i am lost. i am not sure how to write the chemical equation nor get the answer.

Question

i am lost. i am not sure how to write the chemical equation nor get the answer. please help. i really appropriate your time. thanks

7. You may be required to standardize a solution of Na S,03 in this experiment. The approximate molarity of this solution will be 0.025 M. You will take 10.0 mL of a 0.0100 M KIO, solution, dissolve excess amount of KI in it, and, using the Na2S,03 solution, titration to the disappearance of the startch-l2 color. Using the approximate molarity of the Na2S,0, solution, calculate the approximate volume of the solution will be required in the titration.

Explanation / Answer

answer:

According to question, first we add excess amount of KI in KIO3, So, the reaction are-

IO3 –(aq) + 5 I –(aq) + 6 H+(aq) -------> 3 I2(aq) + 3 H2O(l)

3 I2(aq) + 3 I –(aq) --------> 3 I3–(aq)

then we titrate iwth Na2S2O3 solution, so balance reaction is-

3 I3–(aq) + 6 S2O32–(aq) -------> 3 S4O62–(aq) + 9 I –(aq)

Combining these, the net ionic equation for the effective reaction between IO3 – (aq) and S2O3 2– (aq) is:

IO3–(aq) + 6 S2O32–(aq) + 6 H+(aq) I –(aq) + 3 H2O(l) + 3 S4O62–(aq)

For titration

wek now that at equivalence point (point at which titration is completed), the moles of Na2S2O3 is equal to moles of KIO3

therefore M1V1 = M2V2

where M1 and M2 = molarity of Na2S2O3 and KIO3

V1 & V2 = volume of Na2S2O3 and KIO3

So, 0.025 * V1 = 0.01 * 10ml

=> V1 = 4ml

so 4ml of Na2S2 O3 solution is required for titration.

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