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    Okay the analyte is sodium carbonate solution. Now, as a person was adding 25 cm cubed of sodium carbonate, some accidentally spilt.

    If HCL is the titrant (in the burette) and it is added to the sodium carbonate which is now less than 25 cm cubed, how will this affect the HCL concentration compared to if none had been spilt?

    I asked some people and they say the concentration of HCL will be higher...??

    But why or why not?


    Thank you!
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    I know that the concentration of Na2c03 will be less because obviously some has been spilt... so less HCL will be needed to neutralise it... but how would the concentration of this HCL be affected?
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    someone pleaseeeeeeeee answer
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    I think it will be higher because you are adding less liquid in to dilute it with, sorry if I've got the wrong end of the stick here, it's worded slightly confusingly
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    I'm not sure what you're asking but I'll try and clear some things up.

    The concentration of Na2CO3 doesn't decrease because some was spilt. It stays the same. Think about it. If you had a flask of 25cm^3 of 2 M HCl and poured half into another flask, you'd have two 12.5cm^3 solutions of 2 M HCl. The Na2CO3 solution left in the flask is still the same concentration as originally, as is the concentration of the Na2CO3 solution spilt on the bench.

    But because the volume remaining is now less than 25cm^3, according to moles = volume x concentration, the number of moles of Na2CO3 is less. You seem to understand that this means you need fewer moles of HCl to neutralise it.

    What I wasn't sure was whether you're only adding the exact amount of HCl required for neutralisation or adding all of it in the burette. Adding the exact amount doesn't affect the concentration of HCl whatsoever. This is predertermined by whoever made up the HCl solution. If you add all of it, because there is less base to remove the acid, you'll have more acid leftover at the end, hence increasing the ending concentration of HCl. Maybe this is what you meant.
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    (Original post by Cleoleo)
    Okay the analyte is sodium carbonate solution. Now, as a person was adding 25 cm cubed of sodium carbonate, some accidentally spilt.

    If HCL is the titrant (in the burette) and it is added to the sodium carbonate which is now less than 25 cm cubed, how will this affect the HCL concentration compared to if none had been spilt?

    I asked some people and they say the concentration of HCL will be higher...??

    But why or why not?


    Thank you!
    Just to add to Kyri's post above.

    If your HCl is the unknown whose concentration you are trying to determine using a standardised solution of sodium carbonate, then you will end up with titration results with a smaller volume of HCl than expected.

    You will not know about the spillage and assume that you have added 25ml of x mol dm-3 sodium carbonate.

    So, when you carry out your calculation to find the molarity of HCl the value will be higher than the actual value.

    Just for simplicity's sake let's assume it's a 1:1 reaction:

    Molarity HCl = (0.025x)/vol HCl

    and as vol HCl is smaller than it should be, then Molarity HCl is larger.
 
 
 
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