# Question 854b9

Jul 5, 2017

${K}_{s p} = 3.97 \cdot {10}^{- 3}$

#### 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 $\text{AB}$.

In your case, you know that $0.0630$ moles of $\text{AB}$ can be dissolved in $\text{1.00 L}$ of water to form $\text{1.00 L}$ of saturated solution at ${25}^{\circ} \text{C}$.

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

$s = {\text{0.0630 mol L}}^{- 1}$

Now, you know that when $\text{AB}$ is dissolved in water, the following dissociation equilibrium is established

${\text{AB"_ ((s)) rightleftharpoons "A"_ ((Aq))^(+) + "B}}_{\left(a q\right)}^{-}$

Notice that every mole of $\text{AB}$ that dissociates produces $1$ mole of $\text{A}$ and $1$ mole of $\text{B}$, so right from the start, you can say that the saturated solution will contain

$\left[\text{A"] = ["B}\right]$

Since you already know the molar solubility of the salt at ${25}^{\circ} \text{C}$, i.e. the maximum number of moles of $\text{AB}$ that dissociate to produce solvated ions, you can say that

$\left[\text{A"] = ["B}\right] = s$

which means that you have

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

By definition, the solubility product constant, ${K}_{s p}$, is equal to

${K}_{s p} = \left[\text{A"] * ["B}\right]$

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

${K}_{s p} = 0.0630 \cdot 0.0630 = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{3.97 \cdot {10}^{- 3}}}}$

The answer is rounded to three sig figs.