In: Chemistry

# Above what Fe2+ concentration will Fe(OH)2 precipitate from a buffer solution that has a pH of 9.99? The Ksp of Fe(OH)2 is 4.87

Above what Fe2+ concentration will Fe(OH)2 precipitate from a buffer solution that has a pH of 9.99? The Ksp of Fe(OH)2 is 4.87

## Solutions

##### Expert Solution

Concept and reason

The concept used is to calculate the concentration of $$\mathrm{Fe}^{2+}$$.

Fundamentals

The solubility product, $$K_{s p}$$ is the equilibrium constant for a solid that is dissolved in a solution. It is the product of the concentration of the products raised to the power of their coefficients. $$p H$$ is the measure of hydrogen ion concentration. The lower the $$p H$$, the higher is the hydrogen ion concentration and the lower is the hydroxide ion concentration. The higher the $$p H$$, the lower is the hydrogen ion concentration and the higher is the hydroxide ion concentration. The hydrogen ion concentration is calculated as follows:

$$\left[\mathrm{H}^{+}\right]=10^{-p H}$$

The hydroxide ion concentration $$p O H$$ and $$p H$$ are related as follows:

$$\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-14}$$

1

$$\mathrm{pH}=9.99$$

The hydrogen ion concentration is as follows:

\begin{aligned} \left[\mathrm{H}^{+}\right] &=10^{-\mathrm{pH}} \\ &=10^{-999} \\ &=1.023 \times 10^{-10} \mathrm{M} \end{aligned}

The hydroxide ion concentration is as follows:

$$\left(1.023 \times 10^{-10}\right)\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-14}$$

\begin{aligned} \left[\mathrm{OH}^{-}\right] &=\frac{1.0 \times 10^{-14}}{1.02 \times 10^{-10}} \\ &=9.78 \times 10^{-5} \mathrm{M} \end{aligned}

From the given value of $$\mathrm{pH}$$, the hydrogen ion concentration and then the hydroxide ion concentration are calculated.

The given salt is $$\mathrm{Fe}(\mathrm{OH})_{2}$$. Its dissociation in water is as follows:

$$\mathrm{Fe}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{Fe}^{2+}+2 \mathrm{OH}^{-}$$

The expression for the solubility product is as follows:

$$K_{s p}=\left[\mathrm{Fe}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}$$

Substitute the values of solubility product and the hydroxide ion concentration in the formula for solubility product

and calculate the concentration of $$\mathrm{Fe}^{2+}$$ as follows:

$$\begin{array}{l} 4.87 \times 10^{-17}=\left[\mathrm{Fe}^{2+}\right]\left(9.78 \times 10^{-5}\right)^{2} \\ {\left[\mathrm{Fe}^{2+}\right]=\frac{4.87 \times 10^{-17}}{\left(9.78 \times 10^{-5}\right)^{2}}} \\ \quad=5.1 \times 10^{-9} \mathrm{M} \end{array}$$

Part 1

Therefore, the $$\left[\mathrm{Fe}^{2+}\right]$$ is $$5.1 \times 10^{-9} \mathrm{M}$$

From the hydroxide ion concentration and the $$K_{s p}$$ of the salt, the concentration of $$\left[\mathrm{Fe}^{2+}\right]$$ is calculated.