Question
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:
1)0.10M NaOH
2)0.10M HCl
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?
The information known is the following:
1)0.10M NaOH
2)0.10M HCl
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?
Answers
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.
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.
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?
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