The calcium content in a 205.7mg chunk of an antacid tablet (presumed to contain

Question

The calcium content in a 205.7mg chunk of an antacid tablet (presumed to contain no metals except calcium) was determined by digesting (dissolving) it in a few mL of concentrated acid, bringing the resulting mixture to 25.00mL by addition of a pH 10 buffer, adding 3 drops of magnesium solution and a shot of calmagite indicator, and finally titrating to an endpoint with 29.77 mL of 0.01042 M EDTA solution. In a separate titration, the indicator blank correction for the magnesium was determinded by titrating 25.00 mL of calcium free water ( with the magnesium and calmagite added) to an endpoint with 1.15 mL of the EDTA solution.

A.) What is the % calcium (AW 40.078) in the antacid?

B.) Express the percent calcium as % calcium carbonate (MW CaCO3 = 100.087) in the antacid.

Explanation / Answer

amount of EDTA required to titrate Ca = 29.77-1.15 = 28.62mL

moles of EDTA added = 0.01042 moles/L * 28.62 mL/1000mL = 2.98*10^-4

moles of EDTA = moles of Ca^2+ =2.98*10^-4

mass of Ca = 2.98*10^-4* 40.07gm/mole = 0.01195 gm =11.95 mg

mass % = 11.95/205.7 *100 = 5.81 %

(b) 40.078 gm Ca is present in 100.087 gm CaCO3. mass of CaCO3 which contain 0.01195 gm Ca

= 100.087*0.01195/40.078

=0.02984 gm

% = 29.84/205.7 *100 = 14.5 %

(a)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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