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Really annoying chemistry question (AQA A2 Chemistry module 4 June 2009) watch

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    Hi everyone
    first time posting here so dot abuse me too much if the answer is simple! I'm doing my Module 4 next week (old spec) and have been doing every past paper under the sun. I am pretty confident on the paper except for calculations involving the one I'm about to put up, any help with the method to answering these kind of question would be really helpful!

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    QUESTION 8

    An aqueous solutions contained both sodium carbonate and sodium hydrogencarbonate.

    A 25.0cm^3 sampled of the solution was transferred into a conical flask. After the addition of a dew drops of phenolphthalein, 0.150 moldm^-3 hydrochloric acid was then added from a burette.

    The indicator changed colour when exactly 20.80cm^3 of the 0.150 moldm^-3 hydrochloric acid had been added to the conical flask. This end-point showed that the reaction in stage 1 had been completed.

    Stage 1 CO3^2- + H+ ------------> HCO3-

    A few drops of methyl orange were then added to the conical flask. A further 33.25cm^3 of 0.150 moldm^3 hydrochloric acid were required before this second indicator changed colour. This end-point showed that the reaction in stage 2 had been completed.

    Stage 2 HCO3- + H+ -----------> CO2 + H20

    A/ Use the data about stage 1 to calculate the concentration of sodium carbonate in the original solution (3 marks)

    B/ State why the volume of hydrochloric acid used in stage 2 is greater than that used in stage 1. Hence calculate the concentration of sodium hydrogencarbonate in the original solution
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    Right, for Stage 1, figure out many moles of HCl were added to reach first equivalence. All CO3(2-) must have come from Na2CO3. So moles of HCl = moles of Na2CO3 at the end point. Use moles * 1000/25 to find the initial concentration of Na2CO3.

    At Stage 2, we know that we have HCO3- from the NaHCO3, but we also have it from stage 1.
    So figure out how many moles of acid were added and that will equal the number of moles of HCO3- in the soluion. Then take awat the number of moles of HCO3- formed from the first reaction to find the number of moles of HCO3-. Once you've done then, find the volume using moles *1000/25.

    The HCl used in stage 2 is greater because we're neutralising the HCO3- from both the NaHCO3 and the titrated Na2CO3.
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    Random mention- That was the exam I did!
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    (Original post by silva228)
    Hi everyone
    first time posting here so dot abuse me too much if the answer is simple! I'm doing my Module 4 next week (old spec) and have been doing every past paper under the sun. I am pretty confident on the paper except for calculations involving the one I'm about to put up, any help with the method to answering these kind of question would be really helpful!

    ---------------------------------------------------------------------
    QUESTION 8

    An aqueous solutions contained both sodium carbonate and sodium hydrogencarbonate.

    A 25.0cm^3 sampled of the solution was transferred into a conical flask. After the addition of a dew drops of phenolphthalein, 0.150 moldm^-3 hydrochloric acid was then added from a burette.

    The indicator changed colour when exactly 20.80cm^3 of the 0.150 moldm^-3 hydrochloric acid had been added to the conical flask. This end-point showed that the reaction in stage 1 had been completed.

    Stage 1 CO3^2- + H+ ------------> HCO3-

    A few drops of methyl orange were then added to the conical flask. A further 33.25cm^3 of 0.150 moldm^3 hydrochloric acid were required before this second indicator changed colour. This end-point showed that the reaction in stage 2 had been completed.

    Stage 2 HCO3- + H+ -----------> CO2 + H20

    A/ Use the data about stage 1 to calculate the concentration of sodium carbonate in the original solution (3 marks)

    B/ State why the volume of hydrochloric acid used in stage 2 is greater than that used in stage 1. Hence calculate the concentration of sodium hydrogencarbonate in the original solution
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    I did this paper for revision for my Unit 4 next week.

    A) First find moles of HCl:
    Moles of HCl = (20.8^3 / 1000) x 0.150 = 3.12 x 10^-3
    the use that to find carbonate concentration:
    Conc. of carbonate = 3.12 x 10^-3 / (25/1000)
    = 0.125mol/dm^-3

    B) The volume of HCl used in stage 2 is greater than in Stage one because it has to react with the original HCO3- and that which was formed in reaction one from CO3^2- and H+.
    And then work out the value of Hcl needed for the original HCO3-. Then work out the moles of HCl and that is the same as the moles of HCO3-. Work it out for 25cm^3 and sorted

    Sorry if i didn't explain well!
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    (Original post by Lula1991)
    I did this paper for revision for my Unit 4 next week.


    Sorry if i didn't explain well!
    Refrain from posting complete solutions please. Let the OP figure it out for him/herself.
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    (Original post by Hippysnake)
    Right, for Stage 1, figure out many moles of HCl were added to reach first equivalence. All CO3(2-) must have come from Na2CO3. So moles of HCl = moles of Na2CO3 at the end point. Use moles * 1000/25 to find the initial concentration of Na2CO3.

    At Stage 2, we know that we have HCO3- from the NaHCO3, but we also have it from stage 1.
    So figure out how many moles of acid were added and that will equal the number of moles of HCO3- in the soluion. Then take awat the number of moles of HCO3- formed from the first reaction to find the number of moles of HCO3-. Once you've done then, find the volume using moles *1000/25.

    The HCl used in stage 2 is greater because we're neutralising the HCO3- from both the NaHCO3 and the titrated Na2CO3.
    Brilliant explanation thank you very much! that is actually what I would of done if I had to guess but it's good to know the actual answer!! Just as a quick follow up that has annoyed me slightly the question below...

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    A 5cm^3 sample of 0.135 molddm^-3 hydrochloric acid is added to 995cm^3 of water. Calculate the pH of the solution formed
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    you work out (5 x 0.135) / 1000 and then because its made up to 1dm that your concentration in moldm^3 which you then just -log to get pH right? thanks again for all your help ^^
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    (Original post by silva228)
    Brilliant explanation thank you very much! that is actually what I would of done if I had to guess but it's good to know the actual answer!! Just as a quick follow up that has annoyed me slightly the question below...

    --------------------------------------------------------------------
    A 5cm^3 sample of 0.135 molddm^-3 hydrochloric acid is added to 995cm^3 of water. Calculate the pH of the solution formed
    -----------------------------------------------------------------

    you work out (5 x 0.135) / 1000 and then because its made up to 1dm that your concentration in moldm^3 which you then just -log to get pH right? thanks again for all your help ^^
    Correct. The number of moles wll equal the conc in mol dm-3. So just -log it as you said!
 
 
 
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