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When 6 M NaOH is slowly added to a solution containing chromium(III) ion, a precipitate forms. However when an excess of 6 M NaOH is added, the precipitate dissolves, forming a complex ion with a coordination number of four. a. Write the formula of the precipitate.

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Answer:

Cr(OH)3

Explanation:

The addition of 6M sodium hydroxide to a solution that contains chromium(III) ion leads to the net ionic reaction;

3 OH^-(aq) + Cr^3+(aq) -------> Cr(OH)3(s)

Cr(OH)3 is insoluble except an excess of the hydroxide is added then the precipitate dissolves as follows;

Cr^3+(aq) + 4OH^-(aq)------> [Cr(OH)4]^-

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