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    Can someone help me with b) and e)

    I don't understand b) at all. And e), I thought that 2 times as much HCl will have to be used to neutralise twice the amount of carbonate.
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    (Original post by ps1265A)
    Can someone help me with b) and e)

    I don't understand b) at all. And e), I thought that 2 times as much HCl will have to be used to neutralise twice the amount of carbonate.
    Carbon dioxide is only produced after the first stage of the reaction...

    Na2CO3 + HCl --> NaHCO3 + CO2
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    (Original post by charco)
    Carbon dioxide is only produced after the first stage of the reaction...

    Na2CO3 + HCl --> NaHCO3 + CO2
    Thanks, could you give me a hint for e)
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    (Original post by ps1265A)
    Thanks, could you give me a hint for e)
    OK.

    The first end point is when all of the sodium carbonate is changed to sodium hydrogencarbonate.

    The second endpoint is when all of the sodium hydrogencarbonate is changed to sodium chloride.
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    (Original post by charco)
    OK.

    The first end point is when all of the sodium carbonate is changed to sodium hydrogencarbonate.

    The second endpoint is when all of the sodium hydrogencarbonate is changed to sodium chloride.
    For part d) why do we use a 1:2 ratio of moles? I understand that the overall reaction is 1:2 but at point D to E, it is simply focusing on NaHCO3 + HCl ---> NaCl + CO2 + H20 which is a 1:1 reaction?
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    (Original post by charco)
    OK.

    The first end point is when all of the sodium carbonate is changed to sodium hydrogencarbonate.

    The second endpoint is when all of the sodium hydrogencarbonate is changed to sodium chloride.
    Hmmm... the addition of an equimolar amount of sodium hydrogencarbonate wouldn't affect the first end point because there's still sodium carbonate to neutralise. However, now that we have double the amount of sodium hydrogencarbonate in our solution, I would assume that we need twice the volume of NaOH to neutralise the NaHCO3? Therefore, because I original required 15cm3, I will now require 30cm3 and so the new equivalence point is 45cm3?

    PS: We can use the words end-point and equivalence point interchangeably here right?
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    (Original post by ps1265A)
    Hmmm... the addition of an equimolar amount of sodium hydrogencarbonate wouldn't affect the first end point because there's still sodium carbonate to neutralise. However, now that we have double the amount of sodium hydrogencarbonate in our solution, I would assume that we need twice the volume of NaOH to neutralise the NaHCO3? Therefore, because I original required 15cm3, I will now require 30cm3 and so the new equivalence point is 45cm3?

    PS: We can use the words end-point and equivalence point interchangeably here right?
    as there are two end points it might be appropriate to state which one ...
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    (Original post by charco)
    as there are two end points it might be appropriate to state which one ...
    So the first end-point will remain the same

    The second end-point will be at 45cm3
 
 
 
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