If 0.248 mg of MgF2 is the maximum quantity that can be dissolved in 0.5 L of 0.1 Μ KF, what will be the Ksp of MgF2 under the same conditions?

Question

If 0.248 mg of MgF2 is the maximum quantity that can be dissolved in 0.5 L of 0.1 Μ KF, what will be the Ksp of MgF2 under the same conditions?

1 Answer

Impact of this question

1134 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-08T19:41:23

We are provided with a solubility, which means we need an ICE table to solve this. Let's write out the net ionic equation, but with the solid as the reactant.

$M g F_{2} \to M g^{2 +} + 2 F^{-}$ $MgF_2 -> Mg^(2+) + 2 F^-$

Initially, the concentration of Mg is 0M and the concentration of F is already 0.1M

Then at equilibrium, we subtract variable $s$ $s$ from $M g F_{2}$ $MgF_2$ and add $s$ $s$ to both other values.

This gives us

$M g F_{2} = - S$ $MgF_2 = -S$
$M g^{2 +} = + S$ $Mg^(2+) = +S$
$2 F^{- =} 0.1 + S$ $2 F^- = 0.1 + S$

Since we know that S is in mol / L, we must take the given mass and turn it into moles, then divide by L.

So...

$\frac{0.000248 g}{62.3018 (\frac{g}{m o l})} = 0.00000398 m o l M g F_{2}$ $(0.000248g) / (62.3018(g/(mol))) = 0.00000398 mol MgF_2$

$\frac{0.00000398 m o l}{0.5 L} = 0.00000796 M$ $(0.00000398 mol) / (0.5 L) = 0.00000796 M$

And...

$K_{s p} = \frac{P r o d u c t s}{1} = [S \cdot (0.1 + S)] = 0.1 S + S^{2}$ $K_(sp) = [Products] / 1 = [S * (0.1 + S)] = 0.1S + S^2$

S is already 0.00000796 M, so we just plug this in

$0.1 \cdot 0.00000796 M + 0.00000796 M^{2}$ $0.1 * 0.00000796 M + 0.00000796 M^2$

This means that the Ksp is 0.000000796

Or, in scientific notation: $7.96 \cdot 10^{7}$ $7.96*10^7$

Science

Math

Humanities

... and beyond

If 0.248 mg of MgF2 is the maximum quantity that can be dissolved in 0.5 L of 0.1 Μ KF, what will be the Ksp of MgF2 under the same conditions?

1 Answer

Related questions

Impact of this question