Finding concentration of ascorbic acid

**Question:**


A person applied following procedure for the determination of ascorbic acid (Vitamin C) in a sample solution.


A volume of 20.00 cm3 of the ascorbic acid sample was reacted with 25.00 cm3 of 0.04 M KIO3
solution and excess KI in acidic medium. The remaining I2 is titrated against a 0.1 M Na2S2O3 solution until the purple color of I2 disappears.


If the burette reading (volume of Na2S2O3) is 40.00 cm3 then what is the concentration of ascorbic acid in the sample solution? In the presence of Iodine (I2), ascorbic acid gets oxidized to dehydroascorbic acid.


Use the following equations to support your answer.


[![enter image description here](https://i.stack.imgur.com/2i97P.png)](https://i.stack.imgur.com/2i97P.png)


(a) 0.10 M


(b) 0.02 M


(c) 0.025 M


(d) 0.04 M


(e) 0.05 M


**My attempt:**


Moles of Na2S2O3 = 0.1 x (40/1000) = 4 x 10^-3


Moles of I2 remaining = 2 x 10^-3


Moles of KIO3 = 0.04 x (25/1000) = 1 x 10^-3


Moles of I2 formed = 3 x 10^-3


Therfore moles of I2 reacted with ascorbic acid = (3 x 10^-3)-(2 x 10^-3) = 1 x 10^-3 mol = moles of ascorbic acid


Concentration of ascorbic acid = (1 x 10^-3)/(20/1000) = 0.05 mol/dm3


The answer according to the marks scheme is **D) 0.04**. Where did I go wrong?




I got the same answer. You get the answer in the key if you mistakenly think that the ascorbate was in 25 mL, not in 20 mL. Sorry about the inconsistent significant figures in the calculation below.


$V\_{\mathrm{asc}}= 20.00\ \mathrm{mL}$


$n\_{\mathrm{\ce{KIO3}}}= 25.00\ \mathrm{mL} \cdot 0.04\ \mathrm{M}$


$\ \ \ =1.0\times 10^{-3}\ \mathrm{mol}$


$n\_{\mathrm{\ce{I2}}}= 3 \cdot n\_{\mathrm{\ce{KIO3}}}$


$\ \ \ =3.0\times 10^{-3}\ \mathrm{mol}$


$n\_{\mathrm{\ce{Na2S2O3}}}= 40.00\ \mathrm{mL} \cdot 0.1000\ \mathrm{M}$


$\ \ \ =4.000\times 10^{-3}\ \mathrm{mol}$


$n\_{\mathrm{\ce{I2},remain}}= \dfrac{n\_{\mathrm{\ce{Na2S2O3}}}}{2}$


$\ \ \ =2.000\times 10^{-3}\ \mathrm{mol}$


$n\_{\mathrm{\ce{I2},asc}}= n\_{\mathrm{\ce{I2}}} - n\_{\mathrm{\ce{I2},remain}}$


$\ \ \ =1.0\times 10^{-3}\ \mathrm{mol}$


$n\_{\mathrm{asc}}= n\_{\mathrm{\ce{I2},asc}}$


$\ \ \ =1.0\times 10^{-3}\ \mathrm{mol}$


$[\ce{asc}]= \dfrac{n\_{\mathrm{asc}}}{V\_{\mathrm{asc}}}$


$\ \ \ =0.05\ \mathrm{M}$


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