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Ionic Equilibrium Problem

  1. Jun 7, 2014 #1
    1. The problem statement, all variables and given/known data

    Calculate the hydroxide ion concentration:

    16.5 mL of aqueous sulfuric acid at 1.5 M added to 12.7 mL of sodium hydroxide at 5.5M.

    2. The attempt at a solution

    H2SO4 (aq) + 2 NaOH (aq) <--> 2 H2O (l) + Na2SO4 (aq)

    Hydrogen Concentration: (16.5 mL / 1000) (1.5 M) = 0.02475 moles
    Hydroxide Concentration: (12.7 mL / 1000) (5.5 M) = 0.06985 moles

    0.0451 moles of Hydroxide / (16.5 + 12.7 mL) = 1.54 M
     
  2. jcsd
  3. Jun 8, 2014 #2

    Borek

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    Staff: Mentor

    This is hardly an equilibrium, looks like a simple stoichiometry to me.

    You have wrote the reaction equation (good idea) but then you ignored it (bad idea).
     
  4. Jun 8, 2014 #3
    H2SO4 (aq) + 2 NaOH (aq) <--> 2 H2O (l) + Na2SO4 (aq)

    Hydrogen Concentration: (16.5 mL / 1000) (1.5 M) = 0.02475 moles
    Hydroxide Concentration: (12.7 mL / 1000) (5.5 M) = 0.06985 moles x 2 = 0.1397

    0.0.11495 moles of Hydroxide / (16.5 + 12.7 mL) = 3.93 M
     
  5. Jun 8, 2014 #4

    Borek

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    Staff: Mentor

    With some luck third guess will be the correct one.
     
  6. Jun 8, 2014 #5
    Why? I incorporated the NaOH coefficient.
     
  7. Jun 8, 2014 #6

    Borek

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    Staff: Mentor

    Just because you multiplied something by 2 doesn't mean you did it correctly.

    How many moles of acid do you have? How many moles of NaOH will react with this acid?
     
  8. Jun 8, 2014 #7
    H2SO4 (aq) + 2 NaOH (aq) <--> 2 H2O (l) + Na2SO4 (aq)

    Acid: (16.5 mL / 1000) (1.5 M) = 0.02475 moles
    NaOH Concentration: (12.7 mL / 1000) (5.5 M) = 0.06985 moles

    0.0495 moles of NaOH will react with the acid.

    Extra OH: 0.06985 - 0.0495 = 0.02035 moles

    Concentration: (0.02035) / (16.5 + 12.7 mL) = 0.7 M
     
  9. Jun 8, 2014 #8

    Borek

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    Staff: Mentor

    Looks OK now.
     
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