Question #9ad66

For the sake of simplicity, I will use the following notation

• `"CCl"_3"CH"("OH")_2 -> "reactant"`
• `"CCl"_3"CHO" -> "product"`

The balanced chemical equation that describes this equilibrium will be--keep in mind that you have a `1:1` mole ratio between chloral hydrate and chloral!

`"reactant"_ ((sol)) rightleftharpoons "product"_ ((sol)) + "H"_ 2"O"_ ((sol))`

I used `("sol")` to denote the fact that the compounds are dissolved in a solution that does not have water as the solvent.

Now, you know that for every `1` mole of chloral hydrate that dissociates, you get `1` mole of chloral and `1` mole of water.

The initial concentration of chloral hydrate is

`["reactant"]_ 0 = "0.010 moles"/"1 L" = "0.010 M"`

The problem tells you that the equilibrium concentration of water is equal to `"0.0020 M"`. This means that in order for the reaction to produce `"0.0020 M"` of water, it must also produce `"0.0020 M"` of chloral and consume `"0.0020 M"` of chloral hydrate.

So the equilibrium concentration of chloral will be

`["product"] = ["H"_ 2"O"] = "0.0020 M"`

and the equilibrium concentration of chloral hydrate will be

`["reactant"] = ["reactant"]_ 0 - "0.0020 M"`

`["reactant"] = "0.00080 M"`

By definition, the equilibrium constant will be equal to

`K_c = (["product"] * ["H"_2"O"])/(["reactant"])`

Keep in mind that water is a solute here, so its concentration must be included in the expression of the equilibrium constant.

Plug in your values to find--I'll leave the answer without added units.

`color(darkgreen)(ul(color(black)(K_c))) = (0.0020 * 0.0020)/(0.0080) = color(darkgreen)(ul(color(black)(5.0 * 10^(-4))))`

The answer is rounded to two sig figs.