Please answer portion in yellow? If 5 drops of 0.15 M Kl is added to 40 drops of

Question

Please answer portion in yellow?

If 5 drops of 0.15 M Kl is added to 40 drops of Na2S203Na2s203, what is the final concentration of Ki? Do not enter units with your answer, report your answer to 2 significant figures. Which substance(s) is/are used in the experiment to keep ion not appear in the equations? Select all that apply s and volumes constant but do starch Na2 S203 KI KNO3 (NHA)2S2Os (NHA)2SO4 View comments () > Expert Answer o ecure Was this answer helpful? 19 answers Rate answer he Solution : O-1S 2.

Explanation / Answer

Na2S2O3 + 2KI = K2S2O3 +  2NaI

Using M1V1=M2V2

M1=0.15  V1=5 V2=40 We get M2=0.01875

Acid and potassium iodide are added to a solution of potassium iodate getting the following reaction:

KIO3 + 5KI + 3H2SO4 = 3I2 + 3K2SO4 + 3H2O

represented by the following ionic equation:

IO3- + 5I- + 6H+ = 3I2 + 3H2O

Thiosulpathe is titrated against this solution (effectively against iodine):

I2 + 2Na2S2O3 = Na2S4O6 +2NaI

represented by the following ionic equation:

I2 + 2S2O32- = S4O62- + 2I-where the dark brown coloured solution of iodine turns pale yellow and finally colorless as the reaction proceeds (starch is used as indicator after the pale yellow transition forming a black solution due to an iodine-starch complex which turns colourless upon further addition of thiosulphate).

Starch is used in the above reaction but it doesn't come in the reaction.

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