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anonymous

  • one year ago

I am really having trouble with how to set up the equation to solve this, Calculate the number of moles of NaHCO3 required to neutralize the CH3COOH in the vinegar. it took 9.5mL of NaHCO3, and 9.86mL of CH3COOH

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  1. JoannaBlackwelder
    • one year ago
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    The density of acetic acid is 1.01 g/mL

  2. JoannaBlackwelder
    • one year ago
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    Do you have the molarity of the sodium bicarbonate mixture?

  3. anonymous
    • one year ago
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    the molarity is 1.14M

  4. anonymous
    • one year ago
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    how did you find the density of the acetic acid?

  5. JoannaBlackwelder
    • one year ago
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    Google.

  6. JoannaBlackwelder
    • one year ago
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    Molarity=mol solute/L solution

  7. anonymous
    • one year ago
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    so if i multiplied the 1.14M by .0079L would that give me the moles of NaHCO3 used?

  8. JoannaBlackwelder
    • one year ago
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    That is the idea, but I thought we had .0095 L

  9. anonymous
    • one year ago
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    yes that's right, I was reading the number off the wrong paper sorry. if I did that the anser for the overall question should just come out to 0.01083 moles of NaHCO3 required right?

  10. JoannaBlackwelder
    • one year ago
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    Yep :-)

  11. JoannaBlackwelder
    • one year ago
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    And no worries :-)

  12. anonymous
    • one year ago
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    Thanks so much for your help I think I actually understand this now!

  13. JoannaBlackwelder
    • one year ago
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    You're very welcome! :-D

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