A chemist obtains 500.0 mL of a solution containing an unknown concentration of

Question

A chemist obtains 500.0 mL of a solution containing an unknown concentration of calcium iodide. Cal_2- He pipets 20 mL of this solution into a 100 mL volumetric flask and dilutes to the mark. He then pulpits 10 mL of this diluted solution into a 50 mL volumetric flask and dilutes to the mark. He analyzes some of the solution from the final volumetric flask and finds that the iodide ion concentration is 0.0000043 M. (in solution, calcium iodide breaks apart into one Ca^2+ ion for every movie^- ions, so a solution that is 1.0 M in Cal_2 is 2.0 M in I^-.) Determine the molar concentration of calcium iodide the original solution.

Explanation / Answer

Let us assume initial concentration of I- = x

After 1 st dilution,

       [I-] = x . 20/100 = x/5

After 2nd dilution,

   [I-] = (x/5). (10/50) = x/25

But given that

after 2nd dilution, [I-] = 0.0000043 M

Hence,

x/25 = 0.0000043 M

   x = 0.0001075 M

Then,

Initial concentration of I- = 0.0001075 M

Therefore,

Initial concentration of CaI2 = 0.0001075 M/ 2

                                         = 0.00005375 M

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