Question #854b9

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.