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?
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.
For question 6, we need to calculate the number of millimoles of HCl that did not react with the antacid. To do this, we can subtract the number of millimoles of NaOH used in the titration (question 5) from the number of millimoles of HCl added (question 3). So, the calculation would be:
Millimoles of HCl that did not react with the antacid = Millimoles of HCl added - Millimoles of NaOH used in titration
In this case, you mentioned that you had calculated 13.8 millimoles for question 3 and 11.1 millimoles for question 5. Therefore, substituting these values into the formula, we get:
Millimoles of HCl that did not react with the antacid = 13.8 mmol - 11.1 mmol = 2.7 mmol
For question 7, we need to find the millimoles of HCl neutralized by the tablet. This can be calculated using the stoichiometric ratio between HCl and NaOH, as determined by the balanced chemical equation. Assuming a 1:1 ratio (which is often the case in acid-base reactions), the number of millimoles of HCl neutralized by the tablet would be equal to the number of millimoles of NaOH used in the titration (question 5). So, in this case, it would be:
Millimoles of HCl neutralized by tablet = Millimoles of NaOH used in titration = 11.1 mmol
Finally, for question 8, we need to calculate the millimoles of H+ neutralized per gram of antacid. To do this, we'll divide the number of millimoles of HCl neutralized by the tablet (question 7) by the mass of the tablet (question 1). The equation would be:
Millimoles of H+ neutralized per gram of antacid = Millimoles of HCl neutralized by tablet / Mass of tablet
Using the values you provided, this would be:
Millimoles of H+ neutralized per gram of antacid = 11.1 mmol / 1.2173 g = 9.113 mmol/g
I hope this helps clarify questions 6-8 for you! Let me know if you have any further questions.