Answer to Question #97185 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

$CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$ (equation 1)

aa

$NaOH + HCl \rightarrow NaCl + H_2O$ (equation 2)

$n(HCl)_ {total} = c\times V = 0.0981\times 0.025 = 0.00245 mol$

$n(NaOH) = c\times V = 0.104\times 0.00583 = 0.000606 mol$

$n(HCl)_{titration} = n(NaOH) = 0.000606 mol$

Find moles of HCl used for the reaction 1:

$n(HCl) = n(HCl)_{total} - n(HCl)_{titration} = 0.00245 - 0.000606 = 0.00184 mol$

Find moles of CaCO3 (equation 1) :

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

a) Find mass of CaCO3 ;

$m(CaCO_3) = n\times M = 0.000922\times 100.09 = 0.0923 g$

b) Find percent active ingrdient

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