The heavy metal complex salt can theoretically exist as several distinct hydrated coordination isome — Solutions and Colligative Properties Chemistry Question
Question
The heavy metal complex salt $PtCl_4 \cdot 6H_2O$ can theoretically exist as several distinct hydrated coordination isomers. A $1.0\text{ molal}$ aqueous solution of one specific isomer exhibits a measured depression in freezing point of exactly $3.72^\circ\text{C}$. Assuming $100\%$ ionization of the complex and $K_f(\text{H}_2\text{O}) = 1.86^\circ\text{C m}^{-1}$, exactly how many moles of $AgCl$ precipitate will immediately form if $1.0\text{ mole}$ of this exact isomer is treated with excess aqueous $AgNO_3$?
💡 Solution & Explanation
Calculate the van't Hoff factor: $i = \frac{\Delta T_f}{K_f \times m} = \frac{3.72}{1.86 \times 1.0} = 2$. An $i$ value of 2 at $100\%$ ionization indicates the complex dissociates into exactly 2 ions. The only octahedral $Pt(IV)$ isomer that fits this criterion is $[Pt(H_2O)_5Cl]Cl \cdot H_2O$ (Wait, $Pt(IV)$ coordination is 6: $[Pt(H_2O)_3Cl_3]Cl \cdot 3H_2O$ provides 1 complex cation $[Pt(H_2O)_3Cl_3]^+$ and 1 $Cl^-$ anion, totalling 2 ions). Because there is exactly 1 ionizable chloride per molecule outside the coordination sphere, adding excess $AgNO_3$ to 1 mole of the complex yields exactly 1 mole of $AgCl$ precipitate.