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
View Solution FreeMore Ionic Equilibrium Questions
The solubility of Fe(OH)3 would be maximum in...Which does not give iodoform reaction :-...Which of the following is the strongest base :-...pH at which a basic indicator with Kb = 1.0 × 10-10 changes colour when the concentration of indicator is 10-2 M :-...A recent investigation of the complexation of SCN– with Fe3+ led to values of 125, 20 and 1.0 for K1, K2 and K3, respect...