A drop of starch is added to a mixture of 97.30 ml of 0.1096 M I2 and 97.20 ml of 0.1098 M Na2S2O3. I've already calculated the moles of I2 to be 10.66408 moles and the moles of Na2S2O3 to be 10.67256. Will the solution be blue even if their moles aren't the same?
To determine whether the solution will be blue or not, we need to consider the reaction that occurs between I2 (iodine) and Na2S2O3 (sodium thiosulfate). The reaction between them is as follows:
2Na2S2O3 + I2 → Na2S4O6 + 2NaI
This reaction forms a complex called sodium tetrathionate (Na2S4O6) and sodium iodide (NaI). The complex, Na2S4O6, gives a blue color to the solution.
Now, let's analyze the moles of I2 and Na2S2O3 that you've calculated. You have determined that the moles of I2 are 10.66408, and the moles of Na2S2O3 are 10.67256.
Although the moles differ slightly, it's important to note that the balanced equation for the reaction shows a 1:1 stoichiometry between I2 and Na2S2O3. This means that for every 1 mole of I2, we need 1 mole of Na2S2O3 to complete the reaction.
In this case, since both the moles of I2 and Na2S2O3 are similar (with a difference of only 0.00848 moles), it is safe to assume that the Na2S2O3 has been added in excess. This means that all the I2 will react, and there will be excess Na2S2O3 remaining after the reaction is complete.
Therefore, we can conclude that the solution will have a blue color due to the formation of Na2S4O6 (sodium tetrathionate) even if the moles of I2 and Na2S2O3 are not exactly the same.