A 1.345g antacid tablet was dissolved in 75.00mL HCl (0.225M). It was then back

Question

A 1.345g antacid tablet was dissolved in 75.00mL HCl (0.225M). It was then back titrated with 28.76mL Naoh (0.201 M).

1) Calculate the number of moles of HCl that was neutralized by the antacid tablet

2) Assume that the molarity of stomach acid is 0.2M. Calculate the number of mL of 0.2M HCl that would contain the number of moles HCl calculated in the previous question.

3) Assuming the density of 0.2m HCl is 1.00g/mL. Calculate the mass of 0.2M Hcl solution neutralizable (from part 2 above) and compare it to the mass of the tablet used experimentally.

Explanation / Answer

1) Total moles in 75.00 mL of 0.225M of HCl = molarity x volume = 0.225M x 75.00 = 0.0169 mol

Moles of HCl neutralised by antacid = (total moles of HCl - moles of non-neutralised HCl)

Moles of non-nuetralised (unreacted) HCl is equivalent to the moles of HCl that reacted with NaOH. Since HCl and NaOH react in 1:1 ratio,

moles of HCl not nuetralised by antacid = moles of NaOH

= molarity(NaOH) x volume of NaOH = 0.201 x 28.76 = 0.0058 mol

So, number of moles of HCl that was neutralized by the antacid tablet = (0.0169 - 0.0058) mol = 0.0111 moles

2) Moles of HCl = 0.0111 mol

molarity = 0.2 M

volume = moles / molarity = 0.0111mol / 0.2 M = 0.0555 L = 55.5 mL

3) Density = mass/volume = 1.00 g/mL

So, mass of HCl = density x volume = 1.00 g/mL x 55.5 mL = 55.5 g

The mass of the antacid used = 1.345 g.

This value is so low probably because the antacid has a low molar mass compared to HCl

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