Question

What would be the Van’t Hoff factor for a molecule such as I2?

Answer 1

It's not trivial, but it SHOULD just be pretty much `1`... In actuality, it is typically around `1.02`, but you can use `1`.

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Iodine undergoes the following hydrolysis in water at `25^@ "C"` in `"0.2 M NaClO"_4` and a `"pH" = 5.15 - 7.62` phosphate buffer:

`3("I"_2(aq) + "H"_2"O"(l) rightleftharpoons "HOI"(aq) + "I"^(-)(aq) + "H"^(+)(aq))`
`ul(3"HOI"(aq) rightleftharpoons "IO"_3^(-)(aq) + 2"I"^(-)(aq) + 3"H"^(+)(aq))`
`3"I"_2(aq) + 3"H"_2"O"(l) rightleftharpoons "IO"_3^(-)(aq) + 5"I"^(-)(aq) + 6"H"^(+)(aq)`

The reaction has `log K = -47.61`, or `K = 10^(-47.61) = 2.45 xx 10^(-48)`. Suppose that `["I"_2] = 1.8 xx 10^(-4) "M"`. Then we solve the equilibrium:

`2.45 xx 10^(-48) = (["IO"_3^(-)]["I"^(-)]^5["H"^(+)]^6)/(["I"_2]^3)`

`= (x(5x)^5(6x)^6)/(1.8 xx 10^(-4) "M" - 3x)^3`

`= (1.458 xx 10^18 cdot x^12)/(1.8 xx 10^(-4) "M" - 3x)^3`

Since `K` is obviously very small, `3x` `"<<"` `1.8 xx 10^(-4)` so that:

`2.45 xx 10^(-48) = (1.458 xx 10^18 cdot x^12)/(1.8 xx 10^(-4) "M")^3`

Therefore,

`x = (((2.45 xx 10^(-48))(1.8 xx 10^(-4) "M")^3)/(1.458 xx 10^18))^(1//12) = 3.83 xx 10^(-7) "M"`

That means (ignoring the autoionization of water):

`["IO"_3^(-)] = 3.83 xx 10^(-7) "M"`
`["I"^(-)] = 1.91 xx 10^(-6) "M"`
`["H"^(+)] = 2.30 xx 10^(-6) "M"`
`["I"_2] = 1.79 xx 10^(-4) "M"`

The percent ionization of `"I"_2` under these assumptions is

`(3cdot3.83 xx 10^(-7) "M")/(1.80 xx 10^(-4) "M") xx 100% = 0.639%`

Starting with `3` iodine particles, the maximum number of particles obtainable at equilibrium is `12`.

If the percent dissociation was `100%`, then the van't Hoff factor would have been `4.000` and the concentration at equilibrium would have been `7.20 xx 10^(-4) "M"` in total.

In actuality, the total concentration of all species in solution for this equilibrium is

`["IO"_3^(-)] + ["I"^(-)] + ["H"^(+)] + ["I"_2] = 1.84 xx 10^(-4) "M"`,

so the van't Hoff factor is about `color(blue)(1.02)`:

`i = (1.84 xx 10^(-4) "M IO"_3^(-) + "I"^(-) + "H"^(+) + "I"_2)/(1.80 xx 10^(-4) "M I"_2) ~~ 1.02`