Based on the three trials, determine the mean mass of Ca2+. Calculate the percen

Question

Based on the three trials, determine the mean mass of Ca2+. Calculate the percent difference between the mean calculated mass and the manufacturer's reported mass of Ca2+ per tablet: % difference= |calculated mass-reported mass|/reported mass x 100 For this experiment we did three different titrations where we mixed the known solution the titrant with the unknown analyte to create a chemical reaction. To make the titrant we weigh out 1.489g of disodium EDTA needed to make the 250mL of 0.016M EDTA solution. Next, we mixed 100mL of warm distilled water until the EDTA has dissolved. Then we added 9mL of 0.105 M MgCl_2 solution, and 2.8 mL of 3 M NH_4OH to the EDTA solution. For the analyte, we obtain a calcium supplement tablet (Cultrate Calcium and vitamin D_3 supplement) and placed a total of 25mL of 3M HCL to it until it dissolved. Before we began with the titration we added 10mL of ammonia/ammonium chloride buffer solution to make the solution basic, and added 13 drops of calamite indicator. Once the experiment began we titrated with the disodium EDTA from the burette until the last traces of red were gone and the solution became a pure blue color. The volumes of disodium EDTA solution were recorded in a table (Figure 1)

Explanation / Answer

1 mole EDTA reacts with 1 mol ca^2+

moles of EDTA used in 1st titration = molarity * volume in L = 0.016 M * 0.0161 L = 2.58*10^-4 moles

Mass of Ca^2+ = 2.58*10^-4 moles * 40g/mol = 0.0103g

Moles of EDTA used in 2nd titration : =0.016 M * 0.0119 L = 1.904*10^-4 moles

Mass of Ca^2+ = 1.904*10^-4 moles * 40g/mol = 0.00762 g

moles of EDTA used in 3rd titration : 0.016 M * 0.0242 L = 3.872*10^-4 moles

Mass of Ca^2+ used = 3.872*10^-4 moles * 40g/mol = 0.0155 g

Mean mass = 0.0103g + 0.00762g+0.0155g/3 = 0.0111g

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

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