# Question ade43

Jun 16, 2015

${K}_{s p} = 0.5$

#### Explanation:

So, you know that 20 g of your sparingly soluble salt, $M {X}_{2}$, which has a formula weight of 40. g/mol, is dissolved in one liter of water.

This means that you can determine its molar solubility, i.e. the number of moles of salt that can actually be dissolved in one liter of water. Start by figuring out how many moles of salt you have

20cancel("g") * ("1 mole "MX_2)/(40. cancel("g")) = "0.5 moles"

This means that its molar solubility will be

$C = \frac{n}{V} = \text{0.5 moles"/"1 L" = "0.5 M}$

In order to be able to calculate its solubility product constant, you need to determine how many moles of ${M}^{2 +}$ cations and of ${X}^{-}$ anions the salt produces in solution.

$\text{ } M {X}_{2 \left(s\right)} r i g h t \le f t h a r p \infty n s {M}_{\left(a q\right)}^{2 +} + \textcolor{red}{2} {X}_{\left(a q\right)}^{-}$
I.......$-$..................0...............0
C.....$-$.................(+x)...........(+$\textcolor{red}{2}$x)
E.....$-$.................x..................2x

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

K_(sp) = [M^(2+)] * [X^(-)]^(color(red)(2)#

${K}_{s p} = x \cdot {\left(2 x\right)}^{2} = x \cdot 4 {x}^{2} = 4 {x}^{3}$, where

$x$ - the molar solubility of the salt.

Since you know that you can dissolve 0.5 moles in one liter of water, $x$ will be

$x = 0.5$

This means that ${K}_{s p}$ will be

${K}_{s p} = 4 \cdot {0.5}^{3} = \textcolor{g r e e n}{0.5}$

The answer is rounded to one sig fig, the number of sig figs you gave for the mass of the salt.