Ionic EquilibriumHard
Question
Aniline behaves as a weak base. When 0.1 M, 50 ml solution of aniline was mixed with 0.1 M, 25 ml solution of HCl the pH of resulting solution was 8. Then the pH of 0.01 M solution of aniliniumchloride will be (Kw = 10-14)
Options
A.6
B.6.5
C.5
D.5.5
Solution
C6H5NH2 + H+ C6H5NH3+
t = 0 5 2.5
teq 2.5 - 2.5
pOH = pKa = 14 - 8 = 6
Now for the solution of [C6H5NH3+] = 0.01 M
pH = 7 -
pKa - 
log C = 7 - 
log (0.01) = 5
t = 0 5 2.5
teq 2.5 - 2.5
pOH = pKa = 14 - 8 = 6
Now for the solution of [C6H5NH3+] = 0.01 M
pH = 7 -
pKa - log C = 7 - log (0.01) = 5Create a free account to view solution
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