Question

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.

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