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

Question

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

joannablackwelder · Answer

The density of acetic acid is 1.01 g/mL

joannablackwelder · Answer

Do you have the molarity of the sodium bicarbonate mixture?

anonymous · Answer

the molarity is 1.14M

anonymous · Answer

how did you find the density of the acetic acid?

joannablackwelder · Answer

Google.

joannablackwelder · Answer

Molarity=mol solute/L solution

anonymous · Answer

so if i multiplied the 1.14M by .0079L would that give me the moles of NaHCO3 used?

joannablackwelder · Answer

That is the idea, but I thought we had .0095 L

anonymous · Answer

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 · Answer

Yep :-)

joannablackwelder · Answer

And no worries :-)

anonymous · Answer

Thanks so much for your help I think I actually understand this now!

joannablackwelder · Answer

You're very welcome! :-D