# Chemistry A Level Titrations Help

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#1
I did an experiment in class, where I was given 2.00g of an acidic solid 'H3X'. I made a 250cm3 solution of water and the acid and titrated it against 25cm3 of 0.100 mol dm^-3 of NaOH.
The equation is H3X(aq) + 3NaOH(aq) = Na3X(aq) + 3H2O.
I have been given these questions to complete.
1. My average titre. This question was fairly simple, I calculated a mean of all my titres to get an average of 21.54cm^3.
2. Moles of NaOH. For this I did (25/1000)*0.100 to get 2.5*10^-3.
3. Calculate the amount of moles of H3X in your mean titre. For this I simply divided my moles of NaOH by 3 since it is in a 1:3 molar ratio, to get 8.3*10^-4.
4. Calculate the amount, in mol, of H3X present in the 250cm^3 solution that you prepared. For this do I just have to multiply the moles I calculated previously by (250/21.54) since it is the mean titre I got and I multiply it by the factor.
5. Calculate the molar mass, in g mol-1 of H3X (dot) nH2O. This part I am slightly confused on. For this, all I assumed is to use the formula n=m/Mr but this must be wrong since it is a dot formula with water so I obviously can't just use the mass of H3X by itself, so I am unsure about this.

Any help is appreciated, thank you very much.
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#2
Bump
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#3
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4 years ago
#4
Is the acid solid hydrated or not? I think you can use the mass and moles for that to calculate the molecular mass.
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#5
(Original post by PuffyPenguin)
Is the acid solid hydrated or not? I think you can use the mass and moles for that to calculate the molecular mass.
Thank you very much for the reply. The solid acid itself was not hydrated.
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3 years ago
#6
(Original post by Arayan01)
I did an experiment in class, where I was given 2.00g of an acidic solid 'H3X'. I made a 250cm3 solution of water and the acid and titrated it against 25cm3 of 0.100 mol dm^-3 of NaOH.The equation is H3X(aq) 3NaOH(aq) = Na3X(aq) 3H2O.I have been given these questions to complete.1. My average titre. This question was fairly simple, I calculated a mean of all my titres to get an average of 21.54cm^3.2. Moles of NaOH. For this I did (25/1000)*0.100 to get 2.5*10^-3.3. Calculate the amount of moles of H3X in your mean titre. For this I simply divided my moles of NaOH by 3 since it is in a 1:3 molar ratio, to get 8.3*10^-4.4. Calculate the amount, in mol, of H3X present in the 250cm^3 solution that you prepared. For this do I just have to multiply the moles I calculated previously by (250/21.54) since it is the mean titre I got and I multiply it by the factor.5. Calculate the molar mass, in g mol-1 of H3X (dot) nH2O. This part I am slightly confused on. For this, all I assumed is to use the formula n=m/Mr but this must be wrong since it is a dot formula with water so I obviously can't just use the mass of H3X by itself, so I am unsure about this.Any help is appreciated, thank you very much.
number 3 is wrong as you dont use you mean titre
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3 years ago
#7
(Original post by nibber)
number 3 is wrong as you dont use you mean titre
It's from one year ago and it's correct

The equation is H3X(aq) + 3NaOH(aq) = Na3X(aq) + 3H2O.

The number of moles of NaOH is 2.5 x 10 ^-3

The mole ratio of the equation says that 1 mole of H3X reacts with 3 moles of NaOH

so,

The number of moles of H3X in 21.54 cm3 is 8.33 x 10^-4 mol

The number of moles of H3X in 250cm3 is 9.67 x 10^-3 mol

The number of moles of H3X is equal to the number of moles of H3X.nH2O

The number of moles of H3X.nH2O is therefore 9.67 x 10^-3 mol

so the molar mass is 2/9.67x10^-3 = 206.8
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3 years ago
#8
How’d you get the moles for the 250cm^3 question?

Edit: Nevermind I get it lmao
Last edited by HattyHat; 3 years ago
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1 year ago
#9
can you explain this
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1 year ago
#10
(Original post by BobbJo)
It's from one year ago and it's correct

The equation is H3X(aq) + 3NaOH(aq) = Na3X(aq) + 3H2O.

The number of moles of NaOH is 2.5 x 10 ^-3

The mole ratio of the equation says that 1 mole of H3X reacts with 3 moles of NaOH

so,

The number of moles of H3X in 21.54 cm3 is 8.33 x 10^-4 mol

The number of moles of H3X in 250cm3 is 9.67 x 10^-3 mol

The number of moles of H3X is equal to the number of moles of H3X.nH2O

The number of moles of H3X.nH2O is therefore 9.67 x 10^-3 mol

so the molar mass is 2/9.67x10^-3 = 206.8
can you explain 4 and 5 pls
5
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