[xps]

Titrant

[HNO₂]= 0.10 mol/dm³
v= 25ml = 0.025 dm³

+

Analyte

[NaOH]= 0.125 mol/dm³
v= 20ml = 0.02 dm³

Question is: What is the ph after adding a volume 10ml of titrant?
I know the answer is ph= 4,3

I'm not allowed to use Henderson Hasselbach equation.
This is the only type of titration ph calculation i don't know how to do. When you add a certain volume of the titrant and one of them not being strong (the acid in this case).

Hope you can help me with the calculation or at least the steps so that I can try it and confirm.
Thanks

[Borek]

Strong acid, strong base, neutralization goes to the end - stoichiometry and limiting reagent, that's all.

[xps]

HNO₂ is not a strong acid.
Ka= 5,1 x 10^(-4)

[Borek]

Ops, sorry, I misread it as HNO₃.

I wonder what is means you are not allowed to use HH equation - it is just a modfied form of general dissociation equilibrium equation. Just start with general equilibrium and enter known concentrations - then solve for [H⁺].

[AWK]

Usually strong acid or strong base is used as a titrant