posted by Serenity on .
Hi, I posted this question earlier but still need some help:
1.3188g of antacid is weighed and mixed with 75.00mL of excess 0.1746 M HCl. The excess acid required 27.20 mL of 0.09767 M NaOH for back titration. Calculate the neutralizing power of the antacid in terms of mmol H+ per gram of antacid.
So far I have:
mmoles HCl added= 0.1746 x 75.00mL= 13.095 moles
mmles NaOH titrate the excess acid= 27.20 mL x 0.09767 M NaOH= 2.6566 moles
But don't know what to do from here.
Please help asap.
FIRST, 13.095 is mmoles HCl (not moles).
AND 2.6566 is mmoles NaOH (not moles).
You need to understand what is going on to really understand the problem. I suspect you've just copied what I told you to do last night and don't know where to turn now. Here is what is happening. The antacid tablet neutralizes acid and since these tablets often are insoluble, it is inconvenient to titrate them directly with HCl; therefore, one adds a measured EXCESS to completely neutralize all of the tablet AND have plenty left over. Then we simply back titrate (that is we determine how much excess was added) the HCl to see how much of an excess we had. So your 13.095 mmoles is how much TOTAL HCl was added, you back titrated the excess and obtained 2.6566 mmoles NaOH so the difference is how much of the initial HCl added actually reacted. Thus 13.095-2.6566 = mmoles HCl consumed by the tablet at the beginning. The question asks for mmoles H^+ and you have mmoles HCl. But since 1 mmole HCl contains 1 mmole H^+ you have exactly what you want. The question wants it expressed in mmoles H^+ per gram; therefore, divide mmoles H^+ by 1.3188 to obtain mmoles/gram.