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    part b

    please also explain
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    The initial mixture contains two bases, CO32- and HCO3-.

    The CO32- reacts first (forming extra HCO3-). The first end point is reached when all of the CO32- is reacted.

    The second end point is reached when [the HCO3- that was present at the start AND the extra HCO3- that was created in step 1] has reacted.

    Do the math.
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    (Original post by Pigster)
    The initial mixture contains two bases, CO32- and HCO3-.

    The CO32- reacts first (forming extra HCO3-). The first end point is reached when all of the CO32- is reacted.

    The second end point is reached when [the HCO3- that was present at the start AND the extra HCO3- that was created in step 1] has reacted.

    Do the math.
    here is my working
    so 12.45 moles HCL react with HCO3

    C V(HCO3) =C V(HCL)
    C X V = 0.150 X 12.45/1000

    what will be the volume of HCO3
    initial volume was 25 but in step 1 we also added 20.80 HCL so Volume will be 25+20.80?
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    (Original post by Qer)
    here is my working
    so 12.45 moles HCL react with HCO3

    C V(HCO3) =C V(HCL)
    C X V = 0.150 X 12.45/1000

    what will be the volume of HCO3
    initial volume was 25 but in step 1 we also added 20.80 HCL so Volume will be 25+20.80?
    12.45 would be a volume, i.e. cm3 not moles.

    I would use the 20.80 to work out how many mol of CO32- there are (and therefore how many mol of HCO3- made)
    I would use the 33.25 to work out the total number of mol of HCO3- there are.
    I would do a subtraction to work out how many mol of HCO3- there were initially.
    BUT, doing the subtraction first is equally valid.

    The 25 is the volume of buffer and has nothing to do with the 20.80, which is a volume of HCl. The conc of the HCO3- initially is n(HCO3-)/(25/1000).
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    (Original post by Pigster)
    12.45 would be a volume, i.e. cm3 not moles.

    I would use the 20.80 to work out how many mol of CO32- there are (and therefore how many mol of HCO3- made)
    I would use the 33.25 to work out the total number of mol of HCO3- there are.
    I would do a subtraction to work out how many mol of HCO3- there were initially.
    BUT, doing the subtraction first is equally valid.

    The 25 is the volume of buffer and has nothing to do with the 20.80, which is a volume of HCl. The conc of the HCO3- initially is n(HCO3-)/(25/1000).
    that makes sense
    But I was confused
    Yoy mentioned that it is buffer and has nothing to do with the 20.80, which is a volume of HCl.

    but HCL is reacting to buffer then why it has no effect on that
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    There are three numbers in the Q:

    25 cm3 - the volume of the CO32-/HCO3- buffer.
    20.80 cm3 - the volume of HCl required to neutralise just the CO32- in the buffer
    54.05 cm3 - the total volume of HCl added, that neutralises the CO32- first (creating some HCO3-) and then all of the HCO3-.

    Once you work out how many mol of HCO3- there was initially, you're asked to work out the conc. To do that, you need the 25 cm3 - the volume of the buffer.
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    (Original post by Pigster)
    There are three numbers in the Q:

    25 cm3 - the volume of the CO32-/HCO3- buffer.
    20.80 cm3 - the volume of HCl required to neutralise just the CO32- in the buffer
    54.05 cm3 - the total volume of HCl added, that neutralises the CO32- first (creating some HCO3-) and then all of the HCO3-.

    Once you work out how many mol of HCO3- there was initially, you're asked to work out the conc. To do that, you need the 25 cm3 - the volume of the buffer.
    THANKS
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    equ for the reaction between CH3COCL and CH3CH2NH2

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    where CHCH2+ come from?

    Isn't it should be CH3CH2NH3CL?
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    CH3CH2NH3Cl (there is a 3 missing) is called ethylammonium chloride. The + is on the N in NH3, not on a C.
 
 
 
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