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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.

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(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?

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    $\begingroup$ Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger permanent value. You may find useful formatting mathematical/chemical expressions/formulas. $\endgroup$
    – Poutnik
    Commented Feb 4, 2023 at 17:27

1 Answer 1

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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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  • $\begingroup$ Answer calculated with online calculator. $\endgroup$
    – Karsten ♦
    Commented Feb 4, 2023 at 18:05
  • $\begingroup$ Now, why is this answer down voted? What is wrong? $\endgroup$
    – ACR
    Commented Feb 5, 2023 at 3:24

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