Question

evaluating commercial antacids

a student tried this same procedure (with a slight modification) on a gel antacid. The student dissolved 3.00 mL in 25.00 mL of a standardized HCl solution containing1.727 *10^-2 mol of HCl. The back titration of the excess HCl required 4.62 mL of 1.015M NaOH solution.
a) Calculate the molar concentration of the HCl solution that the student used.
b) Calculate the number of moles of HCl neutralized by the antacid tablet.
c) Calculate the volume of the effectiveness of the gel.
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Answer #1

a) Molar concentration = moles/volume = 1.727*10^-2 mole/0.025L = .69M

b)Only a portion of the HCl was neutralized by the antacid so we need to find how much HCl was neutralized by NaOH and then subtract that from thetotal HCl. ,so there is a 1-1 ratio of HCl to NaOH. Now to find the moles of NaOH: . This means thereare of HCLneutralized by NaOH which means there were .025-.0068= 0.0182L neutralized by antacid.

c)I'm not entirely sure what is meant by the this last question, but I am assuming that it is asking for the final volume of the antacid + theneutralized HCL which would be: 3ml+18.2ml = 21.2ml

answered by: Chasidy
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