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    2.50g of an unknown carbonate were dissolved in 100cm3 of 1.00moldm−3 hydrochloric acid (an excess). The resulting solution was made up to 250cm3 in a volumetric flask. 25.00cm3 aliquots of this solution were titrated against 0.250moldm−3 sodium hydroxide. The titre is 24.10cm3. I've calculated amount of NaOH as 0.006025 mol. Calculate the amount of hydrochloric acid remaining after reaction.
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    How many moles of HCl were in the 25cm3 sample?

    This was a 25cm3 sample from 250cm3, so you must have 10 times as many moles HCl in the entire 250cm3 flask as were in the 25cm3 sample pipetted and titrated.
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    (Original post by TutorsChemistry)
    How many moles of HCl were in the 25cm3 sample?

    This was a 25cm3 sample from 250cm3, so you must have 10 times as many moles HCl in the entire 250cm3 flask as were in the 25cm3 sample pipetted and titrated.
    I calculated the moles in in a sample as 0.01, so I then subtracted 0.006025 from that. And it's not right?
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    (Original post by ambershell27)
    I calculated the moles in in a sample as 0.01, so I then subtracted 0.006025 from that. And it's not right?
    How did you calculate the moles in the sample?
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    (Original post by TutorsChemistry)
    How did you calculate the moles in the sample?
    n=cV, which is 0.1. Then divide by 10.
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    (Original post by ambershell27)
    n=cV, which is 0.1. Then divide by 10.
    That was the concentration before the unknown carbonate was added. The carbonate reacted with some of the HCl.

    The titration with NaOH is to find how much remains unreacted after the carbonate.
    The titration with NaOH will tell you the number of moles of HCl in the 25cm3.

    You just need to have a balanced equation of NaOH and HCl. The titration must obey that equation, and you know the number of moles of NaOH as you already stated that.

    How many moles of HCl in the 25cm3 sample?
    So how many in the 250cm3 flask?
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    (Original post by TutorsChemistry)
    That was the concentration before the unknown carbonate was added. The carbonate reacted with some of the HCl.

    The titration with NaOH is to find how much remains unreacted after the carbonate.
    The titration with NaOH will tell you the number of moles of HCl in the 25cm3.

    You just need to have a balanced equation of NaOH and HCl. The titration must obey that equation, and you know the number of moles of NaOH as you already stated that.

    How many moles of HCl in the 25cm3 sample?
    So how many in the 250cm3 flask?
    Ah, I understand now. The question is asking for the amount of HCl after the carbonate is added, which would be 10x0.00625. Thanks for the help!
 
 
 
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