21. Antacids are commonly determined by acid-base back titrations because the lo

Question

21. Antacids are commonly determined by acid-base back titrations because the low solubility of the antacid slows the direct reaction with acid. A 0.3251 g sample of a leading antacid (active ingredient: CaCO3) is added to 50.00 mL of 0.1102 M HCl. After the reaction, the excess acid is titrated with 29.43 mL of 0.1006 M NaOH. Calculate the number of milligrams of CaCO3 per tablet, assuming that an average tablet has a mass of 1.28 g. Also assume the CaCO3 to be totally neutralized. Molar Mass CaCO3 100. g.mol-1 (A) 392 mg (C) 784 mg (B) 502 mg (D) 1004 mg 22. Benzoic acid, C HsCOOH, is a weak, monoprotic, organic acid in aqueous medium. However, in the nonaqueous solvent methylamine, CH3 NH2, benzoic acid is readily ionized. Identify the conjugate acid for this ionization. (A) C6 H5 COOH2+ (C) C6HsCoo- (B) CH3NH3+ (D) CH3NH-

Explanation / Answer

a)Step1. Calculate the moles of HCl and NaOH

HCl = molarity(M) *V(L)

= 0.1102M* 0.050L = 0.00551mol

NaOH = 0.1006M*0.02943L = 0.00296mol

NaOH reacts with HCl in 1:1 ratio

thus0.00296mol NaOH neutralizes 0.00296mol HCl, rest of it is reacted byCO32-

Step2. Calculating the molesof CO32-

molesof HCl reacts with CO32- = 0.00551 - 0.00296=0.00255mol

2H+ + CO32- --> H2CO3

Thus according to the reaction 2mols of HCl reacts with1mol of CO32-

Moles of CO32- = 0.00255mol/2 = 0.001275mol

Step3. Mass of CaCO3

Mass(g) = moles * molar mass(g/mol) of CaCO3

= 0.001275mol* 100g/mol

=0.1275 g

=0.1275*1000/1.28

=1004 mg. Option (D) is correct

b). percent of active ingredient

mass% = mass of CaCO3/ Mass of sample(given) * 100%

Plug in the values to get the answer.

Hope uunderstood!

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