Question

Question #854b9

Answer 1

`K_(sp) = 3.97 * 10^(-3)`

Explanation:

Notice that the problem provides you with the molar solubility of the salt, i.e. the number of moles that can be dissolved per liter of solution in order to have a saturated solution of `"AB"`.

In your case, you know that `0.0630` moles of `"AB"` can be dissolved in `"1.00 L"` of water to form `"1.00 L"` of saturated solution at `25^@"C"`.

You can thus say that the molar solubility of the salt, `s`, is equal to

`s = "0.0630 mol L"^(-1)`

Now, you know that when `"AB"` is dissolved in water, the following dissociation equilibrium is established

`"AB"_ ((s)) rightleftharpoons "A"_ ((Aq))^(+) + "B"_ ((aq))^(-)`

Notice that every mole of `"AB"` that dissociates produces `1` mole of `"A"` and `1` mole of `"B"`, so right from the start, you can say that the saturated solution will contain

`["A"] = ["B"]`

Since you already know the molar solubility of the salt at `25^@"C"`, i.e. the maximum number of moles of `"AB"` that dissociate to produce solvated ions, you can say that

`["A"] = ["B"] = s`

which means that you have

`["A"] = ["B"] = "0.0630 mol L"^(-1)`

By definition, the solubility product constant, `K_(sp)`, is equal to

`K_(sp) = ["A"] * ["B"]`

This means that you have--I won't add the units here

`K_(sp) = 0.0630 * 0.0630 = color(darkgreen)(ul(color(black)(3.97 * 10^(-3))))`

The answer is rounded to three sig figs.