Question

What will be the equivalent weight of `KClO_3` in the following equation ?

`3KClO_3 + 3H_2SO_4 → 3KHSO_4 + HClO_4 + 2ClO_2 + H_2O`

Answer 1

The equivalent weight is183.82 g.

Explanation:

The usual formula for calculating the equivalent weight (`"EW"`) of a substance in a redox equation is

`"EW" = "MW"/n`

where

`"MW ="` the molecular weight of the substance and

`n color(white)(ml)=` the change in oxidation number

For a disproportionation reaction, as you have above, we must modify the formula to

`"EW" = "MW"/N = "MW"/n_1 + "MW"/n_2`

where

`n_1` and `n_2` are the `n`-values for the two half-reactions and
`N` is the equivalent `n`-value for the overall reaction

We can factor out `"MW"` to get

`1/N = 1/n_1 + 1/n_2`

We can solve this equation to get

`color(blue)(bar(ul(|color(white)(a/a)N = (n_1n_2)/(n_1 + n_2)color(white)(a/a)|)))" "`

For the oxidation half-reaction

`"K"stackrelcolor(blue)("+5")("Cl")"O"_3 → "H"stackrelcolor(blue)("+7")("Cl")"O"_4`

`n = 2`

For the reduction half-reaction

`"K"stackrelcolor(blue)("+5")("Cl")"O"_3 → stackrelcolor(blue)("+4")("Cl")"O"_2`

`n = 1`

For the overall reaction

`N = (2 × 1)/(2 + 1) = 2/3`

∴ `"EW" = "MW"/(2/3) = "122.55 g"× 3/2= "183.82 g"`

Answer 2

Here's another way to solve the problem.

Explanation:

The equivalent weight is the molecular weight (`"MM"`) divided by the number of electrons transferred per mole of reactant (`n`).

`color(blue)(bar(ul(|color(white)(a/a)"Equivalent weight" = "MM"/ncolor(white)(a/a)|)))" "`

We need the moles of electrons, so let's balance the equation by the ion-electron method.

Unbalanced equation: `"KClO"_3 → "KClO"_4 + "ClO"_2`

Reduction: `1 × ["KClO"_3 + "H"_2"O" → "KClO"_4 + "2H"^"+" + 2"e"^"-"]`
Oxidation: `2 × ul(["KClO"_3 + "2H"^"+" + "e"^"- "→ "K"^"+" + "ClO"_2 + "H"_2"O"]color(white)(mmm))`
Overall: `color(white)(mmm)3"KClO"_3 + "2H"^"+" → "KClO"_4 + 2"K"^"+" + "2ClO"_2 + "H"_2"O"`

We see that 2 mol of electrons are transferred in the reaction.

Since 3 mol of `"KClO"_3` are involved, the number of electrons transferred per mole of `"KClO"_3` is

`n = ("2 mol e"^"-")/("3 mol KClO"_3) = 2/3color(white)(l) "mol electrons per mole of KClO"_3"`

∴ `"Equivalent weight" = "MM"/n = "122.55 g"/(2/3) = "183.82 g"`

The equivalent weight of `"KClO"_3` in this reaction is 183.82 g.