Question

1. If the HCl used in this experiment was prepared at a concentration of 0.15M. Using your volume of HCl neutralized by your whole tablet, calculate the mass of active ingredient (CaCO3) in the tablet in milligrams.

2. Instead of using ratios for back titrations we can also use molarities, if our solutions are standardized. A 0.196g sample of antacid containing an unknown amount of triprotic base Al(OH)3 was reacted with 25.0mL of 0.111M HCl. The resulting solution was then titrated with 11.05mL of 0.132M NaOH solution. Calculate the mass percent of Al(OH)3 in the antacid sample.

Answer 1

2)

first we have to calculate the no.of moles of remaining Al(OH)3

for that

as the resulting solution was then titrated with 11.05mL of 0.132M NaOH solution

so no . of moles of Al(OH)3 = molarity of NaOH * vol. of NaOH in lt

                                        = 0.132*0.01105 lt = 0.00146 moles

let us assume total no.of moles of Al(OH)3 = x

so (x-0.00146) moles of Al(OH)3 can be react with 5.0mL of 0.111M HCl

ther fore

(x-0.00146) = 0.005 * 0.111

  (x-0.00146) = 0.000555

x = 0.002015 moles

total no. of moles of Al(OH)3 is 0.002015 moles

wt of Al(OH)3 = no.of moles*mol.wt

                      = 0.002015 * 78 = 0.15717

mass % of Al(OH)3 = [mass of Al(OH)3/mass of sample]*100

                             = [0.15717/0.196]*100

                              = 80.2 %

1) for this problem we need volume of HCl so we can calculate but here not mentioned the volume of HCl