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An in-depth exploration of acid-base titrations, focusing on indicators, strong and weak acids, and titration curves. Topics include the importance of matching the transition range with the endpoint pH, the role of weak acids as indicators, and the calculation of titration curves for strong acid-strong base and weak acid-strong base systems. The document also covers the concept of buffer regions and the pH dependence on the conjugate base.
Typology: Lecture notes
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transition range needs to match the endpoint pH as closely as possible in order to minimize titration error
Acid-Base indicators are themselves weak acids…..
e.g. phenolthalein
H 2 In = HIn-^ = In2-
equivalence pt. volume:
50 mL of 0.100 M HCl is titrated with 0.100 M NaOH. Calculate the titration curve for the analysis.
(^1) Initial pH
2 pH before the equivalence pt.
3 pH at the equivalence pt.
4 pH after the equivalence pt.
mL base
[H+^ ] = C (^) HA so pH = -log C (^) HA
Strong Acid - Strong Base Titration (both monoprotic) (analyte) (titrant)
Eq. Pt. pH = 7
[H +^ ] = MaVa - MbVb Vtotal
[OH-] = Mb(Vb beyond eq.pt.) Vtotal
K (^) a
50 mL of a 0.100 M soln of the weak acid HA, Ka = 1.0 x 10 -5^ , is titrated with 0.100 M NaOH. Calculate the titration curve for the analysis.
equivalence pt. volume:
(^1) Initial pH
2 pH before the equivalence pt.
4 pH after the equivalence pt. = same as SA-SB titration
3 pH at the equivalence pt.
Biological Applications - Amino Acids (Sec. 11-1)
low pH high pH
R = (CH 3 ) 2 CHCH 2 -
Finding the pH in Diprotic Systems (Sec. 11-2)
The strength of H 2 L +^ as an acid is much, much greater than HL -
K (^) a1 = 10 -2.328^ = 4.7 x 10 -
K (^) a2 = 10 -9.744^ = 1.8 x 10 -
So assume the pH depends only on H 2 L +^ and ignore the contribution of H+^ from HL.
K (^) a1 = 10 -2.328^ = 4.7 x 10 -3^ K (^) a2 = 10 -9.744^ = 1.8 x 10 -
Strengths of conjugate bases:
for L -^ K (^) b1 = K (^) w/K (^) a2 = 1.01 x 10 -14/1.8 x 10 -10^ = 5.61 x 10 -
for HL K (^) b2 = K (^) w/K (^) a1 = 1.01 x 10 -14/4.7 x 10 -3^ = 2.1 x 10 -
Since the second conj. base HL is so weak, we'll assume all the OH- comes from the L-^ form.
[H +] 2 = K (^) a1 K (^) a
-log [H+] 2 = - log K (^) a1 - log K (^) a
2 pH = pK (^) a1 + pK (^) a
a1 HL
a1 a2 HL w a
assume:
K (^) wK (^) a1 << K (^) a1K (^) a2CHL K (^) a1 << C (^) HL
a1 a HL
pH of a solution of a diprotic zwitterion
Example: pH of the Intermediate Form of a Diprotic Acid
Potassium hydrogen phthalate, KHP, is a salt of the intermediate form of phthalic acid. Calculate the pH of 0.10M KHP and 0.010M KHP.
pt C: 1 st^ eq. pt (HL) =
pt E: 2nd eq. pt (L - ) =
Example p. 233: Titration of Sodium Carbonate (soda ash)
Calculate the titration curve for the titration of 50.0 mL of 0.020 M Na 2 CO 3 with 0.100 M HCl.
equivalence pt. volumes (Ve1 & Ve2) =
Fractional Composition Diagram H 3 PO 4
0 2 4 6 8 10 12 14 pH
alpha
H3PO4 H2PO4- HPO42- PO43-