Question

Question #3d25a

Answer 1

Here's how I would do it.

Explanation:

Problem

What is the equivalent weight of `"H"_3"PO"_2` when it disproportionates to `"PH"_3` and `"H"_3"PO"_3`?

Solution

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: `"H"_3"PO"_2 → "PH"_3 + "H"_3"PO"_3`

Reduction half-reaction: `1 × ["H"_3"PO"_2 + "4H"^+ + "4e"^"-" → "PH"_3 +"2H"_2"O"]`
Oxidation half-reaction: `2 × ["H"_3"PO"_2 + "H"_2"O" → "H"_3"PO"_3 + "2H"^+ + "2e"^"-"]`
Balanced equation: `3"H"_3"PO"_2 → "PH"_3 + 2"H"_3"PO"_3`

We see that 4 mol of electrons are transferred in the reaction.

Since 3 mol of `"H"_3"PO"_2` are involved, the number of electrons transferred per mole is

`n = ("4 mol e"^"-")/("3 mol H"_3"PO"_2) = 4/3color(white)(l) "mol electrons per mole of H"_3"PO"_2"`

∴ `"Equivalent weight" = "MM"/n = "66.00 g"/(4/3) = "66.00 g"× 3/4 = "49.50 g"`

The equivalent weight of `"H"_3"PO"_2` in this reaction is 49.50 g.