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
View Solution FreeMore Ionic Equilibrium Questions
When 20 mL of M/20 NaOH is added to 10 mL of M/10 HCl, the resulting solution will...The reverse process of neutralisation is:...Of the following which change will shift the reaction towards the product ?I2(g) ⇋ 2I(g). ᐃHro(298K) = + 150...Which of the following salts undergoes anionic hydrolysis?...Which of the following is the strongest base :-...