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jbird2704
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In an experiment to find the concentration of some dilute sulphuric acid, a student diluted it exactly 10 times to give a concentration suitable for titration. He made up some standard sodium hydroxide solution by dissolving 1.00g of it in 250cm3 of solution. He then found that 25.0cm3 of the sodium hydroxide solution required 23.5cm3 of the diluted sulphuric acid for neutralisation. Calculate the concentration of the original dilute sulphuric acid in mol dm-3
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Chlorophile
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#2
(Original post by jbird2704)
In an experiment to find the concentration of some dilute sulphuric acid, a student diluted it exactly 10 times to give a concentration suitable for titration. He made up some standard sodium hydroxide solution by dissolving 1.00g of it in 250cm3 of solution. He then found that 25.0cm3 of the sodium hydroxide solution required 23.5cm3 of the diluted sulphuric acid for neutralisation. Calculate the concentration of the original dilute sulphuric acid in mol dm-3
In an experiment to find the concentration of some dilute sulphuric acid, a student diluted it exactly 10 times to give a concentration suitable for titration. He made up some standard sodium hydroxide solution by dissolving 1.00g of it in 250cm3 of solution. He then found that 25.0cm3 of the sodium hydroxide solution required 23.5cm3 of the diluted sulphuric acid for neutralisation. Calculate the concentration of the original dilute sulphuric acid in mol dm-3
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jbird2704
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#3
(Original post by Chlorophile)
Can you give your working so far? We can't do your homework for you. A good starting point would be working out the number of moles of NaOH in the 25ccm.
Can you give your working so far? We can't do your homework for you. A good starting point would be working out the number of moles of NaOH in the 25ccm.
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Chlorophile
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#4
(Original post by jbird2704)
i am not sure how to tackle this question at all sorry, i know i have 1.00g and 25.00cm of 2NAOH and 23.5cm3 of h2so4. Not sure where to begin? thanks
i am not sure how to tackle this question at all sorry, i know i have 1.00g and 25.00cm of 2NAOH and 23.5cm3 of h2so4. Not sure where to begin? thanks
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jbird2704
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how do you go about doing that? which equation do i use, moles = mass / ar or moles=( m x v ) / 1000.
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jbird2704
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#6
(Original post by Chlorophile)
I just gave you a starting point. You should be able to work out how many moles of NaOH there are in 1g and therefore how much there is in 25 ccm of the solution.
I just gave you a starting point. You should be able to work out how many moles of NaOH there are in 1g and therefore how much there is in 25 ccm of the solution.
which equation do i use? moles=mass/ar, if so, what about 25cm then?? surely i need to put that somewhere
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Chlorophile
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#7
(Original post by jbird2704)
which equation do i use? moles=mass/ar, if so, what about 25cm then?? surely i need to put that somewhere
which equation do i use? moles=mass/ar, if so, what about 25cm then?? surely i need to put that somewhere
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Borek
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#8
See if this introduction to titration doesn't help: http://www.titrations.info/titration-calculation
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