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Chapter 16: Problem 64

Find the approximate \(\mathrm{pH}\) range suitable for the separation of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Zn}^{2+}\) ions by precipitation of \(\mathrm{Fe}(\mathrm{OH})_{3}\) from a solution that is initially \(0.010 \mathrm{M}\) in both \(\mathrm{Fe}^{3+}\) and \(\mathrm{Zn}^{2+}\). Assume a 99 percent precipitation of \(\mathrm{Fe}(\mathrm{OH})_{3}\)

Short Answer

Expert verified
The suitable pH range for the separation of Fe3+ and Zn2+ by precipitating Fe(OH)3 is approximately from 11.86 to 12.92.

Step by step solution

01

Calculate [Fe3+] after 99% Precipitation

After 99 percent precipitation of Fe(OH)3, 1 percent, or 0.01, of the initial Fe3+ remains in solution. Therefore, the concentration (denoted as \[Fe^{3+}\] after precipitation) can be calculated as: \[Fe^{3+}\] = 0.01 × 0.010 M = 1.0 × 10-4 M
02

Calculate ion product (Qsp) for Fe(OH)3

The balanced chemical equation for the dissolution of Fe(OH)3 is \[Fe(OH)3_{(s)} \leftrightarrow Fe^{3+} + 3OH^-\]. Using the above reaction, we can evaluate the Ion Product, Qsp. Here, \[Fe^{3+}\] is given from the previous step, and the [OH⁻] can be summarized as [OH⁻] = 3[Fe3+] due to the stoichiometry of the reaction.\n\nQsp = \[Fe^{3+}\] × [OH⁻]^3 = (1.0 × 10-4) × (3 × 1.0 × 10-4)^3 = 2.70 × 10^{-15}\n
03

Calculate the minimum pH required for precipitation of Fe(OH)3

We can calculate [OH⁻] to find the minimum pH required for precipitation to start from the Ion Product calculated above. Using [OH⁻] = ³√Qsp / \[Fe^{3+}\] we find [OH⁻] = ³√(2.70 × 10^{-15} / 1 × 10^{-4} = 0.0138). We can then convert [OH⁻] value to pOH using the formula pOH = - log [OH⁻] and then use pOH to calculate pH, given that pH + pOH = 14 (at 25°C). Thus, we find that pH = 14 - pOH = 14 - (- log 0.0138) = 11.86. This is the minimum pH value required for the precipitation to start.\n
04

Calculate pH range for precipitation of Zn2+ starting

The balanced dissolution equation for Zn(OH)2 is Zn(OH)2_{(s)} \leftrightarrow Zn^{2+} + 2OH⁻\n Here, it is known that Ksp of Zn(OH)2 is 3.0x10^-16 at 25°C. Thus, precipitation will start when Qsp ≥ Ksp.\nFrom the equation, [OH-] = 2[Zn2+], thus we can substitute Zn2+ with 0.01 M which is known from the initial solution and solve for [OH-], then pOH and pH as in step 3. Performing this computation gives a pH of 12.92, which is where Zn2+ will start to precipitate. Therefore, Zn2+ and Fe3+ will both precipitate at this pH.\n

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Most popular questions from this chapter

In a group 1 analysis, a student obtained a precipitate containing both \(\mathrm{AgCl}\) and \(\mathrm{PbCl}_{2}\). Suggest one reagent that would enable her to separate \(\operatorname{AgCl}(s)\) from \(\mathrm{PbCl}_{2}(s)\)

How does the common ion effect influence solubility equilibria? Use Le Châtelier's principle to explain the decrease in solubility of \(\mathrm{CaCO}_{3}\) in a \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution.

(a) Assuming complete dissociation and no ionpair formation, calculate the freezing point of a \(0.50 \mathrm{~m}\) NaI solution. (b) What is the freezing point after the addition of sufficient \(\mathrm{HgI}_{2},\) an insoluble compound, to the solution to react with all the free \(\mathrm{I}^{-}\) ions in solution? Assume volume to remain constant.

Explain, with balanced ionic equations, why (a) \(\mathrm{CuI}_{2}\) dissolves in ammonia solution, (b) AgBr dissolves in \(\mathrm{NaCN}\) solution, (c) \(\mathrm{HgCl}_{2}\) dissolves in KCl solution.

How many milliliters of \(1.0 \mathrm{M} \mathrm{NaOH}\) must be added to a \(200 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{NaH}_{2} \mathrm{PO}_{4}\) to make a buffer solution with a pH of \(7.50 ?\)
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