lecture notes2

Mol hcl added 60 moll00050 l 0030 mol init mol c2h3o2

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: mol/L)(0.0050 L) = 0.030 mol init mol C2H3O2- = (0.201 mol/L)(0.400 L) = 0.0804 mol init mol HC2H3O2 = (0.195 mol/L)(0.400 L) = 0.078 mol init mol mol after rxn H3O+ + C2H3O20.030 0.0804 0 0.0504 6 HC2H3O2 + H2O 0.078 0.108 [HC2H3O2] = 0.108 mol/0.405 L = 0.267 M [C2H3O2-] = 0.0504 mol/0.405 L = 0.124 M HC2H3O2 + H2O init 0.267 M equil ~0.267 X H3O+ + C2H3O2~10-7 M x x = 3.68 x 10-5 M = [H3O+] 0.124 M ~0.124 Y pH = 4.43 Chem 102 D. Miller Solutions to Review Problems for Acid/Base Chemistry 4. (continued) The added NaOH is neutralized by the weak acid and a new buffer is formed. mol NaOH added = (6.0 mol/L)(0.0050 L) = 0.030 mol init mol HC2H3O2 = (0.195 mol/L)(0.400 L) = 0.078 mol init mol C2H3O2- = (0.201 mol/L)(0.400 L) = 0.0804 mol init mol mol after rxn OH- + HC2H3O2 0.030 0.078 0 0.048 6 C2H3O2- + H2O 0.0804 0.1104 [HC2H3O2] = 0.048 mol/0.405 L = 0.12 M [C2H3O2-] =...
View Full Document

This document was uploaded on 03/06/2014 for the course CHEM 102 at CSU Northridge.

Ask a homework question - tutors are online