How come we can't use the equivalence point equation of NV=NV in this problem?

Question

So, the problem is you prepare to standardize a $\ce{Na2S2O3}$ solution. $\pu{32 mL}$ of $\ce{Na2S2O3}$ solution is titrated into $\pu{50 mL}$ of a $\pu{0.01 M} \ \ce{KIO3}$ solution to reach the equivalence point. They first titrate the $\ce{KIO3}$ solution until it loses color, then add a starch indicator until the reaction is complete. The reaction proceeds in these two steps:

$$\ce{IO3- + I- + H+ <=> I3- + H2O} \tag{1}$$ $$\ce{I3- + S2O3^2- <=> I- + S4O6^2-} \tag{2}$$

Determine the concentration of $\ce{Na2S2O3}$ solution at the beginning of the experiment.

The answer is $\pu{0.094 M}$, and basically you need to first balance out the two formulas to create a net ionic equation for the two reactions, and then you use the moles of $\ce{KIO3}$ (it's molarity times it's volume) and then the balanced stoichiometric ratio ($\pu{6 mol}$ of $\ce{S2O3^2-}$/$\pu{1 mol}$ of $\ce{IO3-}$) to get the moles of $\ce{S2O3^2-}$, which is the dissociated version of $\ce{Na2S2O3}$ in a $1:1$ ratio (dissolved $\ce{Na2S2O3- -> 2Na+ + S2O3^2-}$). Then you just divide the moles of $\ce{S2O3^2-}$ by it's volume converted to liters to get $\pu{0.094 M}$.

My question is, since we are looking at the equivalence point in a titration, why can't we use $N_\mathrm{acid} \times \pu{Volume_{acid}} = N_\mathrm{base} \times \pu{Volume_{base}}$, since equivalents of acids and bases are the same at the equivalence point? If we can, how would we do it?

Answer 1

My question is, since are looking at the equivalence point in a titration, why can't we use $N_\text{acid} \times \text{Volume}_\text{acid} = N_\text{base} \times \text{Volume}_\text{base}$, since equivalents of acids and bases are the same at the equivalence point? If we can, how would we do it?

First of all, I would suggest that you stick to the original method using moles and proper mole ratios.

Please forget about this normality equation because it is only historically important and rarely anyone uses this concept in 2020 for research or academic work. Note, industrial analytical calculations still use this equation for very routine work. Still if you want to know why this normality equation fails in your case is as follows.

0) The $N_aV_a = N_bV_b$ relation holds if and only if all the concentration units are in normality, where $a$ and $b$ can be acid-base or oxidizing-reducing agent pair. It will also work for precipitation titrations etc. Corollary: This equation fails in titration calculations when the units are in molarity.

(i) This is a redox titration not an acid-base titration.

(ii) Let us change the subscripts for the sake for indexing, $$N_\text{ox}V_\text{ox}=N_\text{red}V_\text{red}$$

You have $\pu{0.01 M}$ $\ce{KIO3}$, what will be its concentration in normality units?

For that you need to see how many electrons are gained in the half-reaction...

$\ce{KIO3}\longrightarrow\ce{I2}$; Balance the equation and see how many electrons are involved in the redox process. See the example here Balancing this reaction

Multiply the number of electrons involved in the half reaction by the molar concentration to get normal concentration of potassium iodate.

Apply the above equation and you will get the correct result but in normality units i.e. thiosulfate concentration in normality units.

Answer 2

This is a note to M. Farooq's answer: This titration consists of two back to back redox reactions. As Farooq pointed out, your first reductive half reaction is actually $\ce{IO3- <=> I2}$ (or $\ce{IO3- <=> I3-}$). In either way, generated $\ce{I2}$ in the presence of extra $\ce{I-}$ would become $\ce{I3-}$. The presence of $\ce{I3-}$ can be detected by addition of starch, which gives a deep blue color. Thus, you must have initially added excess $\ce{KI}$ solution to your $\ce{KIO3}$ solution so that this redox reaction happens spontaneously before you titrate it with thiosulfate solution. That is the second redox reaction ($\ce{I3- <=> I-}$ and $\ce{S2O3^2- <=> S4O6^2-}$).

Your balanced ionic reductive half-reaction of first redox reaction is:

$$\ce{IO3- + 6 H+ + 6 e- <=> I- + 3 H2O} \quad E^\circ = \pu{1.085 V} \tag{1}$$

I put $E^\circ$ value to show the reaction is spontaneous under standard condition. Your balanced ionic oxidative half-reaction is:

$$\ce{3I- <=> I3- + 2 e-} \quad E^\circ = \pu{-0.536 V} \tag{2}$$

To get the first complete redox reaction, you must add equations $(1)$ and $(2)$ in order to cancel out electrons. Doing so you'd see: $E_\mathrm{rxn}^\circ = E_\mathrm{red}^\circ + E_\mathrm{oxi}^\circ = \pu{0.549 V}$, indicating a spontaneous reaction. After the initial redox reaction has completed, you have $\ce{I3-}$ in your solution. That's when your second redox reaction kicked-off as you add thiosulfate solution. The two relevant balanced ionic reductive and oxidative half-reactions are:

$$\ce{I3- + 2 e- <=> 3I-} \quad E^\circ = \pu{0.536 V} \tag{3}$$ $$\ce{2 S2O3^2- <=> S4O6^2- + 2 e-} \quad E^\circ = \pu{-0.08 V} \tag{4}$$

To get the complete second redox reaction, you must add equations $(3)$ and $(4)$ in order to cancel out electrons. This is easy, because you have only $\ce{2e-}$ in either side of each half-reaction. Thus, net redox reaction is:

$$\ce{2 S2O3^2- + I3- -> S4O6^2- + 3I-} \quad E_\mathrm{rxn}^\circ = \pu{0.456 V} \tag{5}$$

Using these two balanced two redox reactions, you may able to use calculations as M. Farooq explained. However, do not forget to include correct stoichiometric mole ratio $\left(\frac{\pu{mol}_\text{oxi}}{\pu{mol}_\text{red}}\right)$ in each calculation.