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Iodometric Titration Tutorial


-Introduction

-Calculating moles of KIO3 required.

-Calculating concentration of KIO3 solution.

-Calculating mass of KIO3 to prepare solution.

-Conclusion

-Practice Question: Virtual Lab



Experiment 9 Iodometric Titration Online Tutorial >> Preparation of a KIO3 Standard Solution >> Calculating the concentration of the standard KIO3 solution.

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Exp 9: Iodometric Titration Online Tutorial - Preparation of a KIO3 Standard Solution

Step 2 - Calculating the concentration of the standard KIO3 solution.

In this tutor, we will calculate the concentration of KIO3 such that the number of moles calculated in Step 1 is contained in about 25 mL of KIO3 solution. That will ensure that about 25mL of KIO3 is used in each titration and so the 4 significant figures accuracy can be maintained.

In the last tutor we determined that 1.8929x10-5 moles of KIO3 are required to reach the equivalence point in the titration of a 10 mg sample of ascorbic acid. What is the concentration of KIO3 solution that would deliver 1.8929x10-5 moles in 25 mL?

(Please give your answer to 5 significant figures; for very small numbers, you may use scientific notation i.e. 6.3e-8)
Hint
M KIO3
Good job. A standard solution of 7.5716e-4 M KIO3 will reach the equivalence point after 25 mL when titrated with approximately a 10 mg sample.
That's not quite right.
Hint:
To convert from moles of KIO3 to molarity of KIO3, use the definition of molarity (M) and the volume of potasium iodate, KIO3 (25 mL).
 
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Hint:
Molarity = moles/volume in L. Using this conversion, what is the molarity of KIO3?
 
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Hint:
Using this conversion, the result is:
1.8929e-5 mol / 0.02500 L KIO3 = 7.5716e-4 M KIO3
 
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Last Updated: Sunday, November 13th, 2022 @ 07:12:28 pm