AS chemistry
A student dissolves an unknown mass of sodium hydroxide in water to make 200cm3 of an aqueous solution. A 25cm3 sample of this sodium hydroxide is placed in a conical flask and is tut rated with 0.015 moldm-3 sulfuric acid. Titre = 19.58
Calculate mass of sodium hydroxide used to make original solution
2NaOH + h2so4 -> na2so4 + 2h20
For this question I got 2.4g, can someone try this question and see what they get
Calculate mass of sodium hydroxide used to make original solution
2NaOH + h2so4 -> na2so4 + 2h20
For this question I got 2.4g, can someone try this question and see what they get
Hi, for this question I got 0.188g.
Moles of sulphuric acid = 19.58cm3 x 0.015moldm-3 / 1000= 0.0002937mol
Ratio of NaOH to H2SO4 = 2:1 therefore the moles of NaOH = 0.0002937 x 2 = 0.000587mol
That was for the 25cm3 sample so for the 200cm3 sample the moles of NaOH would be: 0.000587 x 8 = 0.0047mol.
Mr of NaOH + 23 + 16 + 1 = 40
Mass = moles x Mr = 0.0047 x 40 = 0.188g
Moles of sulphuric acid = 19.58cm3 x 0.015moldm-3 / 1000= 0.0002937mol
Ratio of NaOH to H2SO4 = 2:1 therefore the moles of NaOH = 0.0002937 x 2 = 0.000587mol
That was for the 25cm3 sample so for the 200cm3 sample the moles of NaOH would be: 0.000587 x 8 = 0.0047mol.
Mr of NaOH + 23 + 16 + 1 = 40
Mass = moles x Mr = 0.0047 x 40 = 0.188g
(edited 2 years ago)
Original post by Dnajs sb
Thank you so much!
However for this question y r we supposed to do 250/26.45 instead of multiplying it by 10
However for this question y r we supposed to do 250/26.45 instead of multiplying it by 10
Because it’s the opposite way round from the question you put before.
This time the unknown acid is the titrant so you use the titre value of 26.45
Thank you so much!!!
Original post by Jaz._bg
Hi, for this question I got 0.188g.
Moles of sulphuric acid = 19.58cm3 x 0.015moldm-3 / 1000= 0.0002937mol
Ratio of NaOH to H2SO4 = 2:1 therefore the moles of NaOH = 0.0002937 x 2 = 0.000587mol
That was for the 25cm3 sample so for the 200cm3 sample the moles of NaOH would be: 0.000587 x 8 = 0.0047mol.
Mr of NaOH + 23 + 16 + 1 = 40
Mass = moles x Mr = 0.0047 x 40 = 0.188g
Moles of sulphuric acid = 19.58cm3 x 0.015moldm-3 / 1000= 0.0002937mol
Ratio of NaOH to H2SO4 = 2:1 therefore the moles of NaOH = 0.0002937 x 2 = 0.000587mol
That was for the 25cm3 sample so for the 200cm3 sample the moles of NaOH would be: 0.000587 x 8 = 0.0047mol.
Mr of NaOH + 23 + 16 + 1 = 40
Mass = moles x Mr = 0.0047 x 40 = 0.188g
hi where did you get the value 19.58cm3 from? there are three values given for the titre (3 trials)
(edited 2 years ago)
Paper is AS chemsitry 7401/1Paper 1 June 2020 AndThe actual answer is 1.88 grams Cuz he did c=n/v But did 0.150 /1000 instead of 19.575 i think
Original post by akb.
hi where did you get the value 19.58cm3 from? there are three values given for the titre (3 trials)
you must find the mean titre and apply to equation
Original post by Jas.acs
Hi, for this question I got 0.188g.
Moles of sulphuric acid = 19.58cm3 x 0.015moldm-3 / 1000= 0.0002937mol
Ratio of NaOH to H2SO4 = 2:1 therefore the moles of NaOH = 0.0002937 x 2 = 0.000587mol
That was for the 25cm3 sample so for the 200cm3 sample the moles of NaOH would be: 0.000587 x 8 = 0.0047mol.
Mr of NaOH + 23 + 16 + 1 = 40
Mass = moles x Mr = 0.0047 x 40 = 0.188g
Moles of sulphuric acid = 19.58cm3 x 0.015moldm-3 / 1000= 0.0002937mol
Ratio of NaOH to H2SO4 = 2:1 therefore the moles of NaOH = 0.0002937 x 2 = 0.000587mol
That was for the 25cm3 sample so for the 200cm3 sample the moles of NaOH would be: 0.000587 x 8 = 0.0047mol.
Mr of NaOH + 23 + 16 + 1 = 40
Mass = moles x Mr = 0.0047 x 40 = 0.188g
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