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Published byDiana Wood Modified over 9 years ago
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Indicators for Acid-Base Titrations (Sec. 9-6)
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transition range needs to match the endpoint pH as closely as possible in order to minimize titration error
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Acid-Base indicators are themselves weak acids….. e.g. phenolthalein H 2 In = HIn - = In 2-
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Ch 10: Acid-Base Titrations phenolthalein 8.0-9.6 Automated titrators determine the endpoint electronically by numerically calculating the 2 nd derivative
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Acid-Base Titrations Curves - pH (or pOH) as a function of mL of titrant added mL base pH analyte = strong acid titrant = strong base mL acid pH analyte = strong base titrant = strong acid 1 2 3 4
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I. Strong Acid-Strong Base Titration Curves (Sec. 10-1) 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
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2 pH before the equivalence pt. 3 pH at the equivalence pt.
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4 pH after the equivalence pt.
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mL base pH [H + ] = C HA so pH = -log C HA Strong Acid - Strong Base Titration (both monoprotic) (analyte) (titrant) Eq. Pt. pH = 7 [H + ] = M a V a - M b V b V total [OH-] = M b (V b beyond eq.pt.) V total
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methyl red 4.2-6.2 phenolthalein 8.0-9.6
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Titration Error phenolthalein 8.0-9.6 0.02 mL/50 mL =0.04% error!
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II. Weak Acid-Strong Base Titration Curve (Sec. 10-2) HA = H + + A - 50 mL of a 0.100 M soln of the weak acid HA, K a = 1.0 x 10 -5, is titrated with 0.100 M NaOH. Calculate the titration curve for the analysis.
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equivalence pt. volume: 1 Initial pH
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2 pH before the equivalence pt.
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4 pH after the equivalence pt. = same as SA-SB titration 3 pH at the equivalence pt.
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mL base pH Weak Acid - Strong Base Titration (both monoprotic) (analyte) (titrant) Eq. Pt. Hydrolysis of the conjugate base [OH-] = M b (V b beyond eq.pt.) V total Buffer region 1/2 eq. pt. pH = pK a
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Ch 11: Titrations in Diprotic Systems Biological Applications - Amino Acids (Sec. 11-1) low pH high pH R = (CH 3 ) 2 CHCH 2 -
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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 -3 K a2 = 10 -9.744 = 1.8 x 10 -10 So assume the pH depends only on H 2 L + and ignore the contribution of H + from HL. 1. The acidic form H 2 L +
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Calculate the pH of 0.050M H 2 L +
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2. The basic form L - K a1 = 10 -2.328 = 4.7 x 10 -3 K a2 = 10 -9.744 = 1.8 x 10 -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 -5 for HLK b2 = K w /K a1 = 1.01 x 10 -14 /4.7 x 10 -3 = 2.1 x 10 -12 Since the second conj. base HL is so weak, we'll assume all the OH- comes from the L - form.
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Example: Calculate the pH of a 0.050M solution of sodium leucinate
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The Intermediate Form The pH of a Zwitterion Solution - Leucine (HL form)
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[H + ] 2 = K a1 K a2 -log [H + ] 2 = - log K a1 - log K a2 2 pH = pK a1 + pK a2 assume: K w K a1 << K a1 K a2 C HL K a1 << C HL pH of a solution of a diprotic zwitterion
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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 pf 0.10M KHP and 0.010M KHP.
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Titration Curve for the Amino Acid Leucine
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equivalence pt. volumes (V e1 & V e2 ) = pts B and D: 1 st and 2 nd half eq. pt's = pt A: init. pH (H 2 L + treat as monoprotic weak acid) =
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pt C: 1 st eq. pt (HL) = pt E: 2nd eq. pt (L - ) =
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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 (V e1 & V e2 ) =
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pt C: 1 st eq. pt (HCO 3 - ) = pt E: 2nd eq. pt (H 2 CO 3 treat as monoprotic weak acid) = pts B and D: 1 st and 2 nd half eq. pt's = pt A: init. pH (CO 3 2- treat as monoprotic weak base) =
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pt E: 2nd eq. pt (H 2 CO 3 treat as monoprotic weak acid) =
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Buffers of Polyprotic Acids and Bases H3PO4HPO42- PO43-H2PO4-
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