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# Chemistry Calculation [A2] watch

1. A sample of ammonia was dissolved in water to produce 100cm3 of a solution , X. 10cm3 of this solution was made up to a volume of 250cm3. 25cm3 of this diluted ammonia solution was then titrated with aqueous HCL of concentration 0.11 mol dm-3 . 37.10cm3 of the acid was required to neutrailise the ammonia solution.

Write and equation for the reaction between ammonia and HCL, and calculate the number of moles of ammonia in solution X. Calculate the concentration of solution X in mol dm-3.

Cant seem to write the equation, if I can im pretty sure I can do the rest [ I hope ]

Can anybody help me...
2. NH3 + HCl --> NH4Cl
3. NH3 + HCl ------ NH4Cl
4. I am so slow lol
5. Yeah I got that one hehe, but the first one..

NH3 + H20 ---> ???
6. (Original post by TerrorSquad)
Yeah I got that one hehe, but the first one..

NH3 + H20 ---> ???
NH3 + H2O --> NH4+ + OH- ?

I didn't actually read the rest of the information you gave, so I could be wrong.
7. Well, after reading it, I don't think you need to worry about that, both OH- and NH3 will react with the acid in the same proportions.

Do you have the final answer to check?
8. No - I got stuck 1/2 way lol. Ive found the moles and concentration of ammonia in the 25cm3 solution. THats where i am up to
9. (Original post by TerrorSquad)
No - I got stuck 1/2 way lol. Ive found the moles and concentration of ammonia in the 25cm3 solution. THats where i am up to
Once you have the moles in 25cm3 then times by 10 to get moles in 250cm3.

This is also the moles in 10cm3 and so you times it by 10 to get moles in 100cm3 (X)
10. I got the mark scheme in front of me so ill tell you answer along with where the marks are allocated:

NH4 + HCl --->NH4Cl or NH4+Cl- (1)

Amount HCl = 0.11 mol dm3 x 0.0371 dm3 = 0.004081 mol (1)
Amount of NH3 in 25.0cm3 = 0.004081 mol
Amount of 250cm3 = 0.04081 mol (1)
Amount of NH3 in 10cm3 of X = 0.04081 mol
[NH3] in X = 0.04081 mol/0.0100 dm3 = 4.08 mol dm-3 or [NH3] in X = 0.4081 mol/0.100dm3 = 4.08 mol dm-3 (1)

(NOTE: Diluting solution does not chage the number moles)
11. You said dilution does not change the number of moles, then how come :

"Amount of NH3 in 25.0cm3 = 0.004081 mol
Amount of 250cm3 = 0.04081 mol"

you multiplied for x10

[thanks for the answer anyway ]
12. i'm confused...surely 250cm3 is more dilute than 25cm3 so should it be 0.0004081 in 250 not 0.04081??
13. Amount of NH3 in 25.0cm3 = 0.004081 mol
Amount of 250cm3 = 0.04081 mol (1)
Amount of NH3 in 10cm3 of X = 0.04081 mol
[NH3] in X = 0.04081 mol/0.0100 dm3 = 4.08 mol dm-3 or [NH3] in X = 0.4081 mol/0.100dm3 = 4.08 mol dm-3 (1)

Can anyone explain these steps as i do understand how the moles of the acid are worked out but then after am Kinda confused!!!!..lol
14. Ok 10cm3 of solution X was made ---yes??
Then with the 10cm3, 250cm3 was made by diluting it (so here the number of moles does not change - as i said dilution does not chnage the number of moles)

Then from the 250cm3 made, they took 25 cm3 for the titration- so it was not diluted (this is the naswer for your question TerrorSquad).

Ok here goes- step by step how to work it out.

1- Work out the number of moles of HCl used in the titration (Amount HCl = 0.11 mol dm3 x 0.0371 dm3 = 0.004081 mol )

2- Then work out the number of moles of NH3 in 25cm3 because 25cm3 of it was used to titrate it against HCl (As its 1:1 ratio - Amount of NH3 in 25.0cm3 = 0.004081 mol)

3- Then work out the number of moles in 250cm3 becuase 25.0cm3 was taken from the 250cm3 sample, so therefore 250cm3 must have a higher number of moles of NH3 then 25cm3 (Amount of 250cm3 =0.004081 x 10 = 0.04081 mol)

4- As the 250cm3 from made from thr 10cm3 of X by diluting it the number of moles of NH3 in 10cm3 of X does not change (Amount of NH3 in 10cm3 of X = 0.04081 mol)

5- And as thr 10cm3 of X was taken from the 100cm3 of ammonia then you will need to find out the number of moles of NH3 in 100cm3 ( Amount of NH3 in 100cm3 = 0.04081 x 10 = 0.4081 mol)

6- Then just work out the concentration of NH3 ( Concentration of NH3 = 0.4081/0.100 dm3 = 4.08 mol dm-3 -- 0.100dm3 is the same as 100cm3--- i just coverted it in dm3)

So theres your answer 4.08 mol dm-3.

IM SORRY IF YOU CANT UNDERSTAND WHAT IM SAYING COZ SUMTIMES I FIND IT HARD TRYING TO EXPLAIN THINGS BY WIRITNG IT DOWN
15. Hey np Shaika thanx a lot anywayz......will look at that solution properly in a min........
16. ok got it now lol - thanks again

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