Answer to Question #97184 in General Chemistry for Sunday Dak

Question

A 0.204 g sample of a CO3 2- antacid is dissolved with 25.0ml of 0.0981 M HCL. The hydrochloric acid that is not neutralized by the antacid is titrated to a bromophenol blue endpoint with 5.83 ml of 0.104 M NaOH.

a) Assuming the active ingredient in the antacid sample is CaCO3, calculate the mass of CaCO3 in the sample.

b) What is the percent active ingredient in the antacid sample?
Report the answers with three significant figures.

Answer 1

HCl is used for two reaction: reaction with antacid and for titration with NaOH

Find moles of HCl

$n(HCl) = 0.0981\times 0.025 = 0.00245 mol$

Find moles of HCl used for titration:

$NaOH + HCl \rightarrow NaCl + H_2O$

$n(HCl) = n(NaOH) = c\times V = 0.104\times0.00583 = 0.000606 mol$

Fnd moles of HCl used for reaction with antacid:

$n(HCl) = 0.00245 - 0.000606 = 0.00184 mol$

Find moles of CaCO3:

$CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$

$n(CaCO_3) \frac{n(HCl)}{2} = \frac{0.00184}{2} = 0.000922 mol$

a) Find mass of CaCO3:

$m=M\times n = 100.09\times 0.000922 = 0.0923 g$

b) find mass percent of CaCO3 in antacid:

$mass\% =\frac{0.0923}{0.204}\times 100\% = 45.2 \%$