Ionic EquilibriumHard
Question
What will be the pH at the equivalence point during the titration of a 100 mL 0.2 M solution of CH3COONa with 0.2 M solution of HCl ? Ka = 2 × 10-5.
Options
A.3 - log√2
B.3 + log√2
C.3 - log 2
D.3 + log 2
Solution
CH3COONa + HCI → NaCI + CH3COOH
t=0 20 m eq. 20 meq.
teq. - - 20 meq.
[CH3COOH] =
= 0.1 M
pH =
[pKa - log C] = 
[5 - log2 + 1] = 
[6 - log2] = 3 - log√2
t=0 20 m eq. 20 meq.
teq. - - 20 meq.
[CH3COOH] =
= 0.1 MpH =
[pKa - log C] = [5 - log2 + 1] = [6 - log2] = 3 - log√2Create a free account to view solution
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