Problem 184 - A Complete Resource Book in Chemistry for JEE Main

Question

When equal volume of the following solutions are mixed, which of the following gives maximum precipitate? \(\left(\mathrm{K}_{s p}\right.\) of \(\left.\mathrm{AgCl}=10^{-12}\right)\) (a) \(10^{-4} \mathrm{M} \mathrm{Ag}^{+}\)and \(10^{-4} \mathrm{M} \mathrm{Cl}^{-}\) (b) \(10^{-3} \mathrm{M} \mathrm{Ag}^{+}\)and \(10^{-3} \mathrm{M} \mathrm{Cl}^{-}\) (c) \(10^{-5} \mathrm{M} \mathrm{Ag}^{+}\)and \(10^{-5} \mathrm{M} \mathrm{Cl}^{-}\) (d) \(10^{-6} \mathrm{M} \mathrm{Ag}^{+}\)and \(10^{-6} \mathrm{M} \mathrm{Cl}^{-}\)

Step-by-Step Solution

Answer

(b) gives maximum precipitate due to the highest ion product that exceeds \( K_{sp} \).

Step 1: Understanding the Problem

To determine which pair of solutions gives the maximum precipitate, we need to calculate the product of the concentrations of \( \text{Ag}^+ \) and \( \text{Cl}^- \) ions when mixed. Compare these products to the solubility product \( \text{K}_{sp} \) of \( \text{AgCl} \). A higher product than \( \text{K}_{sp} \) indicates more precipitation.

Step 2: Calculate Ion Concentrations After Mixing

Since equal volumes of the solutions are mixed, the concentrations of \( \text{Ag}^+ \) and \( \text{Cl}^- \) ions become half in each scenario:- For (a), concentrations become \( \frac{10^{-4}}{2} = 5 \times 10^{-5} \) M.- For (b), concentrations become \( \frac{10^{-3}}{2} = 5 \times 10^{-4} \) M.- For (c), concentrations become \( \frac{10^{-5}}{2} = 5 \times 10^{-6} \) M.- For (d), concentrations become \( \frac{10^{-6}}{2} = 5 \times 10^{-7} \) M.

Step 3: Calculate the Ion Product for Each Scenario

Multiply the concentrations of \( \text{Ag}^+ \) and \( \text{Cl}^- \) for each case:- For (a), \( (5 \times 10^{-5}) \times (5 \times 10^{-5}) = 2.5 \times 10^{-9} \).- For (b), \( (5 \times 10^{-4}) \times (5 \times 10^{-4}) = 2.5 \times 10^{-7} \).- For (c), \( (5 \times 10^{-6}) \times (5 \times 10^{-6}) = 2.5 \times 10^{-11} \).- For (d), \( (5 \times 10^{-7}) \times (5 \times 10^{-7}) = 2.5 \times 10^{-13} \).

Step 4: Compare Ion Products with Ksp

Compare each ion product with the solubility product \( \text{K}_{sp} = 10^{-12} \):- In (a), \( 2.5 \times 10^{-9} > 10^{-12} \), indicating significant precipitation.- In (b), \( 2.5 \times 10^{-7} > 10^{-12} \), indicating even more precipitation.- In (c), \( 2.5 \times 10^{-11} > 10^{-12} \), indicating some precipitation.- In (d), \( 2.5 \times 10^{-13} < 10^{-12} \), indicating no precipitation.

Step 5: Conclusion on Maximum Precipitation

Since the ion product for scenario (b) is the highest and exceeds \( 10^{-12} \) by the greatest amount, it results in the maximum precipitate.

Key Concepts

Precipitation

Precipitation is the process where ions in solution form an insoluble substance known as a precipitate. When two ionic solutions are mixed, the ions may combine to produce a solid, which will no longer dissolve in the solution. The amount of precipitate formed depends on several factors such as concentration and the nature of the solutes.
The onset of precipitation is primarily dictated by the solubility product constant ( K_{sp} ). If the ion product of the reactants in a given solution exceeds the K_{sp} of the potential compound, precipitation occurs. This happens because the solution can no longer hold the amount of solute in dissolved form beyond a certain limit.
Students should understand that the higher the ion product in relation to the solubility product, the greater the extent of precipitation. In experiments, observing a solution turning cloudy often indicates precipitation is occurring.

Ion Product

The ion product is crucial in determining whether precipitation will occur when solutions are mixed. It is the product of the concentrations of the reacting ions, which are typically calculated after the solutions are mixed and dilutions are considered.
For example, if \[c(\text{Ag}^+) = 10^{-3} M\]and \[c(\text{Cl}^-) = 10^{-3} M\]are mixed in equal volumes, when reacting, each concentration reduces by half. Thus, the ion product becomes \((5 \times 10^{-4}) \times (5 \times 10^{-4}) = 2.5 \times 10^{-7}\). This value is used to compare against the compound’s K_{sp} to predict precipitation.

• If the ion product is greater than the K_{sp}, precipitation occurs.
• If the ion product is equal to the K_{sp}, the solution is saturated and equilibrium is achieved.
• If the ion product is less than the K_{sp}, no precipitation occurs, and the solution is unsaturated.

Understanding and calculating the ion product helps students predict the extent of reaction and formation of the solid phase.

AgCl Solubility

Silver chloride (AgCl) is an example of a salt with low solubility in water. Its solubility product constant ( K_{sp} ) helps chemists and students predict how likely AgCl will form a precipitate in solution.
The solubility product ( K_{sp} = 10^{-12} ) indicates that any ion product exceeding this threshold results in AgCl precipitating out of solution. Generally, AgCl solubility demonstrates the principle that less soluble salts precipitate more easily and requires careful calculation of conditions under which it decides to precipitate from solution.
This property makes AgCl a useful compound in qualitative analysis and inorganic chemistry, where selective precipitation is employed, for instance, in identifying the presence of chloride ions in a solution by the formation of a characteristic white precipitate when mixed with Ag^+ ions.