How to do this pH calculation?

Question

What is the pH of a 0.80 M CH3NH3NO3 solution? Kb (CH3NH2) = 5.2 x 10^-4

Answer 1

To solve this problem you need to recognize that CH₃NH₃NO­₃ will completely ionize in solution:

CH₃NH₃NO­₃ → CH₃NH₃⁺ + NO₃^- (CH₃NH₃⁺ is analogous to NH₄⁺, with a methyl group replacing one of the hydrogens bonded to nitrogen)

CH₃NH₃⁺  is a Bronsted Lowry acid (proton donor) and can dissociate in water according to

 (1) CH₃NH₃⁺ ⇌  CH₃NH₂ + H⁺    The acid dissociation constant for this equilibrium is

 K_a = [CH₃NH₂][H⁺]/[ CH₃NH₃⁺]

We are not given the dissociation constant, K_a. However, we know that for a weak acid and its conjugate base, the relationship of K_a to K­_b is    

(2) K_aK­_b=K_w, where K_w = 10^-14 is the dissociation constant of water, H₂O ⇌ H⁺ + OH^- (if you are studying acid-base equilibria, it is a really good idea to memorize K_w !!)

So from eqn (2) above, we obtain K_a = (10_­^-14)/(5.2 x 10^-4) = 1.9 x 10^-11

Now you can set up your ICE table for the acid dissociation in eqn (1) above:

CH₃NH₃⁺ ⇌  CH₃NH₂ + H⁺

 Initial             0.80                 0           0

 Change          -x                   +x           +x

Equilibrium     0.80-x              x              x

[H⁺] at equilibrium is equal to x, so plugging the above values into the K_a expression from eqn (1) and using the K_a value you got from eqn (2), you have:

1.9 x 10^-11 = x²/(0.80-x)

x << 0.80M (use the 5% rule, that is if x is less than 5% of the initial concentration, you can ignore it to simplify the calculation. If it cannot be ignored, you can solve for x with the quadratic formula. But we don’t need to do that here)

so 1.9 x 10^-11 ≈ x²/0.80  and you obtain x = 3.9 x 10^-6 = [H⁺]

Take the -log of [H⁺] and you will have the pH of the solution.

In general, salts where the anion is not a base and the cation is the conjugate acid of a weak base (in this case, CH₃NH₂ is a weak base and CH₃NH₃⁺ is its conjugate acid) will produce an acidic solution. If you have correctly calculated the pH from the above [H⁺], you should see that pH < 7.