Titration ph

  • Thread starter navaq
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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
 

Answers and Replies

  • #2
symbolipoint
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Navaq,

There is a relatively simple form of equation to use which you can find in this board and in the other homework board, buried in some of the messages; I knew the form very well a few days ago but now forgot exactly how it goes. also easy to use...
without my going back to rederive it, the equation is (you'll need to recheck):
Ka = [H][Fs + H - OH]/(Fa - H + OH)

Manage your concentration calculations carefully; rearrange to the equational form you need, quadratic solution for H; Fs means formality of salt, Fa means formality of the weak acid.

One problem in your exercise description: Would your ANALYTE be the HNO2 (nitrous acid, the weak acid), and the TITRANT be the sodium hydroxide?
 
  • #3
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One problem in your exercise description: Would your ANALYTE be the HNO2 (nitrous acid, the weak acid), and the TITRANT be the sodium hydroxide?
Yes, it really is. Ka (nitrous acid) = 5,1 x 10^(-4)
 

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