Question

The salt formed by the reaction of the weak base methylamine, CH3NH2, with the strong acid nitric acid is methylammonium nitrate, CH3NH3NO3. What is the pH of a 0.088 M solution of methylammonium nitrate at 25∘C given that the value of Kb for methylamine is 4.4×10−4?

Answer 1

The salt of a weak base and a strong acid-

CH₃NH₃⁺NO₃^- = CH3NH2 + HNO3

Sice the salt is of strong acid and weak base, hence pH = 1/2 pK_w - 1/2 log C - 1/2 pK_b

where C= conc of the salt = 0.088 (given)

Kw = dissociation constant of water = 10^-14

Kb= dissociation constant of base = 4.4 * 10^-4 (given)

Now, we know- pK_w = -log K_w = -log 1.008 *10 ^-14 = 13.997 = 14

pKb = - log Kb = - log (4.4 * 10^-4) = -log 4.4 + 4 = -0.64 + 4 = 3.35

log C = log 0.88 = - 1.05

Then, pH of salt = 1/2 pK_w - 1/2 log C - 1/2 pK_b

= ( 14/2) - (-1.05/2 ) - ( 3.35/2 )

= 7 + 0.525 - 1.675

= 5.85

Here we find that the pH if 5.85 i.e less than 7.

Since the salt is of weak base and strong acid, the salt solution will be an acidic solution . So the pH will be less than 7.