Question

An antacid tablet was dissolved in 26.00 mL of 0.650 M HCl. The excess acid was back-titrated with exactly 11.34 mL of 1.05 M NaOH. The average weight of a tablet is 0.834 g. The tablet came from a bottle of 150 tablets that cost $3.99.

Calculate the moles of HCl neutralized by the tablet

Calculate the mass effectiveness of the antacid

Calculate the cost-effectiveness of the antacid

Answer 1

a)

moles of HCl = 26 x 0.650 / 1000 = 0.0169

moles of NaOH = 11.34 x 1.05 / 1000 = 0.0119

moles of HCl neutralized = 0.0169 - 0.0119 = 5.00 x 10^-3

moles of HCl neutralized = 5.00 x 10^-3 mol

b)

mass effectiveness = 5.00 x 10^-3 / 0.834

                               = 6.00 x 10^-3 mol

mass effectiveness = 6.00 x 10^-3 mol/ g

c)

cost effectiveness :

3.99 x 100 / 150 = 2.66

cost effectiveness = 5 x 10^-3 / 2,66

                              = 1.9 x 10^-3 mol / cent