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:



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.

  • $\begingroup$ btw please don't down vote because I didn't "show enough thought" Im a beginner chemistry student and I really would like assistance $\endgroup$ – John Rawls Apr 12 '17 at 18:14

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.

  • $\begingroup$ I understand everything except why it has to be 2/3 to satisfy the fact that we want 5.05, if I understand ill check your answer! $\endgroup$ – John Rawls Apr 12 '17 at 18:28
  • 1
    $\begingroup$ @JohnRawls If you use the henderson-hesselbach equation and substitute in known values, then you can solve for the mole ratio of 2/3 $\endgroup$ – Teoc Apr 12 '17 at 18:30

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