[rayman99]

In a titration of 20mL NH3(aq) with 0,1M HCl(aq)
26mL HCl is added in the Equivalence point.

I think that this reaction of correct  (right?): HCl + NH4+ + OH-   --> NH4Cl + H2O

how do I calculate the concentration of NH3(aq), and calculate the pH in the equivalence point?

what kind of chemical compound do you have in the Equivalence point?

[Borek]

http://www.titrations.info/titration-calculation

http://www.titrations.info/titration-equivalence-point-calculation

[rayman99]

more explanation plz.

i have a potnetiometric titration

[Borek]

Show some effort, nobody is going to solve these questions for you.

[rayman99]

n=V*c
n(NH_3 )=n(HCl)=0.026l*0.1mol/l=0.0026mol
c(NH_3 )=0.0026mol/0.02l=0.13mol/l


is this right?

[Borek]

So far so good, you have a correct concentration of ammonia in the titrated solution.

[rayman99]

cool. what about the pH?

do I have a Buffer solution in equivalence point?

If so my guess is:

pH=pk_s+log⁡(/ )
pK_s (NH_4 )=9,25

pH=9,25+log⁡(0.13/0.1)=9,3

[rayman99]

woops !
one more try


pH=pk_s+log(B/S )

pK_s (NH_4 )=9,25

pH =   9,25 +  log (0.13/0.1)  =  9,3

[Borek]

No, you don't have a buffer at equivalence point. All base was protonated.

Check links that I posted. The answer is there.

[rayman99]

Okay. I hope this solution is right.

pK_b (NH_4 )=9.25
K_b (NH_4 )=〖10〗^(-9.25)

[ OH^- ]=√(K_b*c_b )=√(〖10〗^(-9.25)*0.13)=.000009


pOH=- log(.000009)= 5.5≈5

pH=14-pOH=14-5=9

is the pH right?

[Borek]

pK_b (NH_4 )=9.25

Not there yet. pKb listed is not for NH₄⁺. Ammonia is a base, but NH₄⁺ is not.

[rayman99]

hmm.. so

this is the situation: [H_3 O]^2   + K_s* [H_3 O^+ ]-  c_s* K_s  = 0

concentration oxonium:        [ 〖H_3 O 〗^+ ]=√(K_s*c_s )=√(〖10〗^(-9.25)*0.13)=.000009

pH=-log(.000009)= 5.5≈5


?

[Borek]

No idea what Ks and cs are, no idea what the equation is, no idea what you are doing.

Situation you are dealing with is explained on the page two clicks away from this thread - I suppose you have already visited it, but you have not read it carefully.