Fundamentals of Analytical Chemistry

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Alexandrina Wilkins
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1 Homework Fundamentals of Analytical Chemistry Chapter Principles of Neutralization Titrations 3, odd, 1, 19, 1, 5, 7, 3, 3, 3, 3,, 5 Problems in green may or may not be part of the assignment, depending upon the material that is covered. Indicators Acid/Base indicators Acid form has a different color than the base form of the substance x more acid (or greater) acid color x more base (or greater) base color Between these limits color a combination For ideal behavior, need transition through colors quickly Weak acid/base substances From Henderson-Hasselbalch Hasselbalch, = for transition Indicators Titration Error Titrant reacts with and indicator Minimize by using low concentration of indicator MUST BE highly colored for this to work Ideally, less than 0.0 ml of titrant to cause change Can also correct with blank Table -1 List of indicators pka can be used to chose correct indicator Match pka with of equivalence point Strong Acid/Strong Base Titrations In any titration, we can consider three phases Before the equivalence point, the is in excess At the equivalence point, the and titrant are present in stoichiometrically equivalent amounts. After the equivalence point, the titrant is in excess Titrations To determine, we have to look at the composition of the solution after the reaction has occurred. I starting conditions How much stuff do we have before any reaction occurs stoichiometric relationships HOW will the system react Must follow the stoichiometry F final conditions What s s in solution after the reaction has gone to completion 1

2 Strong Acid/Strong Base Titrations Consider the following titration: 50.0 ml of 0. M HCl is titrated Before any titrant is added: Solution is just the strong acid (HCl( HCl) = -log 0. = 1.00 After 5.00 ml of titrant is added: HCl + NaOH H O + Na + + Cl - I lots F.50 ~0 = -log (.50/55) = 1.09 [NOTE: ~~ means we don t t care because these substances do not affect the of the solution] After.00 ml of titrant is added: I ~~ F.00 ~0 = -log (.00/0) = 1.1 After 5.00 ml of titrant is added: I ~~ F.50 ~0 = -log (.50/75) = 1. After 5.00 ml of titrant is added: I ~~ F 0.50 ~0 = -log (0.50/95) =. After 9.9 ml of titrant is added: I ~~ F 0.00 ~0 = -log (0.00/99.9) =.70

3 After 50.0 ml of titrant is added: I lots 0 0 F ~0 ~0 lots ~~ ~~ What do we have affecting the? After 50.0 ml of titrant is added: I ~~ 0 0 F ~ poh = -log (0.00/0.0) =.70 = = 9.30 After 70.0 ml of titrant is added: I ~~ 0 0 F ~0.00 poh = -log (.00/) = 1.7 = = 1. Strong Acid/Strong Base Titrations Summary Prior to the equivalence point excess determined by (strong acid or strong base) At the equivalence point Nothing causing a from neutral = 7.00 ONLY FOR STRONG ACID/BASE TITRATION!!! Past the equivalence point excess titrant determined by titrant Strong acid/base Consider the following titration: 50.0 ml of a 0. M solution of a weak acid (HA, K a = 1.0 x -5 ) titrated with 0. M NaOH. Before any titrant is added Only a weak acid in solution + [ H ] = K C a HA (0.)(1.0 x + 3 [ H ] = 1.0x ; = = 5 ) 3

4 50.0 ml of a 0. M solution of a weak acid (HA, After 5.00 ml of titrant is added HA + NaOH A - + Na + +H O I ~~ ~~ ~~ ~~ F.50 ~ ~~ ~~ Buffer, = pk a + log b/a = log 0.50/.50 =.05 Differences from strong acid/base titration Cannot ignore the products Will always generate the conjugate base of the This produces a buffer solution! Reverse situation (weak base/strong acid) Will always generate the conjugate acid of the This produces a buffer solution! Will always titrate with either a strong acid or a strong base 50.0 ml of a 0. M solution of a weak acid (HA, After 5.0 ml of titrant is added HA + NaOH A - + Na + +H O I ~~ ~~ ~~ ~~ F.50 ~0.50 ~~ ~~ Buffer, = pk a + log b/a = log.50/.50 = 5.00 For this titration, 5.0 ml gets us ½ way to the equivalence point At this point in any weak/strong titration, we have the same amount of the conjugate acid and the conjugate base in solution. At this point will ALWAYS equal pk a! Note that with a weak base/strong acid titration we still have = pk a (and not pk b ) 50.0 ml of a 0. M solution of a weak acid (HA, After 50.0 ml of titrant is added HA + NaOH A - + Na + +H O I ~~ ~~ ~~ ~~ F ~0 ~ ~~ ~~ Only a weak base in solution [ OH [ OH ] = K C b B = 1.0x 1.0x ] = 7.1x ; poh = 5.15; =.5 5 At the equivalence point does not equal 7.00 Because the conjugate base is in solution, the is driven by this component and will be basic For a weak base/strong acid titration, you will have the conjugate acid in solution at the equivalence point, and the will be acidic

5 50.0 ml of a 0. M solution of a weak acid (HA, After 70.0 ml of titrant is added HA + NaOH A - + Na + + H O I ~~ ~~ ~~ ~~ F ~ ~~ ~~ Mixture of a weak base and a strong base Ignore the weak base; [OH - ] =.00/ poh = 1.7; = 1. Past the equivalence point We will assume (and usually properly) that the contribution to the hydroxide concentration from the weak base is negligible relative to the contribution from the strong base LeChatlier s Principle Weak Acid - Strong Base Strong Acid/ Strong Base Weak Acid/ Strong Base Strong acid/base Weak Acid - Strong Base

Strength of Vinegar by Acid-Base Titration

Strength of Vinegar by Acid-Base Titration Test Exercise 100 points 1? QUESTIONS? How are acid/base titrations conducted? What is standardization? How do you standardize a solution of a base? How, and