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    I read on ChemRevise the following statement:
    "Distilled water can be added to the conical flask during a titration to wash the sides of the flask so that all the acid on the side is washed into the reaction mixture to react with the alkali. It does not affect the titration reading as water does not react with the reagents or change the number of moles of acid added. Only distilled water should be used to wash out conical flasks between titrations because it does not add and extra moles of reagents. "

    https://chemrevise.files.wordpress.c...nd-group-7.pdf

    Could somebody please explain to me how this works chemically? Doesn't water - whether distilled or not, dilute any solution (in this case the acid)? And therefore change the molarity?
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    (Original post by I <3 WORK)
    I read on ChemRevise the following statement:
    "Distilled water can be added to the conical flask during a titration to wash the sides of the flask so that all the acid on the side is washed into the reaction mixture to react with the alkali. It does not affect the titration reading as water does not react with the reagents or change the number of moles of acid added. Only distilled water should be used to wash out conical flasks between titrations because it does not add and extra moles of reagents. "

    https://chemrevise.files.wordpress.c...nd-group-7.pdf

    Could somebody please explain to me how this works chemically? Doesn't water - whether distilled or not, dilute any solution (in this case the acid)? And therefore change the molarity?
    Yes it does change the concentration, but the number of moles is still the same as the volume is greater. It's similar to how if you used double the volume of acid with half the conc it would still take the same amount of NaOH to neutralise as it would with twice the conc but half the volume
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    (Original post by samb1234)
    Yes it does change the concentration, but the number of moles is still the same as the volume is greater. It's similar to how if you used double the volume of acid with half the conc it would still take the same amount of NaOH to neutralise as it would with twice the conc but half the volume
    Ahah thank you I see, so if you were to perform the titration calculations later you would have to subtract the extra volume added right?
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    (Original post by I <3 WORK)
    Ahah thank you I see, so if you were to perform the titration calculations later you would have to subtract the extra volume added right?
    No the number of moles hasn't changed so it would just be number of moles =conc of acid*volume of acid
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    I was wondering since our upcoming chemistry exams are coming up if people wanted to be part of a chemistry whatsapp revision group. If you are interested either pm me or responded to this message











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