Given the very low value of K_sp for Cr(OH)_3, a precipitate of Cr(OH)_3 would b

Question

Given the very low value of K_sp 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 III cations in the first step of the procedure. Explain how and why the chromium remains in solution. A solution may contain one or more of the Group III cations. When this solution is combined with NaOH (aq), NaOCI (aq) and NHa (aq) only a colorless solution is obtained with no precipitate evident. Indicate whether each of the following cations is present, absent or undetermined. Explain.

Explanation / Answer

Q1.

Cr+3 in solution may form plenty of complexes which are soluble in aquous solution

examples:

[Cr(H2O)6]3+

[Cr(H2O(3(OH)3]

[Cr(OH)6]3-

all those are formed with OH- ions

mainly r(OH)6-3, as you can see, the more Oh- ions the more we faovur Cr(OH)6-3 formation

Q2.

NaOH --> OH-

NaOCl --> OCl-

NH3 --> OH- + gas

Cr+3 ---> Cr(OH)2 white, therefore, not present

Al3+ ---> Al(OH)3 white, not present

Fe+3 --> undetermined

Ni+2 + OH- --> Ni(OH)2, not present

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