The total number of ions produced from the complex $[Cr(NH_3)_6]Cl_3$ in aqueous solution will be____________.
Dissociation of $[Cr(NH_3)_6]Cl_3$ Complex
When the coordination complex $[Cr(NH_3)_6]Cl_3$ is dissolved in an aqueous solution, it dissociates into ions based on its ionic structure.
The complex consists of a complex cation, $[Cr(NH_3)_6]^{3+}$, and three chloride anions, $Cl^-$.
The dissociation reaction in water can be represented as:
$ [Cr(NH_3)_6]Cl_3 \xrightarrow{H_2O} [Cr(NH_3)_6]^{3+} + 3Cl^- $
Calculating Total Ions Produced
By examining the dissociation equation, we can count the total number of ions formed:
- One complex cation: $[Cr(NH_3)_6]^{3+}$
- Three chloride anions: $3 \times Cl^-$
The total number of ions is the sum of the cation and the anions:
Total ions = (Number of complex cations) + (Number of chloride anions)
Total ions = 1 + 3 = 4
Therefore, a total of 4 ions are produced from the complex $[Cr(NH_3)_6]Cl_3$ in an aqueous solution.
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