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 #7, where'd you get the 13.84 from? 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.
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
No, mmoles NaOH used is mL NaOH x M NaOH = 17.14 x 0.1482 = 2.545.
6.Millimoles of HCl that did not react with the antacid: (0.1482M x 14.18mL NaOH= 2.504 mmoles)
What's the 14.18? I know I copied that from the problem but you have in this post a volume of 17.14 which will give a different number.
7.mmoles of HCl neutralized by tablet:(13.84 HCl-2.504 NaOH = 11.34 mmoles)
Or 13.84-2.545 = ?? depending upon the 17.14 vs 14.18 discrepancy.
8.mmoles of H+ neutralized per gram of antacid:(7/1.2173=5.750 mmoles/g)
Thanks...but for #8, where did the 7 come from?
I used you work to fix my problems but I had a few questions. For #7, where'd you get the 13.84 from? and For #8, where'd you get the 7 from.
The 13.84 mmoles is the initial amount of HCl added to the tablet (that 75.00 mL of 0.1845 M HCl). That was an excess which neutralized all of the tablet and then some. The rest of the lab was determining how much of excess you added.
and...if I were asked to find the average in mmoles of H+/g of antacid tablet, how would I do that?
I assume you had a trial 1, a trial 2, a trial 3, etc. You take the number of mmoles H^/g antacid (I think that's #8 in your table) and average them together.
Where did the 7 come from? The 7 stands for "step 7". I just meant to take the number you had for step 7 and divide by the mass in grams of the tablet.
for question 6 why did you use NaOH its asking for HCl that didn't react
No problem, let me explain!
For question #7, the 13.84 mmoles came from the calculation in step 3, where you determined the millimoles of HCl added. You multiplied the concentration of HCl (0.1845 M) by the volume of HCl added (75.0 mL), giving you 13.84 mmoles.
For question #8, the 7 mmoles came from the value you provided in your question. It seems that you made a mistake or there might be a typo. Please double-check your calculations or the given information to make sure the correct value is used in the calculation for #8.
To find the average in mmoles of H+/g of antacid tablet, you would need to conduct multiple trials and calculate the mmoles of H+ neutralized per gram of antacid in each trial. Then, you can take the average of those values.
For example, if you conducted three trials, you would perform the calculations for each trial to determine the mmoles of H+ neutralized per gram of antacid. Then, you would add up those values and divide the sum by 3 (the number of trials) to find the average.
I hope this clarifies your questions! Let me know if you need further assistance.