Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    How do you work out question 6e?

    The answers are:
    Volume of HCl(aq) added for first end-point 15
    Volume of HCl(aq) added for second end-point 45
    Attached Files
  1. File Type: rtf 4.3 exam questions (1).rtf (35.6 KB, 111 views)
    • Community Assistant
    • Study Helper
    Offline

    14
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by Smcdonald98)
    How do you work out question 6e?

    The answers are:
    Volume of HCl(aq) added for first end-point 15
    Volume of HCl(aq) added for second end-point 45
    The first inflexion is due to the reaction:

    Na2CO3 + HCl --> NaHCO3 + NaCl

    The addition of more sodium hydrogencarbonate is not going to affect this, as it only shows the reaction of sodium carbonate. The inflexion remains at 15ml

    As the first inflexion is at 15ml adding an equal number of moles of sodium hyrogencarbonate will add an extra 15ml to the second inflexion = 45ml
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by charco)
    The first inflexion is due to the reaction:

    Na2CO3 + HCl --> NaHCO3 + NaCl

    The addition of more sodium hydrogencarbonate is not going to affect this, as it only shows the reaction of sodium carbonate. The inflexion remains at 15ml

    As the first inflexion is at 15ml adding an equal number of moles of sodium hyrogencarbonate will add an extra 15ml to the second inflexion = 45ml

    A titration curve is plotted showing the change in pH as a 0.0450 mol dm–3 solution of sodium hydroxide is added to 25.0 cm3 of a solution of ethanedioic acid, H2C2O4 The titration curve obtained has two equivalence points (end points)
    (ii) When the second equivalence point is reached, a total of 41.6 cm3 of 0.0450 mol dm–3 sodium hydroxide has been added. Calculate the concentration of the ethanedioic acid solution.

    In this question why is the moles of H2C2O4 equal to half moles of OH–? Thanks
    • Community Assistant
    • Study Helper
    Offline

    14
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by Smcdonald98)
    A titration curve is plotted showing the change in pH as a 0.0450 mol dm–3 solution of sodium hydroxide is added to 25.0 cm3 of a solution of ethanedioic acid, H2C2O4 The titration curve obtained has two equivalence points (end points)
    (ii) When the second equivalence point is reached, a total of 41.6 cm3 of 0.0450 mol dm–3 sodium hydroxide has been added. Calculate the concentration of the ethanedioic acid solution.

    In this question why is the moles of H2C2O4 equal to half moles of OH–? Thanks
    Have you tried writing out the equation for the reaction?
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.