5 x 10 3 mol acid 003 l 50 x 10 2 m ha oh io

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Unformatted text preview: l = 1.5 x 10-3 mol acid ÷ 0.03 L = 5.0 x 10-2 M = [HA] [OH-] = [IO-] formed = [A-] Henderson-Hasselbalch Equation Chemistry 102 Quiz 5 Name ___Mr. Perfect_______________________________ Date ____F 13________ c. 12.5 mL of base (4 pts) (at equivalence point) NaOH 0.0125 L x 0.200 mol/L = 2.5 x 10-3 mol base molbase = molacid *All of the weak acid HIO has been converted to the conjugate base IO-. Total Volume = 25.0 mL + 12.5 mL = 37.5 mL or 0.0375 L IO- + H2O ⇄ HIO + OHI 0.0667 0 0 C -x +x +x E 0.0667-x x x Check Solve Quadratic: x2 + 4.0 x 10-4x – 2.67 x 10-5 = 0 √ 2 solutions: x = 0.00495 M or x = - 0.00535 M x = [OH-] = 4.95 x 10-3 M pOH = -log (4.95 x 10-3) = 2.31 pH = 14 – 2.31 = 11.69 Chemistry 102 Quiz 5...
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This document was uploaded on 03/19/2014 for the course CHEMISTRY 102 at Los Angeles Harbor College.

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