# How to calculate the equivalent mass of Na2S2O3? [closed]

Let us consider the following reaction

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

Now, in order to calculate the equivalent mass of $$\ce{Na2S2O3}$$, first I need to calculate it's $$n$$-factor which turns out to be $$0.5$$ because the oxidation state of $$\ce{S}$$ in $$\ce{Na2S2O3}$$ is $$+2$$ whereas in $$\ce{Na2S4O6}$$ it is $$+2.5$$.And the $$n$$-factor of $$\ce{I2}$$ is $$1$$. So, the equivalent mass of $$\ce{Na2S2O3}$$ is

$$\frac{\text{molecular mass}}{0.5}$$ And Iam confused with the following half-reactions: $$\ce{I2 +2e->2I-}$$ $$\ce{2S2O3^{2-}->S4O6^{2-} +2e}$$ I appreciate any help in order to solve a problem using milliequivalents?

• The first and third reactions are not balanced in atoms. – user55119 May 14 at 0:53
• @user55119Thanks for correction – Adnan AL-Amleh May 14 at 1:09

Therefore, n-factor of $$\ce{I2}$$ is $$2$$,
n-factor of $$\ce{Na2S2O3}$$ is $$1$$.
So, equivalent mass of $$\ce{Na2S2O3}$$ is equal to its molar mass, for molecular iodine it is half of its molar mass.
• excuse me: Is the equivalent mass of $\ce{KIO3}$ equal 5.$$\ce{KIO3 + 5 KI + 3 H2SO4 -> 3 I2 + 3 K2SO4 + 3 H2O}$$ – Adnan AL-Amleh May 14 at 17:50