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Introductory to chem course Stoichometry question wrong answer, can anyone fix it

  1. May 16, 2010 #1
    Question:

    A 25 mL sample of H2C204 solution required 19.62 mL of 0.341 M NaOH for neutralization. Calculate the molarity of the acid?

    The reaction was :
    H2C204 + 2NaOH ---------------------> Na2C2O4 + 2H2O

    Attempt:
    [Which line has the mistake in it and what is it?]

    converted 25mL to L
    converted 19.62 to L
    multiplied 0.01962 by 0.341 m/L of NaOH
    and multiplying all of that by 1 mol of H2C204 / for every 2 moles of NaOH
    multplied all of that by .025 L to get 0.00323 moles of the H2C204 , they said they wanted the molarity so molarity is moles/liter of solution, you get .00323/0.025L = .129 Molarity of H2C204
    THE ANSWER CAME OUT WRONG??????? WHAT the heck??!?!? Why??? Which line has the mistake in it and what is it?

    The answer is supposed to be 0.134 right?
     
  2. jcsd
  3. May 17, 2010 #2
    What I do is convert everything to moles (mmoles) in order to figure it out.

    19.62 mL * 0.341 M NaOH = 6.69 mmoles OH-

    To completely neutralize that, you need an equivalent in H+, so you need 6.69 mmoles H+. However, H2C2O4 produces 2 H+ per dissociation, so you would need half that amount, or 3.35 mmoles H2C2O4. Then divide by its volume.

    3.35 mmoles H2C2O4 / 25 mL = 0.134 M H2C2O4 solution


    I got 0.134 M. Are you saying you put that and got it wrong? Your answer says that you got 0.129 M. I think the error was in your calculation rather than concept.
     
  4. May 17, 2010 #3

    Borek

    User Avatar

    Staff: Mentor

    If I understand you correctly you have calculated number of moles of oxalic acid and multiplied it by volume to get number of moles of oxalic acid?

    And then you divided it back by the same volume?

    Strangely, numbers you have listed don't confirm your description. Check why you got 0.00323 and not 0.00334.

    --
     
  5. May 24, 2010 #4
    A 25 mL sample of H2C204 solution required 19.62 mL of 0.341 M NaOH for neutralization. Calculate the molarity of the acid?

    H2C204 + 2NaOH ---------------------> Na2C2O4 + 2H2O

    easy way of doing stoich is with unit analysis, so in order to that we need everything in Liters and since I don't like dealing with mL/mmol that's just what I'm going to do
    0.025 L H2C204
    0.01962 L and 0.341 mol/L NaOH

    we want to find the molarity of our acid so we want mol/L of H2C204

    (0.01962 L/1) * (0.341 mol/1L) * (1 mol H2C2O4/2 mol NaOH) * (1/0.025L) = 0.133 mol/L

    basicly you found the mols of your given and multiplied that by your mol ratio (required on top, given on bottom) and then divide that by your volume of the substance you want to find in order to get mol/L

    and they way I just showed you, you can do it in one step.
     
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