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Concentration of a salt produced by neutralization reaction

  1. Jun 6, 2017 #1
    1. The problem statement, all variables and given/known data
    10.00 cm3 of 1.00 mol dm–3 sulfuric acid is fully neutralized by 20.00 cm3 of 1.00 mol dm–3 of sodium hydroxide. What is the concentration, in mol dm–3, of sodium sulfate solution produced by the reaction?
    A 0.33
    B 0.50
    C 0.67
    D 1.00

    Correct answer = A

    2. Relevant equations

    Mol = Concentration * Volume (dm^3)

    3. The attempt at a solution

    I calculated the number of moles of sodium sulfate which are 0.01 mol from the formula and ratio between the moles of sulfuric acid:sodium sulfate or sodium hydroxide:sodium sulfate. We dont have the volume of sodium sulfate. If i take the volume as 30 cm^3 for sodium sulfate then i get the correct answer A but water is present too so thats not possible, please explain!
     
  2. jcsd
  3. Jun 6, 2017 #2

    Borek

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    Staff: Mentor

    Yes, there are minute amounts of water produced, but also some of the ions that were present disappeared. These effects mostly cancel out so you can safely assume the final volume is just sum of volumes.
     
  4. Jun 6, 2017 #3
    Oh okay, how do the ions disappear?
     
  5. Jun 7, 2017 #4

    Borek

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    Staff: Mentor

    What is the neutralization reaction?
     
  6. Jun 7, 2017 #5
    H+1 and OH-1 form water and the other ions are spectator ions so they cancel out. OH OKAY. But then if they're disappearing why do they have a volume?
     
  7. Jun 7, 2017 #6

    Borek

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    Staff: Mentor

    It is not like they are disappearing out of this world - write the reaction equation. Two ions disappear, something appears. Volumes of the reacting ions and the product are not exactly identical, but they are quite similar.
     
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