Question 3d25a

Jun 24, 2016

Here's how I would do it.

Explanation:

Problem

What is the equivalent weight of ${\text{H"_3"PO}}_{2}$ when it disproportionates to ${\text{PH}}_{3}$ and ${\text{H"_3"PO}}_{3}$?

Solution

The equivalent weight is the molecular weight ($\text{MM}$) 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: ${\text{H"_3"PO"_2 → "PH"_3 + "H"_3"PO}}_{3}$

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: $3 {\text{H"_3"PO"_2 → "PH"_3 + 2"H"_3"PO}}_{3}$

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

Since 3 mol of ${\text{H"_3"PO}}_{2}$ 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"#

$\text{Equivalent weight" = "MM"/n = "66.00 g"/(4/3) = "66.00 g"× 3/4 = "49.50 g}$

The equivalent weight of ${\text{H"_3"PO}}_{2}$ in this reaction is 49.50 g.