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.

Well, it seems like you're in quite a pickle there, trying to figure out what to do next! But don't worry, I'm here to help.

To find the neutralizing power of the antacid, we need to determine how many millimoles of H+ are neutralized by the antacid per gram of antacid.

Since the excess acid is back-titrated with NaOH, we can say that the number of millimoles of H+ neutralized is equal to the number of millimoles of NaOH used in the back titration.

You've already calculated the number of millimoles of NaOH used, which is 2.6566 moles. Now, we need to find the mass of the antacid that was used.

You mentioned that 1.3188g of antacid was weighed. So now, we can divide the number of millimoles of NaOH used by the mass of the antacid to find the neutralizing power.

(2.6566 moles / 1.3188 g) = 2.015 mmol/g

Therefore, the neutralizing power of the antacid is approximately 2.015 mmol H+ per gram of antacid.

Voila! I hope this helps, and remember, if you have any more questions, don't hesitate to ask. Happy calculating!

To calculate the neutralizing power of the antacid in terms of mmol H+ per gram of antacid, you need to determine the number of moles of HCl that reacted with the antacid and then divide it by the mass of the antacid used.

Here's what you can do next:

1. Calculate the moles of HCl that reacted with the antacid:
Moles of HCl = 0.1746 M x 75.00 mL = 13.095 mmol (you've already calculated this)

2. Calculate the moles of NaOH that reacted with the excess HCl:
Moles of NaOH = 27.20 mL x 0.09767 M = 2.6566 mmol (you've already calculated this)

3. Use the balanced equation between HCl and NaOH to determine the ratio of moles between them. From the equation:
HCl + NaOH -> NaCl + H2O
The balanced equation shows that 1 mole of HCl reacts with 1 mole of NaOH.

Therefore, the moles of NaOH that react with the moles of HCl are the same.

4. Since the moles of NaOH that reacted with the excess HCl are the same as the moles of HCl that reacted with the antacid, the moles of HCl that reacted with the antacid can be determined as:
Moles of HCl reacted with the antacid = 2.6566 mmol

5. Calculate the neutralizing power of the antacid in terms of mmol H+ per gram of antacid:
Mass of the antacid = 1.3188 g

Neutralizing power = (moles of HCl reacted with the antacid) / (mass of the antacid)
= 2.6566 mmol / 1.3188 g

Calculate the neutralizing power using the given values.

I hope this helps! Let me know if you have any further questions.

To calculate the neutralizing power of the antacid in terms of mmol H+ per gram of antacid, you need to determine the number of moles of H+ reacted with the antacid.

To do that, you can start by finding the number of moles of HCl that reacted with the antacid. You already calculated that there were 13.095 moles of HCl added to the solution. However, since the excess acid requires 27.20 mL of 0.09767 M NaOH, it means that not all of the HCl reacted with the antacid. This excess HCl is what reacted with the NaOH during the back titration.

To find the moles of HCl that reacted with the antacid, you can subtract the moles of NaOH used in the back titration from the moles of HCl added:

Moles of HCl reacted with antacid = Moles of HCl added - Moles of NaOH used

Moles of HCl reacted with antacid = 13.095 - 2.6566 mol

Next, you need to convert the mass of the antacid used (1.3188 g) to moles by dividing it by the molar mass of the antacid. You will need to know the molar mass of the antacid to proceed. Once you have the moles of the antacid, you can calculate the neutralizing power:

Neutralizing power (mmol H+ per gram of antacid) = (Moles of HCl reacted with antacid / mass of antacid) x 1000

Make sure to convert the result to mmol by multiplying by 1000, as stated in the question.