Three part question: 1) A student obtained 80 mL of .01 M Ca(NO3)2, pipeted 25.0

Question

Three part question:

1) A student obtained 80 mL of .01 M Ca(NO3)2, pipeted 25.0 mL of this solution into a 100 mL volumetric flask, added DI water to the "line" at 100 mL to dilute the solution. What is the concentration of this solution?

2) The student then had to pipet 25 mL of the diluted solution into a new 100 mL volumetric flask and add DI water to the "line" at 100 mL. What is the concentration of this solution?

3) Determine the concentration of Ca2+ ions in each solution. (This is the initial concentration of Ca2+ in each solution.

4) What type of reaction takes place between Ca(NO3)2 and KIO3?

Explanation / Answer

1] 25 ml of 0.01 M Ca(NO3)2 contains 0.25 millimoles of Ca(NO3)2

it is diluted by adding 75 ml of DI water

New concentration = Moles / Total volume in L = 0.0025 M

2] 25 ml of 0.0025M Ca(NO3)2 solution contains 0.0625 millimoles

new concentration = total moles / total volume in L = 6.25*10^-4 M

3] Ca(NO3)2 - ------> Ca+2 + 2NO3-

SO Ca+2 concentrations are 0.0025M and 6.25*10^-4M

4] Precipitation reaction

where calcium iodate forms

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