excess H+
0.00, initial point 10.00 ml, 50% titration 19.98 ml, half-drop before equivalence point 20.00 ml, 100% titration (equivalence point) 20.02 ml, half-drop after equivalence point 40.00 ml, 200% titration (far-over)
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
Titration of Strong Acid with Strong Base
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
At 0.00 mL of NaOH added, initial point
[H3O+] = CHCl = 0.1000 M
at 10.00 mL of NaOH added, Va * Ma > Vb * Mb
(Va Ma) - (Vb Mb) [H3O+] = ----------------------------
(Va + Vb)
((20.00mL)*(0.1000M)) - ((10.00mL)*(0.1000M)) = ------------------------------------------------------------
(20.00 + 10.00)mL
= 3.33x10-2M pH =1.48
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
at 19.98 mL of NaOH added, Va * Ma > Vb * Mb
(Va Ma) - (Vb Mb) [H3O+] = ----------------------------
(Va + Vb)
((20.00mL)*(0.1000M)) - ((19.98mL)*(0.1000M)) = ------------------------------------------------------------
(20.00 + 19.98)mL
=5.0x10-5M pH =4.30
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
at 20.00 mL of NaOH added Va * Ma = Vb * Mb , equivalence point at equivalence point of a strong acid - strong base titration pH = 7.00
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
4
at 20.02 mL of NaOH added, Vb * Mb > Va * Ma , post equivalence point
(Vb * Mb) - (Va * Ma) [OH-] = ---------------------------
(Va + Vb) ((20.02mL)*(0.1000M)) - ((20.00mL)*(0.1000M))
=5.0x10-5M pOH =4.30 pH=9.70
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
at 40.00 mL of NaOH added, Vb * Mb > Va * Ma , post equivalence point
(Vb * Mb) - (Va * Ma) [OH-] = ---------------------------
(Va + Vb) ((40.00mL)*(0.1000M)) - ((20.00mL)*(0.1000M))
=3.33x10-2M, pOH =1.48 pH=12.52
EXAMPLE: Derive the titration curve for the titration of 20.00 mL of 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mL of 0.1000 M NaOH.
Strong Acid - Strong Base Titration
0
2
4
6
8
10
12
14
Volume of Base Added
(mL)
The Effect of Concentration • The change in pH at the equivalence point is less pronounced at decreased concentrations. • The pH at the equivalence point is still 7.
(M)
Titration of Strong Acid with Strong Base Titration of Weak Acid with Strong Base
Titration of Weak Acid with Strong Base
Titration of Weak Acid with Strong Base
•The pH change at the equivalence point decreases (the end point become less distinct) as the acid becomes weaker or more dilute
•The Effect of Concentration: the pH at the equivalence point is lower for lower initial weak acid concentrations and initial pH is higher.
5
Titration of Weak Acid with Strong Base
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5 M
0.00, initial point 15.00 ml, before equivalence point 35.00 ml, 100% titration (equivalence point) 50.00 ml, after equivalence point
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5 M
at 0.00 mL of NaOH added, initial point
[H+] Ka Ca
= (1.75 x 10-5 M) x (0.1000 M) = 1.32 x 10-3 M pH = 2.878
at 15.00 mL of NaOH added, Va Ca > Vb Cb
(Va Ca) - (Vb Cb) [HAc]ex. = ------------------------------
(Va + Vb) [(35.00 mL)x(0.1000 M)) - ((15.00 mL)x(0.1000M)]
= ----------------------------------------------------------------- (35.00 + 15.00) mL
= 4.000x10-2 M
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5 M
at 15.00 mL of NaOH added, Va Ca > Vb Cb
[HAc]ex = 4.000x10-2 M
(35.00 + 15.00) mL
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5 M
at 15.00 mL of NaOH added, Va Ca > Vb Cb
[HAc]ex = 4.000 x 10-2 M [Ac-] = 3.000 x 10-2 M
[HAc]ex [H+] = Ka ------------------
[ Ac-] 4.000 x10-2 M
= 1.75 x 10-5 M ------------------- = 2.33 x 10-5 M 3.000 x10-2 M
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5 M
at 15.00 mL of NaOH added, Va Ca > Vb Cb
[HAc]ex = 4.000 x 10-2 M
[Ac-] = 3.000 x 10-2 M
