M3LC/H2LC Week 4

M3LC/H2LC Week 4
David Fabian
April 21, 2014
1
Unknown concentration
2
Acids & Bases, Titrations & Buffers
indicator
NaOH + CH3COOH NaCH3COO + H2O
3
Acids & Bases, Titrations & Buffers
•
A weak acid (HA) dissociates in water according to equation:
(1) HA (aq) + H2O(l) H3O+(aq) + A–(aq) Ka = [H3O+] [A–] / [HA]
•
The acid dissociation constant is algebraically rearranged:
(2) [H3O+] = Ka [HA] / [A‐]
•
Taking the negative log of both sides of equation (2) results in the derivation of the Henderson‐Hasselbalch equation for a weak acid:
(3) pH = pKa + log ([A–] / [HA])
If [HA] = [A–], the ratio [A–] / [HA] = 1 and log (1) is equal to zero simplifying equation (3):
(4) pH = pKa (half‐way to equivalence point)
•
4
Equivalence Point
• Inflection point
• used to pick an indicator for titration
• Enough titrant has been added to react with all of the analyte present
• Reaction has gone to completion – only products remain in flask
• Knowing concentration and volume of titrant added, can determine number of moles needed to stoichiometrically react with analyte
5
Endpoint and Indicators
• pH at which solution changes color
– Property of indicator, not a function of titrant or analyte
• Most indicators are weak acids in solution
• Bromothymol blue indicator (pKa = 7.10)
6
Spectrophotometric Determination of pKa
• The amount of each species can be quantified by visible spectroscopy
Increasing pH
Increasing pH
Isosbestic point – ensures
we are only dealing with two species (In‐ and HIn)
• Determine pKa of acetic acid and bromothymol blue
‐ Henderson‐Hasselbalch equation
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[In‐]/[HIn] = (Aλ1, acidic∙Aλ2)/(Aλ2, basic∙Aλ1)
(1) A λ1 = ε(HIn,λ1)* b * [HIn]
(2) A λ2 = ε(In_,λ2) * b * [In‐] λ1 = HIn absorbs strongly, not In‐; λ2 = In‐ absorbs strongly, not Hin
λ2
λ1
(3) CT = [HIn] + [In‐]
(4) Low pH: A λ1, acidic = ε(HIn,λ1)* b * CT
(5) High pH: A λ2, basic = ε(In_,λ2) * b * CT
Taking a ratio of (1) to (4) and (2) to (5):
(6) A λ1 / A λ1, acidic = [HIn]/ CT
(7) A λ2 / A λ2, basic = [In‐]/ CT
Henderson‐Hasselbalch:
(7) Divided by (6)  [In‐]/[HIn] = (Aλ1, acidic∙Aλ2)/(Aλ2, basic∙Aλ1)
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