Titration calculation Help?? (1 Viewer)

[greenteddy]

25 ml homemade ammonia was diluted to 250 ml with water. 25 ml of diluted solution required 35 ml 0.2M HCl for neutralisation.

Find the concentration of diluted solution and the original solution?


Do you calculate the dilution by first finding the moles of ammonia first and then use the volume to find the concentration?

 

[zhertec]

For this question you need to work backwards I believe, so first you use the titration formula to deduce the concentration of diluted Ammonia.
So C(Acid) x V(acid) = C(base) x V(base)

Since the neutralisation reaction between HCl and Ammonia is 1:1 the ratios dont really change. So: HCl(aq) + NH3(aq) --> NH4Cl(aq)
So to work out the concentration: 0.2M L(-1) x 0.035 = C(base) x 0.025 NOTE: all volumes are converted into litres.
C(base) = 0.28 M L(-1) So this being the concentration of the diluted base solution.

To work out the original solution you use the dilution formula: C(initial) x V(initial) = C(final) x V(final)
C (initial) x 0.025 = 0.28 M L(-1) x 0.250
C(initial) = 2.8 Mol L(-1) This being the concentration of the original solution.

Hope this answer helped!

P.S. If I got anything wrong, in terms of concepts and or calculations, sorry!

 

[faisalabdul16]

Dilute = 0.28mol/L
Original = 2.8mol/L

 

[Switch75]

@zhertec , you've actually confused your volumes. 35 mL of HCl is used to neutralize the DILUTE solution.
So, the working out should read:

NH3 + HCl --> NH4 + Cl-
C=0.028M C=0.2M
V=0.25 V=0.035
n=0.007 n=0.007

or in your working:
C(Base) x V(Base) = C(Acid) x V(Acid)
C(Base) x 0.25 = 0.2 x 0.035
C(Base) x 0.25 = 0.007
C(Base) = 0.028.
Therefore diluted solution has a concentration of 0.028.

C(Dil) x V(Dil) = C(Conc.) x V(Conc.)
0.028 x 0.25 = C(Conc.) x 0.025
C(Conc.) = 0.28

So all in all,
Dilute = 0.028mol/L
Concentrated = 0.28mol/L

 

[zhertec]

Read the question properly.

25 ml homemade ammonia was diluted to 250 ml with water. 25 ml of diluted solution required 35 ml 0.2M HCl for neutralisation.

Find the concentration of diluted solution and the original solution?


Do you calculate the dilution by first finding the moles of ammonia first and then use the volume to find the concentration?

So the person who titrated the base with the known acid only used 25mL, not all (250mL) of the base, hence that is why I used 0.025 rather than 0.25.