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.

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.

Is the acid solid hydrated or not? I think you can use the mass and moles for that to calculate the molecular mass.

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.

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.

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

can you explain this

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

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

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