The following volumes of solution are combined in a test tube: 2.00 mL of 0.20 M potassium iodide 1.00 mL of 1% starch 0.50 mL of 0.20 M ammonium persulfate 0.50 mL of 0.012 M sodium thiosulfate 2.00 mL of 0.20 M potassium nitrate 2.00 mL of 0.20 M ammonium sulfate What is the concentration of iodide in the solution in the test tube?

Question

Chemistry

Giovanni Hansen

15 February, 01:48

The following volumes of solution are combined in a test tube: 2.00 mL of 0.20 M potassium iodide 1.00 mL of 1% starch 0.50 mL of 0.20 M ammonium persulfate 0.50 mL of 0.012 M sodium thiosulfate 2.00 mL of 0.20 M potassium nitrate 2.00 mL of 0.20 M ammonium sulfate What is the concentration of iodide in the solution in the test tube?

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Answers (1)

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Alicia Hoffman · Answer 1

The answer to your question is: Concentration = 0.05 M

Explanation:

Data

2ml of KI 0.2M

1 ml 1% starch

0.5 ml 0.2 M ammonium persulphate

0.5 ml 0.012 M sodium thiosulphate

2 ml 0.2 M potassium nitrate

2 ml 0.2 M ammonium nitrate

Formula to calculate moles of KI

Molarity = moles / volume

moles = Molarity x volume

moles = (0.2) (0.002)

moles = 0.0004

Total volume = 2 + 1 + 0.5 + 0.5 + 2 + 2 = 8 ml

Molarity = 0.0004 / 0.008

Molarity = 0.05