An anctacid tablet was dissolved in 26.00ml of 0.650M HCl. The excess acid was b
ID: 1049178 • Letter: A
Question
An anctacid tablet was dissolved in 26.00ml of 0.650M HCl. The excess acid was back titrated with eactly 11.34ml of 1.05M NaOH. The average weight of a tablet is 0.834g. The tablet came from a bottle 150 tablets that cost $3.99.
a. Calculate the moles of HCl neutralized by the tablet. (moles of HCl(aq) neutralized per tablet = moles of HCl(aq) added to the antacid - moles of NaOH required to backtitrate)
b. Calculate the mass effectiveness of the antacid. (mass effectiveness = moles of HCl(aq) neutralized per tablet / mass of antacid (g))
c. Calculate the cost effectiveness of the antacid. (cost effectiveness = moles of HCl(aq) neutralized per tablet / price of tablet (cent))
Explanation / Answer
a)
mole s of HCl neutralized
moles of backtitration = MV = 11.34*1.05 = 11.907 mmol of OH-
moles of HCl added = MV = 26*0.65 = 16.9 mmol of H+
so...
neutralized by tablet = 16.9 - 11.907 = 4.993 mmol of H+ neutralized by tablet = 4.993*10^-3 mol
b)
mass of effective acid:
mass of HCl neutralized per tablet / mass of antacid = (4.993*10^-3)(36) / (0.834) = 0.21552517 g
c)
cost effective of antacid
$ per tablet = 3.99 /150= 0.0266 $ per tablet
cost = (4.993*10^-3 ) / (0.0266) = 0.1877067 mol / $ --> 18.7 mol/cent