A 50 ml sample of ascorbic solution was titrated to until the first sign of perm

Question

A 50 ml sample of ascorbic solution was titrated to until the first sign of permanent blue/purple colour with 7.706 mM potassium iodite.The titration volumes are shown below. Calculate the number of moles of Ascobic acid in the 50.00 ml Sample. Titration Volumes of KIO3 Initial Buret reading/ml trial 1=12.54 trial 2=8.78 Final buret reading/ trial 1=38.74 trial 2=34.93 Volume of KIO used/ml trial 1=? trial 2=?

Explanation / Answer

This problem is a bit tricky because there are multiple reactions going on at once. However, the important reaction is this: C6H8O6 + I2(aq) ---> C6H6O6 + 2 H+(aq) + 2 I-(aq) In the equation above you are basically looking for a color change of blue to colorless because while there is excess iodine, I2(aq), there will exist an iodine-starch complex. This complex is very dark blue, but at the equivalence point you should have no I2(aq) remaining to complex to starch and the solution should become colorless. Now, here is where it gets tricky. Iodine doesn't dissolve well in water; therefore you can't make a standard solution of aqueous iodine to titrate with. Instead, you make iodine in-situ by the use of KIO3 and KI: (IO3)-(aq) + 5 I-(aq) + 6 H+ ? 3 I2(s) + 3 H2O(l) Thus, you have to have an excess amount of KI in solution for this to work and we'll prove that is the case here. So, in all truth, the overall equation is the summation of these two reactions: 3 C6H8O6 + 3 I2(aq) ---> 3 C6H6O6 + 6 H+(aq) + 6 I-(aq) (IO3)-(aq) + 5 I-(aq) + 6 H+ ---> 3 I2(s) + 3 H2O(l) --------------------------------------… 3 C6H8O6 + (IO3)-(aq) ---> 3 C6H6O6 + I-(aq) + 3 H2O(l) Where the first equation was multiplied by 3 to equal the number of iodine molecules on both sides because all I2(aq) we make has to be completely oxidized by vitamin C in the reaction. Alright, now that we have finally determined the overall equation we can calculate the number of moles of vitamin C present in the 10.00 mL sample: (0.0137 M KIO3) x (0.01896 L KIO3) x [(3 moles C6H8O6)/(1 mole KIO3)] = 7.79x10^-4 moles C6H8O6 The concentration is given by: C = (7.79x10^-4 moles)/(0.01000 L) = 7.79x10^-2 M or 77.9 mM C6H8O6 Finally, we can calculate the number of grams used of KI for this reaction: (0.0137 M KIO3) x (0.01896 L KIO3) x [(5 moles KI)/(1 mole KIO3)] x [(166.0028 g KI)/(1 mole KI)] = 0.216 g KI This is less than half the amount of KI we used; therefore, everything is in order.

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