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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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
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The density of acetic acid is 1.01 g/mL
Do you have the molarity of the sodium bicarbonate mixture?
the molarity is 1.14M

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

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