Please help with this question, I have no clue how to go about it.

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.

mmoles HCl added = 0.1746 x 75.00 mL = ??

mmoles NaOH to titrate the excess acid = 27.20 mL x 0.09767 M NaOH = ?? moles NaOH.
The difference is the mmoles of the antacid in the 1.3188 g. That difference divided by the 1.3188 will be what you are looking for.

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 I don't understand how to find the mmoles in 1.3188g.

Thanks.

Hi,

I also don't understand what you meant by the difference is the mmoles of the antacid in the 1.3188 g.

Please help, when possible.

To solve this question, you will need to follow a series of steps:

Step 1: Calculate the number of moles of HCl used in the reaction.
To do this, you need to use the equation Molarity (M) = Moles (mol) / Volume (L). Rearrange the equation to solve for moles, which will give you Moles = Molarity x Volume (L).
Moles of HCl = 0.1746 M x (75.00 mL / 1000 mL/L) = ?

Step 2: Calculate the number of moles of NaOH used in the back titration.
Follow the same process as Step 1, using the given molarity and volume of NaOH.
Moles of NaOH = 0.09767 M x (27.20 mL / 1000 mL/L) = ?

Step 3: Calculate the number of moles of HCl that reacted with NaOH.
Since HCl and NaOH react in a 1:1 ratio, the number of moles of HCl that reacted will be equal to the number of moles of NaOH used in the back titration.

Step 4: Calculate the number of moles of HCl that reacted with the antacid.
To do this, you need to subtract the moles of HCl that reacted with NaOH (from Step 3) from the moles of HCl used in the reaction (from Step 1).
Moles of HCl reacted with antacid = Moles of HCl used in the reaction - Moles of HCl reacted with NaOH.

Step 5: Calculate the mass of the antacid used in the reaction.
You were given the mass of antacid as 1.3188g.

Step 6: Calculate the neutralizing power of the antacid in mmol/g.
To do this, divide the moles of HCl reacted with the antacid by the mass of the antacid used, then multiply by 1000 to convert moles to millimoles.
Neutralizing power = (Moles of HCl reacted with antacid / Mass of antacid) x 1000.

Now, using these steps, you can perform the calculations and find the neutralizing power of the antacid in terms of mmol H+ per gram of antacid.