[H+] = 2.33 x 10-5 M pH = 4.632
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5 M
632.4 10000.4 10000.3
6
at 35.00 mL of NaOH added, Va Ca = Vb Cb , equivalence point
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc 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 Ca = Vb Cb , equivalence point
Kw VaCa[OH-] = ----------------- Ka (Va + Vb)
Kb Cb
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5M
1.00 x10-14 M2 * (35.00mL)(0.1000M) [OH-] = -----------------------------------------------
1.75 x10-5 M * (35.00 + 35.00)mL
= 5.34x10-6 M
at 35.00 mL of NaOH added, Va * Ca = Vb * Cb , equivalence point
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc 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 Ca = Vb Cb , equivalence point
[OH-] = 5.34 x 10-6 M
pOH = 5.272 pH = 8.728
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5M
at 50.00 mL of NaOH added, Vb Cb > Va Ca , post equivalence point
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5M
at 50.00 mL of NaOH added, Vb Cb > Va Ca , post equivalence point
(Vb Cb) - (Va Ca)[OH-] = -------------------------- (Va + Vb)
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5M
7
at 50.00 mL of NaOH added, Vb Cb > Va Ca
(Vb Cb) - (Va Ca)[OH-] = -------------------------- (Va + Vb)
[(50.00mL)*(0.1000M)) - ((35.00mL)*(0.1000M)] = -------------------------------------------------------------
(35.00 + 50.00) mL = 1.765 x 10-2 M
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5M
at 50.00 mL of NaOH added, Vb Cb > Va Ca
[OH-] = 1.765 x 10-2M pOH = 1.7533 pH = 14.00 - 1.7533 = 12.25
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5M
(a) Which of the two curves represents the titration of a strong acid, and which a weak acid? (b) What is the approximate pH at the equivalence point for each of the acids? (c) What is the approximate pKa of the weak acid?
The following plot shows two titration curves, each representing the titration of 50.0 mL of 0.100 M acid with 0.100 M NaOH:
Vb (mL)
Acid - Strong Base Titration Curve
Much 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
Volume of Acid Added
(mL)
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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
Diprotic Systems: Titration of Weak Acid with Strong Base
0.00, Initial point 15.00 ml, Before the 1st equivalence point 35.00 ml, At the 1st equivalence point 50.00 ml, After the 1st equivalence point 70.00 ml, At the 2nd equivalence point 90.00 ml, After the 2nd equivalence point
8
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 0.00 mL of NaOH added, initial point
pH = 1.632
Ka
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 Ca > Vb Cb, the 1st buffer region
(Va Ca) - (Vb Cb) [H2A]ex = ----------------------------
(Va + Vb)
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca > Vb · Cb, the 1st buffer region
[(35.00 mL)·(0.1000 M) - (15.00 mL)·(0.1000 M)] [H2A]ex = ----------------------------------------------------------------
(35.00 + 15.00) mL = 4.000x10-2M
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca > Vb · Cb, the 1st buffer region
[H2A]ex = 4.000x10-2 M
(Va + Vb) (35.00 + 15.00)mL
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca > Vb · Cb, the 1st buffer region
[H2A]ex = 4.000 x 10-2 M; [HA-] = 3.000 x 10-2 M [H2A]ex 4.000 x 10-2 M
[H+] = Ka1 x -------------- = 7.11x10-3 M x ------------------ [HA-] 3.000 x 10-2 M
= 9.48 x 10-3 M pH = 2.023
023.2 10000.4 10000.3
M M
AH HA
pKpH a
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca = Vb · Cb, the 1st equivalence point
[H+] Ka1 · Ka2
= (7.11 x 10-3 M)(6.34 x 10-8 M) = 2.12 x 10-5 M pH = 4.673
9
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca) > Vb · Cb, the 2nd buffer region
2 (Va · Ca) - (Vb · Cb)
[HA-]ex = ------------------------------- (Va + Vb)
