JEE MAIN - Chemistry (2023 - 15th April Morning Shift - No. 16)
The volume (in $\mathrm{mL}$) of $0.1 \mathrm{M} ~\mathrm{AgNO}_{3}$ required for complete precipitation of chloride ions present in $20 \mathrm{~mL}$ of $0.01 \mathrm{M}$ solution of $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$ as silver chloride is __________.
Answer
4
Explanation
The given solution contains a complex ion $\left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) \mathrm{Cl}_2$ which contains one chloride ion per complex. Therefore, when this complex is treated with silver nitrate, each mole of the complex will consume 2 moles of silver nitrate to form two moles of silver chloride.
The first step is to write the balanced chemical equation for the reaction between the complex and silver nitrate, as follows:
$$\left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) \mathrm{Cl}_2 + 2 \mathrm{AgNO}_3 \longrightarrow \left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) + 2 \mathrm{AgCl} + 2 \mathrm{NO}_3^-$$
From the equation, we can see that 1 mole of the complex consumes 2 moles of silver nitrate to form 2 moles of silver chloride. Therefore, the number of millimoles of chloride ions present in the given solution can be calculated as:
$$\text{Millimoles of } \mathrm{Cl}^- \text{ ions} = \text{concentration} \times \text{volume} = 0.01\ \mathrm{M} \times 2 \times 10\ \mathrm{mL} = 0.2\ \mathrm{mmol}$$
To calculate the volume of 0.1 M silver nitrate required, we can use the formula:
$$\text{Millimoles of silver nitrate required} = \text{Millimoles of chloride ions} \times 2 = 0.4\ \mathrm{mmol}$$
We can then use the formula:
$$\text{Volume of silver nitrate solution} = \frac{\text{Millimoles of silver nitrate required}}{\text{Molarity of silver nitrate solution}}$$
Substituting the values, we get:
$$\text{Volume of silver nitrate solution} = \frac{0.4\ \mathrm{mmol}}{0.1\ \mathrm{M}} = 4\ \mathrm{mL}$$
Therefore, the volume of 0.1 M silver nitrate required for complete precipitation of chloride ions present in 20 mL of 0.01 M $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_{2}$ solution as silver chloride is $\boxed{4\ \mathrm{mL}}$.
The first step is to write the balanced chemical equation for the reaction between the complex and silver nitrate, as follows:
$$\left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) \mathrm{Cl}_2 + 2 \mathrm{AgNO}_3 \longrightarrow \left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) + 2 \mathrm{AgCl} + 2 \mathrm{NO}_3^-$$
From the equation, we can see that 1 mole of the complex consumes 2 moles of silver nitrate to form 2 moles of silver chloride. Therefore, the number of millimoles of chloride ions present in the given solution can be calculated as:
$$\text{Millimoles of } \mathrm{Cl}^- \text{ ions} = \text{concentration} \times \text{volume} = 0.01\ \mathrm{M} \times 2 \times 10\ \mathrm{mL} = 0.2\ \mathrm{mmol}$$
To calculate the volume of 0.1 M silver nitrate required, we can use the formula:
$$\text{Millimoles of silver nitrate required} = \text{Millimoles of chloride ions} \times 2 = 0.4\ \mathrm{mmol}$$
We can then use the formula:
$$\text{Volume of silver nitrate solution} = \frac{\text{Millimoles of silver nitrate required}}{\text{Molarity of silver nitrate solution}}$$
Substituting the values, we get:
$$\text{Volume of silver nitrate solution} = \frac{0.4\ \mathrm{mmol}}{0.1\ \mathrm{M}} = 4\ \mathrm{mL}$$
Therefore, the volume of 0.1 M silver nitrate required for complete precipitation of chloride ions present in 20 mL of 0.01 M $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_{2}$ solution as silver chloride is $\boxed{4\ \mathrm{mL}}$.
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