The physiological mechanism for renal tubule exchange of ammonia (NH3 + NH4+) between the blood and the urine is illustrated in the figure above. The equilibrium dissociation reaction for ionized ammonia is written as NH4+ ↔ NH3 + H+ and is defined by the equilibrium dissociation constant:
Kdn = [NH3]*[H+]/[NH4+] = 10^(-9.25)

Determine the CU/CB ratio for the total ammonia concentration in urine, CU – where CU = [NH3]U + [NH4+]U – divided by the total ammonia concentration in blood, CB – where CB = [NH3]B + [NH4+]B

In formulating your answer, make the following assumption:
1) Only unionized ammonia (NH3) freely diffuses across the membranes separating blood from urine;
2) Only unionized ammonia (NH3) is in equilibrium between both sides of the membranes separating blood from urine such that [NH3]U = [NH3]B
Explain and justify
CU/CB = ?

What I've got so far is:
CU/CB = [NH3]U + [NH4+]U / [NH3]B+ [NH4+]B

At pKdn (NH3) = 9.25

Ya @ pH 5.5 = 1 / 1 + 10^(5.5 - 9.25) = .9998 NH4+ in Urine
Ya @ pH 7.4 = 1 / 1 + 10^(7.4 - 9.25) = .9861 NH4+ in Blood

CU/CB = .9998 (NH3) / .9861 (NH3) = 1.0139

But I'm pretty sure this isn't complete or correct.
Any help would be greatly appreciated.