Question
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.
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.
Answers
DrBob222
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
<b>No, mmoles NaOH used is mL NaOH x M NaOH = 17.14 x 0.1482 = 2.545. </b>
6.Millimoles of HCl that did not react with the antacid: (0.1482M x 14.18mL NaOH= 2.504 mmoles)
<b>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.</b>
7.mmoles of HCl neutralized by tablet:(13.84 HCl-2.504 NaOH = 11.34 mmoles)
<b>Or 13.84-2.545 = ?? depending upon the 17.14 vs 14.18 discrepancy.</b>
8.mmoles of H+ neutralized per gram of antacid:(7/1.2173=5.750 mmoles/g)
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
<b>No, mmoles NaOH used is mL NaOH x M NaOH = 17.14 x 0.1482 = 2.545. </b>
6.Millimoles of HCl that did not react with the antacid: (0.1482M x 14.18mL NaOH= 2.504 mmoles)
<b>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.</b>
7.mmoles of HCl neutralized by tablet:(13.84 HCl-2.504 NaOH = 11.34 mmoles)
<b>Or 13.84-2.545 = ?? depending upon the 17.14 vs 14.18 discrepancy.</b>
8.mmoles of H+ neutralized per gram of antacid:(7/1.2173=5.750 mmoles/g)
Akhed-To DrBob222
Thanks...but for #8, where did the 7 come from?
DrBob222
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.
<b>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.</b>
and...if I were asked to find the average in mmoles of H+/g of antacid tablet, how would I do that?
<b>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.
</b>
<b>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.</b>
and...if I were asked to find the average in mmoles of H+/g of antacid tablet, how would I do that?
<b>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.
</b>
DrBob222
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.
meltem
for question 6 why did you use NaOH its asking for HCl that didn't react