How to determine which combination of substances will give a buffered solution given target pH and Kb of each substances


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> 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 = $1.7 × 10^{–9}$)
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> a)1.0 mole NH3 and 1.5 mole NH4Cl
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> b)1.5 mole NH3 and 1.0 mole NH4Cl
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> c)1.0 mole C5H5N and 1.5 mole C5H5NHCl
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> d)1.5 mole C5H5N and 1.0 mole C5H5NHCl
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According to my text book it is choice C but I do not understand why and here was my thought process:




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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.


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