3 x 10 10 aniline 0025 l x 015 moll 375 x 10 3 mol

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Unformatted text preview: of HCl: (15 pts) Kb = 4.3 x 10-10 aniline: 0.025 L x 0.15 mol/L = 3.75 x 10-3 mol base a. 10.00 mL of HCl. (before equivalence point) HCl: 0.010 L x 0.10 mol/L = 1.00 x 10-3 mol acid 3.75 x 10-3 mol – 1.00 x 10-3 mol = 2.75 x 10-3 mol base remains b. 37.5 mL of HCl. (at equivalence point) molesa = molesb 0.0375 L x 0.10 mol/L = 3.75 x 10-3 mol acid I C E C6H5NH3+ 0.06 -x 0.06 – x ⇄ C6H5NH2 + H+ 0 0 +x +x x x Check: c. 45.00 mL of HCl. (past equivalence point) 0.045 L x 0.10 mol /L = 4.50 x 10...
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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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