Asked by Akhed-To DrBob222

Sorry DrBob, it's me again!

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:(0.1845M x 75.0mL= 13.84 mmoles

4.Volume of NaOH used: 17.14 mL

5.Millimoles of NaOH used in titration: (0.1482M x 75.0mL= 11.12 mmoles

6.Millimoles of HCl that did not react with the antacid: (0.1482M x 14.18mL NaOH= 2.504 mmoles)

7.mmoles of HCl neutralized by tablet:(13.84 HCl-2.504 NaOH = 11.34 mmoles)

8.mmoles of H+ neutralized per gram of antacid:(7/1.2173=5.750 mmoles/g)

I used you work to fix my problems but I had a few questions. For #6, where'd you get the 14.18mL? and For #8, where'd you get the 7 from.

and...if I were asked to find the average in mmoles of H+/g of antacid tablet, how would I do that?

Sorry, my brain isn't functioning at its best.
Answered by Akhed-To DrBob222
sorry, i posted twice but I don't understand where the 7 came from in #8.
Answered by DrBob222
The 7 stands for step #7. You were to use the number you had in step 7 and divide that by mass antacid tablet to arrive at the value for step 8.
There are no AI answers yet. The ability to request AI answers is coming soon!