# How much ferrous sulfide will dissolve in one liter of water if its ksp is 3.7x10^-19?

May 5, 2018

$6.1 \times {10}^{- 10}$ moles.

#### Explanation:

Here's our balanced equation:

$F e S \left(s\right) r i g h t \le f t h a r p \infty n s F {e}^{2 +} \left(a q\right) + {S}^{2 -} \left(a q\right)$

From this, we can see that $1$ mole of $F e S$ dissolves to form $1$ mole of $F {e}^{2 +}$ and ${S}^{2 -}$. The mole ratio is $1 : 1 : 1$.

From the balanced equation, we can also know that the expression for ${K}_{s p}$ is:

${K}_{s p} = \left[F {e}^{2 +}\right] \times \left[{S}^{2 -}\right]$

(For more information on how this expression was derived, check out this awesome webpage on ChemTeam!)

The question tells us that the value of ${K}_{s p}$ is $3.7 \times {10}^{-} 19$. Now, we can relate the value and the expression:

$\left[F {e}^{2 +}\right] \times \left[{S}^{2 -}\right] = 3.7 \times {10}^{-} 19$

The mole ratio of $1 : 1 : 1$ tells us that $\left[F {e}^{2 +}\right]$ is equal to $\left[{S}^{2 -}\right]$, which is equal to $\left[F e S\right]$.
This is because concentration, here, is measured in moles per liter. They're in the same solution—in other words, the number of liters is the same. Therefore, we can use the mole ratio to relate their concentrations.

So, the expression could be rewritten as:

$\left[F e S\right] \times \left[F e S\right] = 3.7 \times {10}^{-} 19$
${\left[F e S\right]}^{2} = 3.7 \times {10}^{-} 19$

To solve for $\left[F e S\right]$, we'll just need to take the square root of $3.7 \times {10}^{-} 19$:

$\left[F e S\right] = \sqrt{3.7 \times {10}^{-} 19}$
$\left[F e S\right] = 6.1 \times {10}^{- 10} M$

This is the same as $6.1 \times {10}^{- 10}$ $\text{mol/L}$, so the number of moles of $F e S$ which would dissolve in $\text{1 L}$ of water is $6.1 \times {10}^{- 10}$ moles.