Finding molarity of hyposulfite solution used in titration

Question

$\pu{0.1 g}$ of $\ce{KIO3}$ is treated with excess $\ce{KI}$ solution. The liberated iodine required $\pu{44 ml}$ of $x$ molar hypo solution. Calculate the value of $x$.

I have used the equivalent concept and got the answer as $\pu{0.026 M}$. But the correct answer to this question is $\pu{0.063 M}$. Can you tell me where I have gone wrong in my solution?

$$ \begin{array}{cc} \ce{&KIO3 &+ KI -> &I2}\\ &\frac{0.5}{214}~\mathrm{meq} & &\frac{0.5}{214}~\mathrm{meq} \end{array} $$

$$ \begin{array}{cc} \ce{&I2 &+ &2 Na2S2O3 &-> 2 NaI + Na2S4O6}\\ &\frac{0.5}{214}~\mathrm{meq} & &\frac{0.5}{214}~\mathrm{meq} \end{array} $$

$$∴\frac{0.5}{214} = 44\times10^{-3}\times x\times 2 \to n\text{-factor of hypo}$$

$$∴x = \pu{0.026 M}$$

Answer 1

To find the molarity $c$, you need to know the amount $n$ required for a titration:

$$c(\ce{Na2S2O3}) = \frac{n(\ce{Na2S2O3})}{V(\ce{Na2S2O3})}$$

Unknown amount $n(\ce{Na2S2O3})$ can be found from the second balanced reaction:

$$\ce{I2 + 2 Na2S2O3 -> 2 NaI + Na2S4O6}$$

$$n(\ce{Na2S2O3}) = 2\cdot n(\ce{I2})$$

Finding $n(\ce{I2})$ is trivial if you would've written the first redox equation correctly, and that's where the problem in your solution arises. The incomplete first equation resulting in wrong stoichimetry and false ratio between $\ce{KIO3}$ and $\ce{I2}$. The reacton takes place in acidic media (I added [diprotic] sulfuric acid, but basicity of acid doesn't matter), and the ratio between $\ce{KIO3}$ and $\ce{I2}$ won't be $1:1$:¹

$$\ce{KIO3 + 5 KI + 3 H2SO4 -> 3 I2 + 3 K2SO4 + 3 H2O}$$

$$n(\ce{I2}) = 3\cdot n(\ce{KIO3}) = \frac{3\cdot m(\ce{KIO3})}{M(\ce{KIO3})}$$

Now, with all the steps sorted out, the final expression for the molarity of hyposulfite can be rewritten algebraically and solved by plugging in the values:

$$ \begin{align} c(\ce{Na2S2O3}) &= \frac{2\cdot 3\cdot m(\ce{KIO3})}{M(\ce{KIO3})\cdot V({\ce{Na2S2O3})}}\\ &= \frac{2\cdot 3\cdot \pu{0.1 g}}{\pu{214.0 g mol-1}\cdot\pu{44e-3 L}} \\ &\approx \pu{0.064 mol L-1} \end{align} $$

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¹ All you need to do to prove that is to write half-reactions for the redox process between iodate and iodide:

$$ \begin{align} \ce{2\overset{+5}{I}O3- + 12 H+ + 10 e- &-> \overset{0}{I}_2 + 6 H2O} & \tag{red}\\ \ce{2\overset{-1}{I}^- &-> \overset{0}{I}_2 + 2 e-} &|\cdot 5 \tag{ox}\\ \hline \ce{2 IO3- + 10I- + 12 H+ &-> 6 I2 + 6 H2O} \tag{redox} \end{align} $$

or, after dividing the total redox reaction by 2, finally

$$\ce{IO3- + 5I- + 6 H+ -> 3 I2 + 3 H2O}$$

Answer 2

Your method in using the equivalent concept is correct, but the miscorrect in your solution is in determining the n-factor.

Refer to the above equations and the calculation N-factorPoutnik

N-factor Is the change of atom oxidation state multiplied by the number of atoms changing this state. Because the iodine is just an intermediate product to be titrated by $\ce{Na2S2O3}$, the oxidation number change is by $6$.Therefore, n-factor of $\ce{KIO3}$ is $6$.So, the equivalent mass is M/6.And , n-factor of $\ce{Na2S2O3}$ is $1$.So, the equivalent mass of $\ce{Na2S2O3}$ is equal to its molar mass.

$$ \begin{array}{cc} \ce{&KIO3 &+ 5KI + 3H2SO4 -> &3I2 +&3K2SO4 + 3H2O}\\ &\frac{0.6}{214}~\mathrm{meq} & &\frac{0.6}{214}~\mathrm{meq} \end{array} $$

$$ \begin{array}{cc} \ce{&I2 &+ &2 Na2S2O3 &-> 2 NaI + Na2S4O6}\\ &\frac{0.6}{214}~\mathrm{meq} & &\frac{0.6}{214}~\mathrm{meq} \end{array} $$

$$∴\frac{0.6}{214} = 44\times10^{-3}\times x\times 1 \to n\text{-factor of hypo}$$

$$∴x \approx{\pu{0.064 M}}$$