Titration question
2.10g of sodium hydrogencarbonate was dissolved in water and the solution made up to 250cm3.
25cm3 of this solution was pipetted into a conical flask and some methyl orange indicator added.
The solution was neutralised by 25.9cm3 of dilute hydrochloric acid added from a burette.
Calculate the concentration of the acid in g dm -3?
[6]
My book says the units of the concentration of the sodium hydrogen carbonate is not right, you have to ch age the volume and unit if concentration
Someone help?
Posted from TSR Mobile
25cm3 of this solution was pipetted into a conical flask and some methyl orange indicator added.
The solution was neutralised by 25.9cm3 of dilute hydrochloric acid added from a burette.
Calculate the concentration of the acid in g dm -3?
[6]
My book says the units of the concentration of the sodium hydrogen carbonate is not right, you have to ch age the volume and unit if concentration
Someone help?
Posted from TSR Mobile
When calculating concentration remember that in n = c * v, volume should be in dm3 so divide the volume by 1000 if it's in ml or cm3
Do you understand how to do the rest of the question, was it just the NaHCO3 conc you were struggling with?
Do you understand how to do the rest of the question, was it just the NaHCO3 conc you were struggling with?
Original post by Stiff Little Fingers
When calculating concentration remember that in n = c * v, volume should be in dm3 so divide the volume by 1000 if it's in ml or cm3
Do you understand how to do the rest of the question, was it just the NaHCO3 conc you were struggling with?
Do you understand how to do the rest of the question, was it just the NaHCO3 conc you were struggling with?
By book goes through the answer and says that before you can treat it like a straightforward calculation you have convert the grams to moles and the 250cm3 to 1dm3. They do it like this:
250cm3 is quarter of 1000dm3 so to make up the solution with same concentration you would need four times as much as in 1 dm3, so you do 2.10 x 4 g dm3
= 8.40 g dm3
You now have the volume units right so convert grams to moles, so the concentration of NaHCO3 is 0.100 mol dm3, as 1 mole weighs 84g.
I don't why you do this?
Posted from TSR Mobile
Original post by Jimmy20002012
By book goes through the answer and says that before you can treat it like a straightforward calculation you have convert the grams to moles and the 250cm3 to 1dm3. They do it like this:
250cm3 is quarter of 1000dm3 so to make up the solution with same concentration you would need four times as much as in 1 dm3, so you do 2.10 x 4 g dm3
= 8.40 g dm3
You now have the volume units right so convert grams to moles, so the concentration of NaHCO3 is 0.100 mol dm3, as 1 mole weighs 84g.
I don't why you do this?
Posted from TSR Mobile
250cm3 is quarter of 1000dm3 so to make up the solution with same concentration you would need four times as much as in 1 dm3, so you do 2.10 x 4 g dm3
= 8.40 g dm3
You now have the volume units right so convert grams to moles, so the concentration of NaHCO3 is 0.100 mol dm3, as 1 mole weighs 84g.
I don't why you do this?
Posted from TSR Mobile
You calculate the moles of NaHCO3 so that you can use the reacting ratio to find the moles of acid and work back to find the concentration.
Posted from TSR Mobile
Original post by Stiff Little Fingers
You calculate the moles of NaHCO3 so that you can use the reacting ratio to find the moles of acid and work back to find the concentration.
Posted from TSR Mobile
Posted from TSR Mobile
I don't get the conversion parts of g to get concentation?
Posted from TSR Mobile
From grams to concentration?
You calculate the number of moles in the mass using n=m/mr, then apply that to n=c*v.
When the volume is 1dm3, v just equals one, so the number of moles is equal to the concentration in moldm-3
Posted from TSR Mobile
You calculate the number of moles in the mass using n=m/mr, then apply that to n=c*v.
When the volume is 1dm3, v just equals one, so the number of moles is equal to the concentration in moldm-3
Posted from TSR Mobile
(edited 11 years ago)
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