The student decided to do another experiment with his leftover copper(II) sulfate (CuSO4) solution. He divided the solution up i
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Answer:
Yes, precipitation of barium iodate will occur.
Explanation:
Molarity of barium nitrate solution = 0.050 M
Volume of barium nitrate solution =15.0 mL = 0.0150 L
1 mL = 0.001 L
Moles of barium nitrate = n
Moles of barium ions:
Molarity of potassium iodate solution = 0.10 M
Volume of potassium iodate solution =100.0 mL = 0.1000 L
1 mL = 0.001 L
Moles of potassium iodate = n'
Moles of iodate ions =
After mixing of both solution in 250 mL in erlenmeyer flask
Volume of the final solution = 250 mL = 0.250 L
Concentration of barium ions in 250 mL solution :
Concentration of iodate ions:
Solubility product of barium iodate,
Ionic product of the barium iodate in solution :
( precipitation)
As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.
Empirical formula: MoO3
