The volume (in $\mathrm{mL}$ ) of $0.1 \mathrm{M} \mathrm{AgNO}_3$ required for complete precipitation of…
The volume (in $\mathrm{mL}$ ) of $0.1 \mathrm{M} \mathrm{AgNO}_3$ required for complete precipitation of chloride ions present in $30 \mathrm{~mL}$ of $0.01 \mathrm{M}$ solution of $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}_{)_5} \mathrm{Cl}\right] \mathrm{Cl}_2\right.$, as silver chloride is close to
Solution
$\mathrm{mmol}$ of complex $=30 \times 0.01=0.3$ Also, 1 mole of complex $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$ gives only two moles of chloride ion when dissolved in solution
$
\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}^{2+}+2 \mathrm{Cl}^{-}\right.
$
$\Rightarrow \mathrm{mmol}$ of $\mathrm{Cl}^{-}$ion produced from its $0.3 \mathrm{mmol}=0.6$
Hence, $0.6 \mathrm{mmol}$ of $\mathrm{Ag}^{+}$would be required for precipitation.
$
\begin{aligned}
& \Rightarrow 0.60 \mathrm{mmol} \text { of } \mathrm{Ag}^{+}=0.1 \mathrm{M} \times \mathrm{V}(\text { in } \mathrm{mL}) \\
& \Rightarrow \quad V=6 \mathrm{~mL} .
\end{aligned}
$