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, CaI2. He pipets 15 mL of this solution into a 100 mL volumetric flask and dilutes to the mark. He then pipets 10 mL of this diluted solution into a 25 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 0.00000736 M. (Note: in solution, calcium iodide breaks apart into one Ca2+ ion for every two I- ions, so a solution that is 1.0 M in CaI2 is 2.0 M in I-.) Determine the molar concentration of calcium iodide in the original solution.

Explanation / Answer

Since, there are two dilutions. If the initial [I-] is x then after the first dilution,

[I-] = x * 15/100 = 0.15 x

after the second dilution

[I-] = (0.15x) *10/25= 0.06x

the final [I-] = 0.00000745 M so

0.06 x= 0.00000736 and x = 0.000122 M

and the initial [CaI2] = x/2 = 0.000122/2 M = 0.0000613 M

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