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A solution of 0.220 g of KIO3 is prepared in 50 ml volumetric flask. 2.00 mL of this solution is pipetted into a beaker with 8.00 ml distilled water, to give 10 ml of diluted KIO3. To this, 10.00 mL of starch sulfite was added. What would be the initial concentration of IO3- in this final solution

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maacastrobr

Answer:

0.00044 g/mL

Explanation:

First, 0.220 g of KIO₃ are dissolved in 50 mL, so the concentration at this point is:

  • 0.220 g / 50 mL = 0.0044 g/mL

Then 2.00 mL of this solution are added to 8.00 mL of water (final volume = 10.00 mL). We calculate the concentration of the resulting solution:

  • 0.0044 g/mL * 2.00mL / 10.00mL = 0.00088 g/mL

Finally, to this 10 mL solution, another 10.00 mL were added. Thus the KIO₃ (or IO₃⁻) concentration is:

  • 0.00088 g/mL * 10.00mL / 20.00mL = 0.00044 g/mL

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