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca) > Vb · Cb, the 2nd buffer region
[2(35.00mL)*(0.1000M)) - ((50.00mL)*(0.1000M)] [HA-]ex = ---------------------------------------------------------------
= 2.353x10-2M
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca) > Vb · Cb, the 2nd buffer region [HA-]ex = 2.353x10-2M
(Vb · Cb) - (Va · Ca) [A2-] = ---------------------------
(Va + Vb)
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca) > Vb · Cb, the 2nd buffer region [HA-]ex = 2.353x10-2M
(50.00 mL)(0.1000 M) - (35.00 mL)(0.1000 M) [A2-] = ---------------------------------------------------------
(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 H2A 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 · Ca) > Vb · Cb, the 2nd buffer region [HA-]ex = 2.353x10-2 M [A2-] = 1.764x10-2 M
Ka2 [HA-]ex 6.34 x 10-8 M x 2.353x10-2 M [H+] = ------------------ = ------------------------------------
[A2-] 1.764x10-2 M = 8.46x 10-8 M
pH = 7.073
pKpH
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca) = Vb · Cb, the 2nd equivalence point
Kw · Va · Ca [OH-] = ---------------------
10
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 · Ca) = Vb · Cb, the 2nd equivalence point
1.00 x 10-14 M2 x (35.00 mL)x(0.1000 M) [OH-] = ---------------------------------------------------- = 7.25x10-5 M
6.34 x 10-8 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 H2A 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 90.00 mL of NaOH added, Vb · Cb > 2(Va · Ca), after 2nd equivalence point
Vb · Cb - 2(Va · Ca) [OH-] = -----------------------------------
(Va + Vb)
EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A 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 90.00 mL of NaOH added, Vb · Cb > 2(Va · Ca), after 2nd equivalence point
(90.00 mL)(0.1000 M) - 2 x (35.00 mL)(0.1000 M) [OH-] = ---------------------------------------------------------------
(35.00 + 90.00) mL
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 Diprotic Systems: Titration of Weak Acid Base with Strong Base
(mL)
Standard Solutions for Acid/Base Titrations
Standard acid or base reagents are always strong acids or bases • HCl, HClO4, H2SO4, NaOH, and KOH • They yield sharper end points. • Weak reagents react incompletely. • Nitric acid and hot sulfuric and perchloric yield undesirable
ox/red side reactions.
• Tris (THAM) – tris(hydroxymethyl)aminomethane, (HOCH2)3CNH2
• Sodium tetraborate decahydrate, Na2B4O7.10H2O
1. Indicators 2. Derivatives of titration curve 3. Gran Plot
Acid/Base Indicators • Weak acids or bases themselves • The conjugate forms differs in color
HIn In- + H+
In + H2O ⇔ InH+ + OH-
base acid Color1 Color2
base acid Color1 Color2
[InH+][OH-] Kb = --------------- (2)
[In-] Ka------ = ---------- [HIn] [H+]
Indicators Indicator Behavior acid color shows when [In-] 1 [H+][In-] 1 ------ <= ---- ------------- = ---*[H+] = Ka[HIn] 10 [HIn] 10
base color shows when [In-] 10 [H+][In-] ------ >= ----- ------------- = 10*[H+] = Ka[HIn] 1 [HIn]
12
Indicators
and base color shows when pH - 1 = pKa
Color change range is pH = pKa + 1 (Transition range)
Choosing the Proper Indicator
• Color change range should be in area where titration curve is most vertical.
• Indicator error: the difference between the observed end point (color change) and the true equivalence point.
Common Indicators
Using Derivatives of Titration Curve to Determine the End Point
Titration curve for a polyprotic acid - strong base titration.
First derivative of a weak acid- strong base titration
Second derivative of a weak acid - strong base titration
Using Derivatives of Titration Curve to Determine the End Point
Using Derivatives of Titration Curve to Determine the End Point
13
)(10
bea
A
)( 1
10
)(10
ae
BH
B
a