Question

If excess of ${\mathrm{AgNO}}_3$ solution is added to 100 mL of a 0.024 M solution of dichlorobis (enthylenediamine) cobalt (III) chloride, how many moles of $\mathrm{AgCl}$ be precipitated?

Answer 1

Step 1: Forming the reaction

We know that charge of Enthylenediamine is 0 as it's neutral.

Also, charge of Cobalt is +3 and that of Chlorine is -1

Thus the complex will be like $\left[\mathrm{Co}{\left(\mathrm{en}\right)}_2{\mathrm{Cl}}_2\right]\mathrm{Cl}$

$\therefore$The charge of the cation is:

$$\begin{gathered} +3+2\times \left(0\right)+2\times (-1) \\ =+3-2 \\ =+1 \end{gathered}$$

Hence the compound would dissociate as:

$\left[\mathrm{Co}{\left(\mathrm{en}\right)}_2{\mathrm{Cl}}_2\right]\mathrm{Cl}\to {\left[\mathrm{Co}{\left(\mathrm{en}\right)}_2{\mathrm{Cl}}_2\right]}^{+}+{\mathrm{Cl}}^{-}$

$\therefore$The complete reaction of $\left[\mathrm{Co}{\left(\mathrm{en}\right)}_2{\mathrm{Cl}}_2\right]\mathrm{Cl}$ with Silver nitrate(${\mathrm{AgNO}}_3$) is:

$\underset{\mathrm{Dichlorobis}\left(\mathrm{enthylenediamine}\right)\mathrm{cobalt}\left(\mathrm{III}\right)\mathrm{chloride}}{\left[\mathrm{Co}{\left(\mathrm{en}\right)}_2{\mathrm{Cl}}_2\right]\mathrm{Cl}\left(\mathrm{aq}\right)}+\underset{\mathrm{Silver}\mathrm{nitrate}}{{\mathrm{AgNO}}_3\left(\mathrm{aq}\right)}\to \underset{\mathrm{Dichlorobis}\left(\mathrm{enthylenediamine}\right)\mathrm{cobalt}\left(\mathrm{III}\right)\mathrm{nitrate}}{\left[\mathrm{Co}{\left(\mathrm{en}\right)}_2{\mathrm{Cl}}_2\right]{\mathrm{NO}}_3\left(\mathrm{aq}\right)}+\underset{\mathrm{Silver}\mathrm{chloride}}{\mathrm{AgCl}\left(\mathrm{s}\right)(\downarrow )}$

Step 2: Finding the moles of $\left[\mathrm{Co}{\left(\mathrm{en}\right)}_2{\mathrm{Cl}}_2\right]\mathrm{Cl}$

From the above reaction we can clearly see that, 1 mole $\left[\mathrm{Co}{\left(\mathrm{en}\right)}_2{\mathrm{Cl}}_2\right]\mathrm{Cl}$ produces 1 mole of Silver chloride $\mathrm{AgCl}$ precipitate.

Here molarity$\left(\mathrm{M}\right)$ is given as $0.024\mathrm{M}$ and the volume$\left(\mathrm{V}\right)$ of solution is given as $100\mathrm{mL}$

$\therefore$the number of moles of $\left[\mathrm{Co}{\left(\mathrm{en}\right)}_2{\mathrm{Cl}}_2\right]\mathrm{Cl}$ present in this solution is:

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}=\mathrm{Molarity}\times \left(\frac{\mathrm{V}}{1000}\right) \\ \Rightarrow \mathrm{Number}\mathrm{of}\mathrm{moles}=0.024\times \frac{100}{1000} \\ \Rightarrow \mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{0.024}{10} \\ \Rightarrow \mathrm{Number}\mathrm{of}\mathrm{moles}=0.0024 \end{gathered}$$

Step 3: Finding the moles of Silver chloride

Now if, 1 mole $\left[\mathrm{Co}{\left(\mathrm{en}\right)}_2{\mathrm{Cl}}_2\right]\mathrm{Cl}$ produces 1 mole of Silver chloride $\mathrm{AgCl}$ precipitate

$\therefore$$0.0024\mathrm{moles}$ of $\left[\mathrm{Co}{\left(\mathrm{en}\right)}_2{\mathrm{Cl}}_2\right]\mathrm{Cl}$ will produce

$$\begin{gathered} \frac{1}{1}\times 0.0024 \\ =0.0024\mathrm{moles}\mathrm{of}\mathrm{AgCl} \end{gathered}$$

Therefore, $0.0024\mathrm{moles}$ of Silver chloride($\mathrm{AgCl}$) will be precipitated.