Problem 275 - A Complete Resource Book in Chemistry for JEE Main

Question

Solubility product of silver bromide is \(5.0 \times 10^{-13} .\) The quantity of potassium bromide (molar mass taken as \(120 \mathrm{~g} \mathrm{~mol}^{-1}\) ) to be added to 1 litre of \(0.05 \mathrm{M}\) solution nitrate to start the precipitation of \(\mathrm{AgBr}\) is [2010] (a) \(1.2 \times 10^{-10} \mathrm{~g}\) (b) \(1.2 \times 10^{-9} \mathrm{~g}\) (c) \(6.2 \times 10^{-5} \mathrm{~g}\) (d) \(5.0 \times 10^{-8} \mathrm{~g}\)

Step-by-Step Solution

Answer

The correct answer is (b) \(1.2 \times 10^{-9} \text{ g}\).

Step 1: Determine the Precipitation Condition

To determine when precipitation begins, we need to compare the ion product (Qsp) to the solubility product (Ksp). For precipitation to start, the ion product should be equal to the Ksp, i.e., \( Q_{sp} = K_{sp} \) for \( \text{AgBr} \).

Step 2: Write the Dissociation Equations

Write the dissociation equation for silver bromide (AgBr): \[ \text{AgBr}_{(s)} \rightleftharpoons \text{Ag}^+ + \text{Br}^- \]Write the reaction for potassium bromide (KBr) dissociation: \[ \text{KBr} \rightarrow \text{K}^+ + \text{Br}^- \]

Step 3: Calculate the Required Bromide Ion Concentration

From the solubility product equation of AgBr, we have:\[ K_{sp} = [\text{Ag}^+] \times [\text{Br}^-] = 5.0 \times 10^{-13} \]Since the silver nitrate solution's concentration provides [Ag\(^+\)] = 0.05 M, set up the equation:\[ 0.05 \times [\text{Br}^-] = 5.0 \times 10^{-13} \]Solve for \([\text{Br}^-]\):\[ [\text{Br}^-] = \frac{5.0 \times 10^{-13}}{0.05} = 1.0 \times 10^{-11} \text{ M} \]

Step 4: Convert Bromide to Potassium Bromide Mass

Calculate the moles of KBr required to achieve this bromide concentration in 1 liter:\[ \text{Moles of } \text{Br}^- = [\text{Br}^-] \times \text{Volume in Liters} = 1.0 \times 10^{-11} \text{ mol} \]Calculate the mass of KBr using its molar mass (120 g/mol):\[ \text{Mass of KBr} = 1.0 \times 10^{-11} \text{ mol} \times 120 \frac{\text{g}}{\text{mol}} = 1.2 \times 10^{-9} \text{ g} \]

Step 5: Choose the Correct Option

Upon calculating, the required mass of KBr is \(1.2 \times 10^{-9} \text{ g}\), matching option (b).

Key Concepts

Precipitation

Precipitation is a process in which solute particles come out of a solution to form a solid. In chemistry, it's like having a separated phase appear. This usually occurs when the amount of solute exceeds its solubility limit in a particular solvent under given conditions.
To understand when and why precipitation occurs, it is important to know about the solubility product constant. But before diving into that, consider this: Precipitation happens when ions in a solution combine to form an insoluble compound. This can be likened to when sugar cubes don't dissolve in a filled cup of coffee any further.

• It's essential for many laboratory techniques and environmental processes.
• Precipitation is influenced by factors like temperature and the concentration of ions in solution.

Ksp

The solubility product constant, or Ksp, is a numerical value that indicates the solubility of a compound in water. It is specific to each compound and refers to the equilibrium constant for its dissolution.
Simply put, if you have a substance that hardly dissolves in water, its Ksp will be quite small. This constant provides information required to predict whether a precipitate will form in a given solution.
The Ksp expression for a salt like silver bromide (AgBr) can be written as:

• When AgBr dissolves, it releases Ag⁺ and Br^-.
• Ksp equation is \[ K_{sp} = [Ag^+] [Br^-] \].

Comparing the product of the ionic concentrations, often called the ion product, to the Ksp value helps determine whether a precipitate will form. If the ion product exceeds the Ksp, precipitation occurs.

AgBr precipitation

Silver bromide (AgBr) is a compound known for its low solubility in water. Due to its low Ksp value, AgBr needs careful conditions for it to dissolve or precipitate.
AgBr's precipitation from an aqueous solution is a great representation of how solubility equilibrium works.

• It is commonly studied in chemistry to understand concepts of solubility and precipitation.
• The precipitation process involves mixing ions until the product of their concentrations matches or exceeds the solubility product.

The equation for the precipitation of AgBr is:\[\text{AgBr}_{(s)} \rightleftharpoons \text{Ag}^+ + \text{Br}^-\]To find out when AgBr starts to precipitate, calculations involving the equilibrium expression and the ion concentrations are needed. The equilibrium point depends greatly on these calculations.

Bromide ion concentration

The bromide ion concentration in a solution determines whether particular reactions or precipitates involving bromide will occur. To start precipitation of AgBr, the concentration of bromide ions must be in a delicate balance.
In a solution where silver ions are in excess, introducing bromide ions will eventually lead to saturation, at which point precipitation begins.
Here's how to calculate what bromide ion concentration is necessary for AgBr to start precipitating:

• Given the Ksp for AgBr, determine necessary [Br^-] that ensures the product of [Ag⁺] and [Br^-] matches Ksp.
• Use the equation \[ K_{sp} = [Ag^+] \times [Br^-] \] to find the precise concentration of Br^- required.

When concentration conditions are met, adjusting further will push the system beyond equilibrium, causing solid to form as a precipitate.