Can a buffer solution be obtained by mixing-

A) $\ce{NH4Cl}$ + $\ce{CH3COONa}$

B) $\ce{CH3COOH}$ + $\ce{NH4OH}$

My Thoughts

I know that a buffer solution can be obtained if we mix a acid and its salt of conjugate base or a base and its salt of conjugate acid.

But I was wondering whether a buffer solution can be prepared if we mix a weak acid and weak base or if we mix an acidic salt and basic salt?

  • 3
    $\begingroup$ To be precise, $\ce{NH4Cl}$ and $\ce{CH3COONa}$ are solids and $\ce{NH4OH}$ doesn't exist. $\endgroup$
    – user7951
    Aug 18, 2020 at 18:33
  • 1
    $\begingroup$ I meant there aqueous solutions. And you can assume $\ce{NH4OH}$ to be $\ce{NH3}$ dissolved in water. $\endgroup$ Aug 18, 2020 at 18:55
  • $\begingroup$ Not really. NH4OH is kind of an immortal myth. $\endgroup$
    – Poutnik
    Aug 18, 2020 at 19:12

1 Answer 1


Both cases are equivalent in the case of molar ratio 1:1. It would lead to a double buffer of $\ce{HAc/Ac-}$ and $\ce{NH4+/NH3}$. But the buffer capacity would be inferior, as the resulting $\mathrm{pH}$ is too far from the both respective $\mathrm{p}K_\mathrm{a}$.

  • $\begingroup$ That is correct but I was wondering since we are mixing an acidic and basic substance would they neutralise each other or not (as they are weak acid and base) $\endgroup$ Aug 18, 2020 at 18:13
  • 2
    $\begingroup$ They would almost neutralise each other. For there would remain less than 1% of free acetic acid and less than 1% of ammonia, as both pKa are more than 2 units from pH 7 ( 4.75 and 9.25 ). $\endgroup$
    – Poutnik
    Aug 18, 2020 at 18:17
  • $\begingroup$ One more question : will it be a buffer if the molar ratio is not 1:1? If yes, then will it be a better buffer? $\endgroup$ Aug 18, 2020 at 18:21
  • 1
    $\begingroup$ If it was a mixture acid : base 2 : 1, that it would be the ideal 1:1 acetic acid/acetate buffer of pH 4.75. If it was a mixture acid : base 1 : 2, that it would be the ideal 1:1 NH3/NH4+ ammonia buffer of pH 9.25, both at their maximum buffer capacity. $\endgroup$
    – Poutnik
    Aug 18, 2020 at 19:17

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