anonymous
  • anonymous
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
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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JoannaBlackwelder
  • JoannaBlackwelder
The density of acetic acid is 1.01 g/mL
JoannaBlackwelder
  • JoannaBlackwelder
Do you have the molarity of the sodium bicarbonate mixture?
anonymous
  • anonymous
the molarity is 1.14M

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anonymous
  • anonymous
how did you find the density of the acetic acid?
JoannaBlackwelder
  • JoannaBlackwelder
Google.
JoannaBlackwelder
  • JoannaBlackwelder
Molarity=mol solute/L solution
anonymous
  • anonymous
so if i multiplied the 1.14M by .0079L would that give me the moles of NaHCO3 used?
JoannaBlackwelder
  • JoannaBlackwelder
That is the idea, but I thought we had .0095 L
anonymous
  • anonymous
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?
JoannaBlackwelder
  • JoannaBlackwelder
Yep :-)
JoannaBlackwelder
  • JoannaBlackwelder
And no worries :-)
anonymous
  • anonymous
Thanks so much for your help I think I actually understand this now!
JoannaBlackwelder
  • JoannaBlackwelder
You're very welcome! :-D

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