DeathTest.com--Hydrolysis Problem

Problem: One day, while giving Fido a walk, you trip on a garden hose and an open 10g packet of ammonia acetate crystals (NH₄C₂H₃O₂, abbreviated NH₄OAc) falls out of your pocket into a 1.0 L puddle of water and dissolves completely. Before you can stop the dumb mutt, Fido laps up the puddle. Before rushing Fido to the vet, you remember that dogs can withstand pH levels of 4 and higher. You pull out your trusty pocket calculator and you go to work. What is the pH level of the puddle and do you need to rush Fido to the vet? K_a of HOAc = 1.8 E-5, K_b of NH₃ = 1.8 E-5

Solution: For the sake of Fido, let's get going. Okay, first, you gotta recognize that this is straight up hydrolysis problem. Why? Because a salt of a weak acid and a weak base have been added to water and both will react with the water, producing [OH¯] or [H+]. The trick to saving Fido's life is recognizing what reactions take place once the crystals have been added to water and then deciding which reactions are important. So you start by listing the obvious reactions:

NH₄OAc ---> NH₄⁺ + OAc¯ (dissociation reaction - not import)
NH₄⁺ + H₂O ---> NH₃ + H₃O⁺ (the hydration of the conjugate acid of NH₃...therefore, K_a = K_w / 1.8 E-5 = 5.6 E-10
OAc¯ + H₂O ---> HOAc + OH¯ (the hydration of the conjugate base of HOAc...therefore, K_b = K_w / 1.8 E-5 = 5.6 E-10)

In order to figure out which reaction to use, you remember Mr. McAfoos's advice to go with the reaction with the bigger K (along with some excuse that you didn't really understand). But wait a darned tootin' second! The K's are the same! Before your head explodes, however, you have a moment of inspiration and realize that you have forgotten about yet another reaction, this one:

OAc¯ + NH₄⁺ ---> HOAc + NH₃

But now you have yet another problem. What's the equilibrium constant of this mysterious reaction? Yet another light bulb goes off and you decide to add reactions 2 and 3, multiplying the K's. Unfortunately, this gives you a messy equation with a spare OH¯ and a spare H₃O⁺ on the right, and two spare waters on the left. To cancel these suckers out, you add this reaction, thus factoring in another K:
OH¯ + H₃O⁺ ---> 2H₂O (K = 1 / K_w)

So now you have a beast of an equilibrium constant that looks like this (meanwhile Fido is starting to whimper):

K = K_a * K_b / K_w =
(5.6 E-10)2 / (1 E-14) = 3.1 E-5

With a K so huge, you now know the path to Fido's salvation and you decide to go with reaction 4. But before plugging and chugging you gotta figure out how much stuff you already have in the puddle:

Gmm NH₄OAc = 14 + 7(1) + 2(12) + 2(16) = 77 g / mol (yeah, we were lazy on the Gmm, so sue us)
Conc. of OAc¯ and NH₄⁺ = 10g / (77 g/mol) / 1 L = 0.13 M

So:

OAc¯	+	NH₄⁺	-->	HOAc	+	NH₃
0.13 - X		0.13 - X		X		X

Using reaction 4, you use the solver function on your trusty caluladora and you enter the following (X is the concentration of both NH₃ and HOAc when the reaction shifts to the right):

0 = X² / (0.13 - X)² - 3.1 E-5 ---> X = [NH₃] & [HOAc] = 7.2 E-4

Now what? Well, you call up your friend Heiniken Hasselbach for help and he gently reminds you of this equation he wrote with his friend Ricky Henderson that you can use whenever you know the concentration of an acid and its conjugate base (NH₃ and NH₄⁺) :

pOH = pK_b - log ( [NH₃] / [NH₄⁺] ) = -log (1.8 E-5) - log ( [7.2 E-4] / [0.13] )
pOH = 7.0
pH = 7.0

So you can pat yourself on the back; your dog might have a little stomach ache in the morning, but Fido is gonna make it.

Great job! Why don't you head on over to Buffers?