I suggested this good information about calculating changes in buffer solution:

https://courses.lumenlearning.com/introchem/chapter/calculating-changes-in-a-buffer-solution/

Ikram Madani Ahmed Yeah, unfortunately in this case the problem cannot be tackled the same way as a normal buffer solution as we are adding an excess of a weak base to a weak acid and therefore we only have a partial dissociation of the base. I have solved the task but unfortunately I don`t know whether the solution ph ist correct. Should be slightly acidic though. I hope the few german explanatory words won’t be a problem.

Since the pKa's of acid and base are far apart, you probably start with a near neutral solution before you add the extra weak base. Then you pH goes up into the alkaline area.

Normally you prepare a buffer by either using a weak acid and adding about half of its concentration of a strong base or vica versa. I do not understand that you call a mix of a weak acid and a weak base a buffer.

What is the purpose of your 'buffer' ?

Let us consider the case of NH4F/HF system (aq. sol.) as analogous case study:

This is the solution of an acid and a salt of its conjugate base.

HF behaves as weak acid in dilute solution while ammonia (NH4OH) is a weak base.

The salt can be taken as a strong electrolyte, hence nearly entirely dissociated: NH4F → NH4+ + F-.

For NH4OH dissociation: NH4OH⇌ NH4+ + OH-, Kb = [NH4+]·[OH-]/[NH4OH] = 1.80·10-5 M.

For NH4+ hydrolysis: NH4+ + 2H2O ⇌ NH4OH + H3O+. Kh = [NH4OH]·[H3O+]/[NH4+] = Kw/Kb = 5.56·10-10 M (Kw = 1.00 x 10-14 M2). So: [NH4OH] =Kh·[NH4+]/[H3O+].

For HF dissociation: HF + H2O ⇌  H3O+ + F-, Ka = [H3O+]·[F-]/[HF] = 6.6·10-4 M.

So: [HF] = [H3O+]·[F-]/Ka.

The salt is predicted to be acidic (Ka > Kb).

Let C denote nominal (i.e. formal) concentrations, which do not account for speciation.

Fluoride balance is: CHF + CNH4F ≈ [HF] + [F-]. Hence: [F-] = (CHF + CNH4F)/{1 + [H3O+]/Ka}.

Ammonium balance yields: CNH4F ≈ [NH4+] + [NH4OH]. Hence: [NH4+] = CNH4F /{1 + Kh/[H3O+]}.

Charge balance writes: [NH4+] + [H3O+] = [F-] + [OH-] ≈ [F-].

It follows that: CNH4F /{1 + Kh/[H3O+]} + [H3O+] ≈ (CHF + CNH4F)/{1 + [H3O+]/Ka}.

Since Kh is small enough for us to expect [H3O+] > 100Kh:

CNH4F + [H3O+] ≈ (CHF + CNH4F)/{1 + [H3O+]/Ka}.

As long as the salt is not too dilute (CNH4F > 100[H3O+]):

CNH4F ≈ (CHF + CNH4F)/{1 + [H3O+]/Ka}.

This gives:

[H3O+] = Ka·(CHF/CNH4F) > 10-7 M, where CNH4F ≠ 0.

And we obtain:

pH ≈ pKa + log10(CNH4F/CHF)

where pH = -log10{[H3O+]/(1 M)}, pKa = 3.18, and CHF ≠ 0.

This conforms with the Henderson–Hasselbalch equation.

Note: speciation here considered did not include the bifluoride anion (HF2-), or its ammonium salt, i.e., ammonium bifluoride (NH4HF2 or NH4F·HF). This limits the accuracy of this pH prediction; particularly for CHF ≈ CNH4F ― close to the ammonium bifluoride stoichiometry.