[21385]

Will equal volumes of 0.05 M HNO2 and 0.05 M NH3 make a buffer solution? Why?

I know that a buffer solution consists of a weak acid and its corresponding salt but I don't know what will happen if a weak acid is added to a weak base.

[Borek]

What will be the pH of the solution? What will be concentrations of conjugated acids and bases for both substances? Remember, that to work buffer solution must contain both acid and conjugated base in comparable quantities.

[21385]

I think HNO2 + NH3 => NH4+ + NO2-
Krxn=KaKb/Kw
So, if i substitute 4.5E-4 for Ka of HNO2, and 1.8E-5 for Kb of NH3,
Krxn=810000
This means that the rxn lies far to the right
From a previous question (the question of what is the pH of NH4F), [H+]=sqr(KwKa/Kb)
so, [H+]=5E-7 and pH=6.3

So, because the rxn lies far to the right, the final concentrations will be near 0.05 M for both NH4+ and NO2- and the final pH will be a bit acidic. So, since there are no pair of acid and its conjugated base in comparable quantities, it is not a buffer?

Is this right?

[Borek]

From a previous question (the question of what is the pH of NH4F), [H+]=sqr(KwKa/Kb)
so, [H+]=5E-7 and pH=6.3

OK

So, because the rxn lies far to the right, the final concentrations will be near 0.05 M

0.025 - you forgot about dilution

for both NH4+ and NO2- and the final pH will be a bit acidic. So, since there are no pair of acid and its conjugated base in comparable quantities, it is not a buffer?

Is this right?

That's how I see it

Note, that Henderson-Hasselbalch equation gives us very important hint here. pH=pKa + log(ratio) - it means that ratio = 10^pH-pKa. pH is 6.3, we are about 3 pH units from both acid and base pKa - which means that in both cases amount of one form is about 10³=1000 times higher than amount of the second form.

[AWK]

This is a completely other chemical problem.
NH4NO2 specifically decompose, even at RT
NH4NO2 = N2+2H2O
Since equal amounts of HNO2 and NH3 were used, pH will be neutral soon