Question

1-Solid calcium iodide is slowly added to 175 mL of a 0.394 M sodium carbonate solution until the concentration of calcium ion is 0.0284 M. The percent of carbonate ion remaining in solution is  %.

2- Solid potassium sulfite is slowly added to 175 mL of a 0.211 M barium chloride solution until the concentration of sulfite ion is 0.0324 M. The percent of barium ion remaining in solution is  %.

Answer 1

Solution

1)

Total number of moles of carbonate ions = total number of moles of sodium carbonate

= volume (L) x molarity = 0.175 L x 0.394 M = 0.06895 moles

on addition of calcium iodide, it reacts with sodium carbonate and forms calcium carbonate.

Thus,

number of moles of carbonate reacted = total number of moles of calcium ion reacted

= Volume (L) x molarity = 0.175 L x 0.0284 M = 0.00497 moles

now, the unreacted carbonate in solution = 0.06895 moles - 0.00497 moles = 0.06398 moles

The percent of carbonate ion remaining in solution = (0.06398 moles/0.06895 moles) x 100

= 92.79 % (Ans)

2)

Total number of moles of barium ions = total number of moles of barium chloride

= volume (L) x molarity = 0.175 L x 0.211 M = 0.036925 moles

on addition of potassium sulfite, it reacts with barium chloride and forms barium sulfite

Thus,

number of moles of barium ions reacted = total number of moles of sulfite ions reacted

= Volume (L) x molarity = 0.175 L x 0.0324 M = 0.00567 moles

now, the unreacted barium ion in solution = 0.036925 moles - 0.00567 moles = 0.031255 moles

The percent of carbonate ion remaining in solution = (0.031255 moles/0.036925 moles) x 100

= 84.64 % (Ans)