Question about titration calculations in analysis of antacids lab. We did a lab yesterday with antacids, HCl and NaOH and I'm having trouble doing the calculations.
The information known is the following:
3)Mass of Antacid .22g
4)Volume of NaOH titrated from burette 41mL
So I figured the inital moles of HCl by multiplying the M and volume in liters, which came out to be .005 moles of HCl.
Next I have to find moles of HCl neutralized by NaOH which I'm confused on how to do, can anyone help me out?
chemistry - DrBob222, Thursday, March 11, 2010 at 6:03pm
You need to explain more about what you did. It sounds as if you performed a back-titration; something like adding a known quantity of HCl to the antacid tablet, then titrating the excess HCl with NaOH. If that is what you did.
moles HCl added intially = M x L = ?? (I assume this is the 0.005 moles).
Then M x L NaOH is the amount of excess HCl present (that is, not neutralized by the antacid). Moles initially - moles excess = moles antacid present at the beginning.
chemistry - Mindy, Thursday, March 11, 2010 at 8:05pm
Well we basically took antacid tablets dissolved them in HCl, heated it and then added phenolthalien and then titrated until we saw a pink color. At 41mL we saw the pink and stopped. So I guess like you said, I think I have to figure out the excess HCl present but how do I do that?