exam_3_practice_exam_questions_answers

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1 This document contains extra practice problems involving equilibrium, acid/base, buffer, and titration calculations. The exam will also include short answer conceptual questions on these topics. 1. A 0.0560 g quantity of CH 3 COOH is dissolved in enough water to make 50.0 mL of solution. Calculate the concentrations of H + , CH 3 COO - and CH 3 COOH at equilibrium. K a =1.8 x 10 -5 . CH 3 COOH=60 g/mol 0.0560 g=9.33 x 10 -4 mol 9.33 x 10 -4 mol/0.050 L=0.01867 M CH 3 COOH CH 3 COOH CH 3 COO - H 3 O + I 0.01867 0 0 C -x +x +x E 0.01867-x x x Ka=[ CH 3 COO - ][ H 3 O + ]/[CH 3 COOH] X=5.786 x 10 -4 [H 3 O + ]=[ CH 3 COO - ]=5.786 x 10 -4 M [CH 3 COOH]=0.0181 M 2. A titration in conducted where exactly 500 mL of 0.167 M NaOH was titrated into 500 mL 0.100 M CH 3 COOH. Calculate the pH of the solution at this point in the titration. CH 3 COOH Ka=1.8 x 10 -5 . Initial moles NaOH: 0.5 L x 0.167 mol/L=0.0835 mol CH 3 COOH : 0.5 L x 0.10 mol/L=0.050 mol CH 3 COOH + - OH Æ CH 3 COO - + H 2 O All acid is neutralized, 0.0335 mol - OH remaining 0.0335 mol/L=0.0335 M pOH=1.47 pH=12.53
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This note was uploaded on 04/15/2009 for the course LB 171L taught by Professor All during the Spring '08 term at Michigan State University.

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