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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Sis290025 on March 24, 2006, 06:56:13 AM

Title: What is pH at titration's equivalence point?
Post by: Sis290025 on March 24, 2006, 06:56:13 AM

Consider an experiment where 54.100 mL of 0.17500M acetic acid (CH3CO2H) is titrated with 0.61100 M NaOH. What is the pH at the equivalence point of this titration? Ka(CH3CO2H)=1.8 x 10-5


At equivalence point:
(MV)_CH3CO2H = (MV)_NaOH

Volume_NaOH = (54.1 mL*0.175 M)/(0.611 M) = 15.49509002 mL NaOH at point

---------CH3COOH + NaOH --> NaCH3COO + H2, where:
final-----0 mmol-----0 mmol---9.4675 mmol

54.1 mL*0.175 M = 9.4675 mmol

CH3COO- + H2O <-> OH- + CH3COOH

K_b = [OH-][CH3COOH]/[CH3COO-] = 5.555E-10

K_b = K_w/K_a

 5.555E-10 = X^2/[0.136232646-x], where initial [CH3COO-] = 9.4675 mmol/(54 ml + 15.49509002 mL) = 0.136232646 M

x = sqrt(5.555E-10*0.136232646) = 8.699701336E-6 = [OH-] Based on approximation

pOH = -log[8.699701336E-6 ] = 5.060495657
pH = 14 - 5.060495657 = 8.9395 = 8.94

Thanks.

Title: Re:What is pH at titration's equivalence point?
Post by: AWK on March 24, 2006, 07:40:32 AM

OK