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)

• A. 6
• B. 6.5
• C. 5
• D. 5.5

Answer 1

Answer: (C)

5

Answer 2

C6H5NH2 + H+ $\longrightarrow$ C6H5NH3+ a

t=0 5 2.5

teq 2.5 – 2.5

pOH = pKa = 14 – 8 = 6

$\therefore$ pKa = 6

Now for the solution of [C6H5NH3+] (vc [C6H5NH3+]

= 0.01 M

pH = 7 – $\frac{1}{2}$pKa – $\frac{1}{2}$ log C = 7 – $\frac{6}{2}$ – $\frac{1}{2}$ log (0.01) = 5