No, I'm sorry I didn't see it. Those get buried after a page or two.
You added 75.0 mL HCl initially. It reacted with the tablet and used up all of its neutralizing power. So mmoles HCl added initially = mL x M which we've done.
But you had some HCl left over unreacted. That was titrated with NaOH. So mmoles NaOH added to titrate HCl = mLNaOH x M NaOH. That subtracted from the mmoles HCl initially gives you the amount of base/alkalinity/neutralizing power in the tablet.
I wouldn't round numbers as you did. You have four significant figures in the HCl M (0.1845) and if that is 75.00 mL that = 13.8375 (which I would keep in my calculator but it rounds to 13.84 mmoles). For mmoles NaOH to titrate the excess HCl I have 0.1482M x 14.18 mL NaOH = 2.504 mmoles NaOH used to tirate the unused HCl (that's number 6). (I think 13.84-2.504 = 11.34 mmoles for the HCL used by the tablet (that's number 7).
8 is mmoles from 7/1.2173.
Check my work.
Sorry DrBob22, I wasn't sure if you saw my last post. But, I am still a little confused.
Here's the info:
Antacid Brand: Life
Concentration of HCl: 0.1845 M
Concentration of NaOH: 0.1482 M
Trial 1:
1.Mass of Table: 1.2173g
2.Volume of HCl added: 75.0 mL
3.Milliomoles of HCl added:mmoles ? Help
4.Volume of NaOH used: 17.14 mL
5.Millimoles of NaOH used in titration:mmoles ? Help.
6.Millimoles of HCl that did not react with the antacid:mmoles ? Help.
7.mmoles of HCl neutralized by tablet:? Help.
8.mmoles of H+ neutralized per gram of antacid:mmoles/g ? Help.
I understand 3 and 5 now but still don't understand 6-8.
I got for 3. 13.8 mmoles and for 5. 11.1 mmoles.
Help with 6-8 anyone?
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