1. If the HCl used in this experiment was prepared at a concentration of 0.15M.

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.

Explanation / Answer

1.   There must be .15 M CaCO3 present. M.W of CaCO3 is 100. so, 100*.15 = .15 g = 150 mg CaCO3 present

2. let V mL .111 M HCl will nutralised by NaOH so V = .132 * 11.05 /.111 = 13.14 mL

that means 25 - 13.14 = 11.86 mL HCl requir to nuralised Al(OH)3.

Molarity of Al(OH)3 in the tablet is ( 11.86 * .111 ) /25 = .053M. since it is a triprotic acid the molarity of Al(OH)3 = .053 / 3 = .018 M

now M.W of Al(OH)3 = 78

so mass of Al(OH)3 = 78 * 0.018 = 1.36 g present in 1000 mL so in 25 ml .034 gm present

parcentage =[ .034 / .196] * 100 = 17.34% Al(OH)3 PRESENT

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.