Question

1. A student was given an antacid tablet that contained the active ingredient calcium carbonate. The tablet weighed 4.2810 g. He dissolved it in 25.00 mL of 0.6200 M HCl and heated it gently for several minutes. After cooling, it took 37.12 mL of 0.1414 M NaOH to titrate the excess acid to the phenolphthalein end point.

a. Write a balanced equation for the reaction of the antacid with HCl.

b. Write a balanced equation for the reaction of HCl with NaOH.

c. How many moles of the HCl were used to dissolve the antacid tablet? 0.0155 mol d

. How many moles of NaOH were needed to titrate the excess acid? 0.0052 mol

e. How many moles of acid per gram of tablet were neutralized?

f. How many grams of CaCO3 were in the antacid tablet?

Answer 1

a. Write a balanced equation for the reaction of the antacid with HCl.

Answer: CaCO_3(s) + 2 HCl_(aq) CaCl_2(aq) + H₂O_(l) + CO_2(g)

b. Write a balanced equation for the reaction of HCl with NaOH.

Answer: NaOH_(aq) + HCl_(aq) NaCl_(aq) + H₂O_(l)

c. How many moles of the HCl were used to dissolve the antacid tablet?

Answer:

Thus, moles of acid = Molarity Volume of Solution = 0.62 M 0.025 L = 0.0155 mol.

d. How many moles of NaOH were needed to titrate the excess acid?

Answer:

Thus, moles of NaOH = Molarity Volume of Solution = 0.1414 M 0.03712 L = 0.0052 mol.

e. How many moles of acid per gram of the tablet were neutralized?

Answer: From the chemical equation of HCl and NaOH, it is clear that one mole of HCl reacts with one mole of NaOH. Thus,

Moles of acid (HCl) neutralized by NaOH = 0.0052 mol.

Moles of acid neutralized by 4.281 g of tablet = 0.0155 mol - 0.0052 mol = 0.0103 mol.

Moles of acid neutralized by per gram of tablet =

f. How many grams of CaCO₃ were in the antacid tablet?

Answer: From the chemical equation, it is clear that one mole of CaCO₃ neutralizes two moles of HCl. Thus, to neutralize 0.0024 moles of HCl,

0.0012 moles of CaCO₃ is required.

Now, Molar mass of CaCO₃ = 100 g/mol

A gram of CaCO₃ in the antacid tablet = Molar mass of CaCO₃ moles of CaCO₃ in the antacid tablet

A gram of CaCO₃ in the antacid tablet = 100 g/mol 0.0012 mol = 0.12 gram