0.688g of antacid was treated with 50.00ml of 0.200M HCl. The excess acid required 4.21 mL of 1.200 M NaOH for back-titration. What is the neutralizing power of this antacid expressed as mmol of HCl per gram of antacid?

I think the answer should be:
(50.00ml HCl)(0.200M HCl)= 10mmol HCl
10mmol HCL/0.688g antacid= 14.535mmol/g.

I don't know what needs to done with 4.21ml NaOH?

Caroline has a problem just above yours. Here is a link.

http://www.jiskha.com/display.cgi?id=1428332431

To determine the neutralizing power of the antacid expressed as mmol of HCl per gram of antacid, you need to consider the reaction that took place between the antacid, HCl, and NaOH during the titration.

The balanced equation for the reaction is:

HCl + NaOH -> NaCl + H2O

Based on the information given, the excess HCl was determined by back-titrating with NaOH. The reaction between the excess HCl and NaOH involves a 1:1 mole ratio. Therefore, the number of moles of excess HCl can be calculated using the concentration and volume of NaOH used in the back-titration.

First, convert the volume of NaOH used into liters:

4.21 mL = 4.21 * (1/1000) L = 0.00421 L

Next, calculate the number of moles of NaOH used:

0.00421 L NaOH * 1.200 mol/L NaOH = 0.005052 mol NaOH

Since the reaction is 1:1, the number of moles of excess HCl is also 0.005052 mol.

Now, determine the number of moles of HCl that reacted with the antacid. From the given data, it is known that 0.688 g of antacid was used.

To convert the grams of antacid to moles of HCl, you will need to use the molecular weight (MW) of the antacid. However, this information is not provided, so without it, it's not possible to calculate the number of moles of HCl that reacted with the antacid.

To obtain the neutralizing power expressed as mmol of HCl per gram of antacid, you would need the molecular weight of the antacid. With that information, you can perform the calculations as indicated in your previous attempt:

(50.00 mL HCl) * (0.200 M HCl) = 10 mmol HCl
10 mmol HCl / 0.688 g antacid = 14.535 mmol/g

However, since the molecular weight of the antacid is not provided, the final answer cannot be calculated. Additionally, the information about the volume and concentration of NaOH used in the back-titration does not contribute to finding the neutralizing power in this specific case.