Animated Solution for Chemistry - Ionic Equilibrium: Solubility product of silver bromide is $5.0\times 10^{-13}$. The quantity of potassium bromide (molar mass taken as $120{\text{ g mol}}^{-1}$) to be added to $1\text{ L}$ of $0.05\text{ M}$ solution of silver nitrate to start the precipitation of $\text{AgBr}$ is

The problem of finding the exact moment a precipitate forms is like finding the tipping point of a perfectly balanced scale. In this journey, we will explore the delicate equilibrium between dissolved ions and solid precipitates, using the principles of ionic equilibrium.

A Delicate Balance

Imagine you are in a laboratory, standing in front of a beaker. Inside this beaker is exactly $1\text{ L}$ of a $0.05\text{ M}$ silver nitrate (${\text{AgNO}}_3$) solution. Because silver nitrate is a strong electrolyte, it completely dissociates. This means our beaker is swimming with silver ions (${\text{Ag}}^{+}$) and nitrate ions (${\text{NO}}_3^{-}$), with the concentration of ${\text{Ag}}^{+}$ being exactly $0.05\text{ M}$.

Now, we hold a dropper filled with potassium bromide ($\text{KBr}$). We are going to add it, drop by drop, into the beaker. The question asks: exactly how much $\text{KBr}$ do we need to add to just start the precipitation of silver bromide ($\text{AgBr}$)?

Ionic Product vs Solubility Product

To answer this, we need to understand the concept of the Solubility Product Constant ($K_{sp}$). For a sparingly soluble salt like $\text{AgBr}$, the $K_{sp}$ represents the maximum capacity of the solution to hold its constituent ions before they combine and fall out as a solid precipitate.

The actual product of the ion concentrations at any given moment is called the Ionic Product ($Q_{sp}$). - If $Q_{sp}<K_{sp}$, the solution is unsaturated. No precipitate forms. - If $Q_{sp}>K_{sp}$, the solution is supersaturated, and precipitation occurs. - If $Q_{sp}=K_{sp}$, the solution is exactly saturated. This is the tipping point—the exact moment precipitation is about to begin.

For our reaction, the equilibrium is:

And the tipping point condition is:

Finding the Missing Piece

We are given the $K_{sp}$ of $\text{AgBr}$ as $5.0\times 10^{-13}$. We also know the concentration of silver ions, $[{\text{Ag}}^{+}]=0.05\text{ M}$. Let's substitute these values into our master equation:

Now, we solve for the unknown bromide ion concentration:

This tells us that the moment the concentration of ${\text{Br}}^{-}$ hits $10^{-11}\text{ M}$, the solution can no longer hold any more ions, and $\text{AgBr}$ will start to precipitate.

The Final Stretch

We have the required molarity of bromide ions, but the question asks for the mass of potassium bromide ($\text{KBr}$) to be added.

First, let's find the number of moles. Since molarity is moles per liter, and our beaker contains exactly $1\text{ L}$ of solution, the calculation is straightforward:

Finally, we convert these moles into mass using the given molar mass of $\text{KBr}$ ($120{\text{ g mol}}^{-1}$):

Conclusion: We need to add an incredibly tiny amount—just $1.2\times 10^{-9}\text{ g}$—of potassium bromide to trigger the precipitation of silver bromide. This elegant calculation perfectly matches option (b).