Determine the total number of ions generated from the dissociation of \( [Cr(NH_3)_6]Cl_3 \) in an aqueous environment.
The complex in question is \( [Cr(NH_3)_6]Cl_3 \). This compound comprises a central chromium ion, \( Cr^{3+} \), bonded to six ammonia molecules (\( NH_3 \)) and three chloride ions (\( Cl^- \)) as counterions.
Step 1: Dissolution Process
Upon introduction to water, \( [Cr(NH_3)_6]Cl_3 \) dissociates. The chemical equation for this dissociation is: \[ [Cr(NH_3)_6]Cl_3 \rightarrow [Cr(NH_3)_6]^{3+} + 3Cl^- \] In this equation:
- \( [Cr(NH_3)_6]^{3+} \) represents the complex ion.
- The 3 \( Cl^- \) entities are the counterions.
Step 2: Ion Count
The dissociation yields the following ions:
- One \( [Cr(NH_3)_6]^{3+} \) ion.
- Three \( Cl^- \) ions. Therefore, the cumulative count of ions produced is calculated as: \[ 1 \, \text{(complex ion)} + 3 \, \text{(chloride ions)} = 6 \, \text{ions} \] Summary:
The complete dissociation of \( [Cr(NH_3)_6]Cl_3 \) in water results in the formation of 6 ions.