Understanding Acid-Base Titrations: Indicators, Acids, and Curves, Lecture notes of Chemistry

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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-^ = In2-

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

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

II. Weak Acid-Strong Base

Titration Curve (Sec. 10-2)

HA = H +^ + A-

[HA]

[H][A ]

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.

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 = 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.

1. The acidic form H 2 L +

Calculate the pH of 0.050M H 2 L+

2. The basic form L -

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

K C

K K C K K

[H ]



assume:

K (^) wK (^) a1 << K (^) a1K (^) a2CHL K (^) a1 << C (^) HL

a1 a HL

a1 a2 HL so [H ] K K

C

K K C

[H ]  



pK pK

pH a1^ a

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) =

Buffers of Polyprotic Acids and Bases

Fractional Composition Diagram H 3 PO 4

0 2 4 6 8 10 12 14 pH

alpha

H3PO4 H2PO4- HPO42- PO43-