# Please can you check my titration calculations...Watch

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#1
I have a feeling I've done the last part of this wrong- can you please check this for me:

Ammonia sample was dissolved in water to produce 100cm3 of a solution, X. 10cm3 of this was made up to volume of 250cm3. 25cm3 of this diluted ammonia solution was titrated with hydrochloric acid of concentration 0.11moldm-3. 37.1cm3 of acid was required to neutralise.

1) Find number of moles of ammonia in solution X

Moles HCl = 37.1/1000 x 0.11 = 0.00408mol
Moles in 250cm3 = 0.0408mol
Moles in 10cm3 = 0.0408/25 = 0.00163mol
Moles in 100cm3 = 0.0163mol

2) Calculate concentration of solution X

Conc. = 0.0163/ (100/1000) = 0.163moldm-3

Does this look right? Please correct me if it isn't! Thanks
0
13 years ago
#2
(Original post by Mimo)
I have a feeling I've done the last part of this wrong- can you please check this for me:

Ammonia sample was dissolved in water to produce 100cm3 of a solution, X. 10cm3 of this was made up to volume of 250cm3. 25cm3 of this diluted ammonia solution was titrated with hydrochloric acid of concentration 0.11moldm-3. 37.1cm3 of acid was required to neutralise.

1) Find number of moles of ammonia in solution X

Moles HCl = 37.1/1000 x 0.11 = 0.00408mol
Moles in 250cm3 = 0.0408mol
Moles in 10cm3 = 0.0408/25 = 0.00163mol
Moles in 100cm3 = 0.0163mol

Here you just want to multiply 0.0408mol by 10, to give 0.408 mol in the original 100ml of solution.

2) Calculate concentration of solution X
Then change this bit to work with the new value
Conc. = 0.0163/ (100/1000) = 0.163moldm-3

Does this look right? Please correct me if it isn't! Thanks
I'm pretty sure that's right now
0
#3
Brilliant- thanks thanks thanks!
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