Ka by Titration

Ka by Titration - Part B: Unknown Acid Equivalence Pt....

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10 November 2010 Determination of Ka by Titration Part A: Acetic acid Equivalence Pt. (20.50, 8.55) Half-way Pt. (10.25, 4.90) 1. pH=pKa ( 4.90), [Ka]=10 -4.90 => 1.26e-5 2. Ka (Acetic Acid = 1.7e-5) & ( pka= -log( 1.7e-5)= 4.77 %error= ((|4.77-4.90|)/ (4.77))*100%= 2.73% 3. M acid * ml acid = M NaOH *ml NaOH X * 20.00 = 0.1059 * 20.50 X= 0.108548 M X= .1085 M
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Unformatted text preview: Part B: Unknown Acid Equivalence Pt. (20.50, 8.55) Half-way Pt. (10.25, 5.00) 1. pH=pKa ( 5.00) , [Ka]=10-5.00 => 1.00e-5 2. Ka (Propioic Acid= 1.3e-5) & ( pka= -log(1.3e-5)= 4.89 %error= ((|4.89-5.00|)/ (4.89))*100%= 2.33% 3. M acid * ml acid = M NaOH *ml NaOH X * 20.00 = 0.1059 * 20.50 X= 0.108548 M X= .1085 M...
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This note was uploaded on 03/19/2011 for the course CHM 152 taught by Professor Allen during the Fall '08 term at Wake Tech.

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Ka by Titration - Part B: Unknown Acid Equivalence Pt....

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