Question

at what ph will fe(oh)₃(s) begin to precipitate from 8.15 × 10⁻³ m fecl₃? ksp for fe(oh)₃ is 2.79 × 10⁻³⁹.

Answer 1

Fe(OH)₃ will begin to precipitate from the solution when the pH reaches approximately 4.72.

To determine the pH at which Fe(OH)₃(s) will begin to precipitate from a 8.15 × 10⁻³ M FeCl₃ solution, you can use the solubility product constant (Ksp) expression for Fe(OH)₃:

Fe(OH)₃(s) ⇌ Fe³⁺(aq) + 3OH⁻(aq)

The Ksp expression for this equilibrium is:

Ksp = [Fe³⁺][OH⁻]³

Given that the Ksp for Fe(OH)₃ is 2.79 × 10⁻³⁹, and initially, there are no OH⁻ ions present, the reaction will proceed until the concentration of OH⁻ ions in the solution reaches a certain level where it begins to exceed the solubility product (Ksp).

Let x represent the concentration of OH⁻ ions produced when Fe(OH)₃ begins to precipitate. At this point, the concentration of Fe³⁺ ions will also be x M because of the 1:1 stoichiometry of Fe(OH)₃ dissociating into Fe³⁺ and OH⁻ ions.

So, the Ksp expression becomes:

Ksp = (x)(x)³

Ksp = x⁴

Now, we can calculate the value of x:

2.79 × 10⁻³⁹ = x⁴

Taking the fourth root of both sides:

x ≈ (2.79 × 10⁻³⁹)^(1/4)

x ≈ 5.24 × 10⁻¹⁰ M

This concentration of OH⁻ ions will cause Fe(OH)₃ to begin to precipitate from the solution.

To find the pH at this point, we need to calculate the pOH first:

pOH = -log10(OH⁻ concentration)

pOH = -log10(5.24 × 10⁻¹⁰)

pOH ≈ 9.28

Now, calculate the pH:

pH = 14 - pOH

pH ≈ 14 - 9.28

pH ≈ 4.72

So, Fe(OH)₃ will begin to precipitate from the solution when the pH reaches approximately 4.72.

Answer 2

Fe(OH)₃ will begin to precipitate from a 8.15 × 10⁻³ M FeCl₃ solution at a pH of 13.29.

The precipitation of Fe(OH)₃(s) occurs when the concentration of hydroxide ions in the solution reaches the solubility product constant (Ksp) for Fe(OH)₃. The balanced equation for the dissociation of Fe(OH)₃ is:

`\[ Fe(OH)_3 \rightleftharpoons Fe^(3+) + 3OH^- \]`

The Ksp expression is given by:

`\[ Ksp = [Fe^(3+)][OH^-]^3 \]`

Given that the concentration of FeCl₃ is 8.15 × 10⁻³ M and the ionization of FeCl₃ in water is:

`\[ FeCl₃ \rightarrow Fe^(3+) + 3Cl^- \]`

This implies that the concentration of
`Fe^(3+)` is also 8.15 × 10⁻³ M.

Now, using the stoichiometry from the dissociation equation, the concentration of
`\(OH^-\)` is three times the concentration of
`Fe^(3+)`.

`\[ [OH^-] = 3 * [Fe^(3+)] \]`

Substitute this into the Ksp expression:

`\[ Ksp = (8.15 * 10^(-3))(3[Fe^(3+)])^3 \]`

`\[ [Fe^(3+)] = \sqrt[3]{(Ksp)/(27)} \]`

`\[ [Fe^(3+)] = \left((2.79 * 10^(-39))/(27)\right)^(1/3) \]`

`\[ [Fe^(3+)] \approx 5.13 * 10^(-14) \, \text{M} \]`

Now, calculate the concentration of \(OH^-\) using the relationship mentioned earlier:

`\[ [OH^-] = 3 * [Fe^(3+)] \]`

`\[ [OH^-] = 3 * 5.13 * 10^(-14) \, \text{M} \]`

Finally, calculate the pH using the concentration of hydroxide ions:

`\[ \text{pH} = -\log(3 * 5.13 * 10^(-14)) \]`

`\[ \text{pH} \approx 13.29 \]`

So, at a pH of approximately 13.29, Fe(OH)₃ will begin to precipitate from a 8.15 × 10⁻³ M FeCl₃ solution.