Indicators for Acid-Base Titrations (Sec. 9-6)

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 - = In 2-

Ch 10: Acid-Base Titrations phenolthalein Automated titrators determine the endpoint electronically by numerically calculating the 2 nd derivative

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

I. Strong Acid-Strong Base Titration Curves (Sec. 10-1) equivalence pt. volume: 50 mL of M HCl is titrated with 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  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

methyl red phenolthalein

Titration Error phenolthalein mL/50 mL =0.04% error!

II. Weak Acid-Strong Base Titration Curve (Sec. 10-2) HA = H + + A - 50 mL of a M soln of the weak acid HA, K a = 1.0 x 10 -5, is titrated with 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.

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

Ch 11: Titrations in Diprotic Systems 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 = = 4.7 x K a2 = = 1.8 x 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 +

Calculate the pH of 0.050M H 2 L +

2. The basic form L - K a1 = = 4.7 x K a2 = = 1.8 x Strengths of conjugate bases: for L - K b1 = K w /K a2 = 1.01 x /1.8 x = 5.61 x for HLK b2 = K w /K a1 = 1.01 x /4.7 x = 2.1 x Since the second conj. base HL is so weak, we'll assume all the OH- comes from the L - form.

Example: Calculate the pH of a 0.050M solution of sodium leucinate

The Intermediate Form The pH of a Zwitterion Solution - Leucine (HL form)

[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

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.

Titration Curve for the Amino Acid Leucine

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) =

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 M Na 2 CO 3 with M HCl. equivalence pt. volumes (V e1 & V e2 ) =

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) =

pt E: 2nd eq. pt (H 2 CO 3 treat as monoprotic weak acid) =

Buffers of Polyprotic Acids and Bases H3PO4HPO42- PO43-H2PO4-