Question

What is the pH of 100 mL of 0.10 M C6H5NH3+Cl-? Kb = 7.4 x 10-10 for C6H5NH2.   Kw = 1.0 x 10-14

I think it would dissociate into C6H5NH2 and HCl, but HCl is a strong acid, so it would just dissociate again....?

Answer 1

Ka = Kw/Kb

= 1.0*10^-14 / (7.4*10^-10)

= 1.35*10^-5

C6H5NH3+Cl- would dissociate into C6H5NH2 and HCl, but HCl is a strong acid, so it would just dissociate again into H+ and Cl-

So, I will just focus on H+ and ignore Cl-

C6H5NH3+ <——> C6H5NH2 + H+

0.10 0 0. (initial)

0.10-x x x (at equilibrium )

Ka = x*x/(0.10-x)

1.35*10^-5 = x^2 / (0.10-x)

Since Ka is small, x will be small and can be ignored as compared to 0.10

Above expression thus becomes,

1.35*10^-5 = x^2 / (0.10)

x = 1.16*10^-3

So, [H+] = x = 1.16*10^-3 M

pH = -log [H+]

= -log (1.16*10^-3)

= 2.93

Answer: 2.93