Given the very low value of Ksp for Cr(OH)3, a precipitate of Cr(OH)3 would be e

Question

Given the very low value of Ksp for Cr(OH)3, a precipitate of Cr(OH)3 would be expected if only 6 M NaOH were added to the mixture of Group Ill cations in the first step of the procedure. Explain how and why the chromium remains in solution 1. A solution may contain one or more of the Group IIl cations. When this solution is combined with NaOH (ag), NaOCI (aq) and NH3 (aq) only a colorless solution is obtained with no precipitate evident. Indicate whether each of the following cations is present, absent or undetermined. 2. Cr3 Fe3- Ni2 Explain.

Explanation / Answer

Q1

Cr3+ will precipitate as stated

Cr3+ + 3OH- <-> Cr(OH)3(s)

note that ecess OH- will form complex:

[Cr(H2O)6]+3 forms

[Cr(H2O)6]+3 --> [Cr(H2O)3(OH)3] --> [Cr(OH)6]-3

note that all these complex are "aqueous" i.e. will remain in solution

no precipitate is "seen" or observed

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.