[emissivity]

analyte : 0.2M NH₄OH      20mL

titrant : 0.1M HCl

Calculate the concentration of NH₄OH(or NH₃) in solution at equivalent point.




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I used
0.2M × 20mL    =    0.1M × xmL
x=40mL


its initial concentration is 0.2M
its initial volume is 20mL
its final volume is 20+40mL = 60mL


So NH₄OH 's concentration is

0.2M × (20mL/60mL)

=0.2/3

=0.06₇M

is this right?

I don't know how to calculate this..

help me..

[Borek]

You correctly took care of the dilution, but you forgot NH₃ reacts with the acid - so it is not only the dilution, but also an acid-base equilibrium.

[emissivity]

You correctly took care of the dilution, but you forgot NH₃ reacts with the acid - so it is not only the dilution, but also an acid-base equilibrium.

How can I consider acid-base equilibrium?
I totally don't know how to start it..

Actually I don't even understand why the problem gave me NH₃ as well as NH₄OH..

[Borek]

NH₃ and NH₄OH are basically the same thing, just written differently. Some say ammonia in water solutions is present just as NH₃, some say it is present as NH₄OH. It doesn't matter for most practical purposes.

What is the reaction taking place during titration?

[emissivity]

NH₃ and NH₄OH are basically the same thing, just written differently. Some say ammonia in water solutions is present just as NH₃, some say it is present as NH₄OH. It doesn't matter for most practical purposes.

What is the reaction taking place during titration?

NH₄OH + HCl  NH₄Cl + H₂O

but next I don't know.. 

[Borek]

OK, so what is the substance present in the solution at the equivalence point?