Animated Solution for Chemistry - Basic Concepts in Chemistry: The volume (in $\text{mL}$) of $0.125\text{ M}$ ${\text{AgNO}}_3$ required to quantitatively precipitate chloride ions in $0.3\text{ g}$ of $[\text{Co}({\text{NH}}_3{)}_6]{\text{Cl}}_3$ is ......... . $M_{[\text{Co}({\text{NH}}_3{)}_6]{\text{Cl}}_3}=267.46\text{ g/mol}$ $M_{{\text{AgNO}}_3}=169.87\text{ g/mol}$

Imagine you are standing in a laboratory, holding a beaker filled with a solution of a fascinating cobalt complex, $[\text{Co}({\text{NH}}_3{)}_6]{\text{Cl}}_3$. Your mission? To completely precipitate all the chloride ions hidden within this solution using silver nitrate (${\text{AgNO}}_3$).

This problem is a beautiful blend of two fundamental concepts: understanding the structure of coordination compounds and applying the principles of volumetric analysis. Let's break it down step by step.

Before we can calculate anything, we need to understand the nature of our reactant. The formula $[\text{Co}({\text{NH}}_3{)}_6]{\text{Cl}}_3$ tells a story. The cobalt ion is surrounded by six ammonia (${\text{NH}}_3$) ligands, tightly bound within the square brackets. This is the coordination sphere.

The three chloride (${\text{Cl}}^{-}$) ions, however, reside outside this sphere. They satisfy the primary valency of the central metal atom and are ionizable. When this complex dissolves in water, it dissociates as follows:

This is the crucial catch! One mole of the complex releases exactly three moles of chloride ions.

Now, let's quantify what we have. We are given $0.3\text{ g}$ of the complex. To find the number of moles, we divide the given mass by its molar mass ($267.46\text{ g/mol}$).

Since each mole of the complex gives three moles of chloride ions, the total moles of ${\text{Cl}}^{-}$ available for precipitation will be:

We intentionally keep this as a fraction to avoid rounding errors early in our calculation.

Enter silver nitrate. The precipitation reaction is a simple one-to-one dance between silver ions and chloride ions:

For complete precipitation, the moles of ${\text{AgNO}}_3$ added must perfectly match the moles of ${\text{Cl}}^{-}$ present in the solution.

We know that the number of moles of a solute in a solution is the product of its molarity ($M$) and its volume in liters ($V$).

We are given the molarity of the ${\text{AgNO}}_3$ solution as $0.125\text{ M}$. Substituting this into our equation:

Rearranging the equation to solve for the volume:

Since the question specifically asks for the volume in milliliters, we multiply our result by $1000$:

Notice how the molar mass of ${\text{AgNO}}_3$ ($169.87\text{ g/mol}$) was provided in the question but never used? This is a classic trap set by examiners to test your conceptual clarity. Because we were given the molarity directly, we didn't need the molar mass to find the volume. Always trust your fundamental equations!