when 6 M NaOH is slowly added to a solution containing chromium (III) ion, a precipate forms. However when an excess of 6 M NaOH is added, the precipated dissolves forming a complex ion with a coordination number of four.
1) write the formula of the precipitate.
2) write the formula of the complex ion.
i don't know how to do it
To answer these questions, you need to understand the properties of the chromium (III) ion (Cr3+) and the reactions that occur when it reacts with sodium hydroxide (NaOH).
1) Writing the Formula of the Precipitate:
When 6 M NaOH is slowly added to a solution containing chromium (III) ion, a precipitate forms. This indicates that chromium (III) ion has a hydroxide ion (OH-) precipitate.
To determine the formula of the precipitate, consider the charges of the ions involved. The chromium (III) ion has a charge of +3, while the hydroxide ion has a charge of -1.
Since the charges must balance in a compound, we need three hydroxide ions to balance the charge of one chromium (III) ion. Therefore, the formula of the precipitate is Cr(OH)3.
2) Writing the Formula of the Complex Ion:
When an excess of 6 M NaOH is added, the precipitate dissolves forming a complex ion. This suggests that the hydroxide ions (OH-) from the excess NaOH combine with the chromium (III) ion to form a complex.
To determine the formula of the complex ion, consider the coordination number of four mentioned in the question. This indicates that the chromium (III) ion forms a complex with four hydroxide ligands.
The complex ion formed is called a tetraaquachromium (III) ion. The formula for this complex is [Cr(OH2)4]3+.
To summarize:
1) The formula of the precipitate formed when adding 6 M NaOH to the solution containing chromium (III) ion is Cr(OH)3.
2) The formula of the complex ion formed when an excess of 6 M NaOH is added is [Cr(OH2)4]3+.