Question

A student tried this same procedure (with slight modification) on a gel antacid. The student dissolved 3.00mL in 25.00mL of standardized HCI solution containing 1.727x10^-2 mol of HCI. The back-titration of the excess HCI required 4.62mL of 1.015M NaOH solution.

A) Calculate the molar concentration of the HCI solution that the student used.
B) Calculate the number of miles of HCI neutralized by the antacid tablet.
C) Calculate the volume effectiveness of the gel.

Answer 1

A) The molar concentration of the HCl solution used by the student is 1.013 M. B) The number of moles of HCl neutralized by the antacid tablet is 0.0077263 mol. C) The volume effectiveness of the gel is 254%.

Step-by-step explanation:

A) To calculate the molar concentration of the HCl solution, we can use the equation:

Moles of HCl = Moles of NaOH

Given that 25.00 mL of the standardized HCl solution containing 1.727x10^-2 mol of HCl was required to neutralize 4.62 mL of 1.015M NaOH, we can calculate the volume of HCl solution used:

Volume of HCl solution used = 3.00 mL + Volume of standardized HCl solution used in back-titration = 3.00 mL + 4.62 mL = 7.62 mL

Moles of HCl = Volume of HCl solution used × Molar concentration of NaOH = 7.62 mL × 1.015M = 0.0077263 mol

Molar concentration of HCl = Moles of HCl / Volume of HCl solution used = 0.0077263 mol / 0.00762 L = 1.013 M

B) To calculate the number of moles of HCl neutralized by the antacid tablet, we can use the equation:

Number of moles of HCl = Molar concentration of HCl × Volume of HCl solution used = 1.013 M × 0.00762 L = 0.0077263 mol

C) The volume effectiveness of the gel can be calculated using the equation:

Volume effectiveness = Volume of HCl solution used / Volume of gel used × 100%

Given that the volume of HCl solution used is 7.62 mL and the volume of gel used is 3.00 mL, we can calculate the volume effectiveness:

Volume effectiveness = (7.62 mL / 3.00 mL) × 100% = 254%