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?
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.
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?
To find the moles of HCl neutralized by NaOH, you need to use the stoichiometry of the balanced chemical equation between HCl and NaOH.
The balanced chemical equation between HCl and NaOH is:
HCl + NaOH -> NaCl + H2O
From this equation, you can see that for every 1 mole of HCl, 1 mole of NaOH is needed to neutralize it.
Given that the initial moles of HCl is 0.005 moles, you know that the same amount of moles of NaOH will be used to neutralize it.
Therefore, the moles of HCl neutralized by NaOH is also 0.005 moles.
Based on the information you provided, you titrated 41 mL of the 0.10 M NaOH solution. To find the moles of NaOH used, you can use the following formula:
Moles of NaOH = concentration (M) x volume (L)
Moles of NaOH = 0.10 M x 0.041 L = 0.0041 moles
Since the stoichiometry of the equation is 1:1, the moles of NaOH used to neutralize HCl is the same as the moles of HCl neutralized by NaOH, which is 0.0041 moles.
I hope this helps! Let me know if you have any further questions.