How to determine which combination of substances will give a buffered solution given target pH and Kb of each substances What combination of substances will give a buffered solution that has a pH of 5.05? (Assume each pair of substances is dissolved in 5.0 L of water.) (Kb for NH3 = $1.8 × 10^{–5}$; Kb for C5H5N = $1.7 × 10^{–9}$)
  
  a)1.0 mole NH3 and 1.5 mole NH4Cl
  
  b)1.5 mole NH3 and 1.0 mole NH4Cl
  
  c)1.0 mole C5H5N and 1.5 mole C5H5NHCl
  
  d)1.5 mole C5H5N and 1.0 mole C5H5NHCl


According to my text book it is choice C but I do not understand why and here was my thought process:



For starters I change the $K_b$ into $K_a$ to get : $5.6*10^{-10}$ for Ammonia and for the other I got: $5.9*10^{-6}$

It seems to me that this has to do with the Henderson Hasselbach Equation so I set up the following:

$5.05=5.9*10^{-6}+log{\frac{Base}{acid}}$

$5.05=5.9*10^{-10}+log{\frac{Base}{acid}}$

because I already have two substances.

But from here I feel like that wasn't the best route I should take because I got stuck right here and would like some assistance.
Using the Henderson-Hesselbach equation is a good idea. As you want the buffer's pH to be acidic, then using the $\ce{NH3/NH4+}$ buffer is not a good idea, as that would be basic, so we should use the $\ce{C5H5N/C5H5NHCl}$. Then, by Henderson-Hesselbach, we have that $\ce{5.05 = pH = pKa + \log \left(\frac{[C5H5N]}{[C5H5NH+]}\right)}\implies \ce{5.05 - pKa = 5.05 - 5.23 = \log \left(\frac{[C5H5N]}{[C5H5NH+]}\right)} \implies \ce{\frac{[C5H5N]}{[C5H5NH+]} = 10^{-0.18}\implies \frac{[C5H5N]}{[C5H5NH+]} \approx 0.66\approx \frac 23}.$ Thus, we want a buffer solution where the ratio of the base to the acid is approximately 2/3, which answer choice C satisfies.