Problem 87 - A Complete Resource Book in Chemistry for JEE Main

Question

Equal volumes of \(0.1 \mathrm{M} \mathrm{AgNO}_{3}\) and \(0.2 \mathrm{M} \mathrm{NaCl}\) are mixed. The concentration of nitrate ions in the mixture will be (a) \(0.1 \mathrm{M}\) (b) \(0.05 \mathrm{M}\) (c) \(0.2 \mathrm{M}\) (d) \(0.15 \mathrm{M}\)

Step-by-Step Solution

Answer

(b) 0.05 M

Step 1: Understand the Reaction

When equal volumes of \(0.1 \text{ M} \text{ AgNO}_{3}\) and \(0.2 \text{ M} \text{ NaCl}\) are mixed, \(\text{AgNO}_{3}\) and \(\text{NaCl}\) react to form \(\text{AgCl}\) and \(\text{NaNO}_{3}\). This is a precipitation reaction where \(\text{AgCl}\) forms as a solid precipitate.

Step 2: Initial Concentrations

Initially, both solutions are equal in volume, say \(V\) each. Therefore, the initial moles before mixing for \(\text{AgNO}_{3}\) is \(0.1 \times V\) and for \(\text{NaCl}\) is \(0.2 \times V\).

Step 3: Concentration After Mixing

Upon mixing, the total volume becomes \(2V\). Since \(\text{AgNO}_{3}\) and \(\text{NaCl}\) react in a 1:1 ratio, all \(\text{AgNO}_{3}\) will be consumed, becoming the limiting reactant.

Step 4: Nitrate Ion Concentration Calculation

Each mole of \(\text{AgNO}_{3}\) produces one mole of nitrate ions \((\text{NO}_3^-)\). With \(0.1 \times V\) moles of \(\text{AgNO}_{3}\), the moles of \(\text{NO}_3^-\) produced are the same. Thus, the concentration of \(\text{NO}_3^-\) ions in the final mixture is \(\frac{0.1V}{2V} = 0.05 \text{ M}\).

Key Concepts

Precipitation Reaction

A precipitation reaction is a chemical process wherein two solutions react to form a solid, called a precipitate. When mixing aqueous solutions of chemicals, certain combinations result in substances that are insoluble in water. This means that they form a distinct solid that settles out of the solution.
In our exercise, silver nitrate (\(\text{AgNO}_3\)) reacts with sodium chloride (\(\text{NaCl}\)) to generate silver chloride (\(\text{AgCl}\)) as the precipitate, while sodium nitrate (\(\text{NaNO}_3\)) remains dissolved in the solution.

• The equation for this reaction can be represented as \(\text{AgNO}_3 (aq) + \text{NaCl} (aq) \rightarrow \text{AgCl} (s) + \text{NaNO}_3 (aq)\).
• Since \(\text{AgCl}\) doesn't dissolve in water, it appears as a cloudy solid, indicating the occurrence of a precipitation reaction.

Recognizing when a precipitation occurs is crucial in experiments, especially when tracking ion concentrations.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is entirely consumed first, stopping the reaction from continuing further. This concept is crucial when calculating the amounts of products formed or leftover reactants.
Here, \(\text{AgNO}_3\) and \(\text{NaCl}\) mix at concentrations of \(0.1 \text{ M}\) and \(0.2 \text{ M}\) respectively.

• Both react in a molar ratio of 1:1, meaning they combine equally until one is used up.
• Given equal volumes, with \(0.1 \text{ M} \text{ AgNO}_3\) and \(0.2 \text{ M} \text{ NaCl}\), \(\text{AgNO}_3\) becomes the limiting reactant.

Since \(\text{AgNO}_3\) runs out first, we use it to calculate the concentration of products and leftover ions.

Molar Concentration

Molar concentration, often called molarity, is a measure of the number of moles of a solute per liter of solution, expressed in molarity (\(\text{M}\)). It provides important information about how concentrated a solution is.
To compute molarity after mixing:

• Combine volumes of the two initial solutions, doubling the volume and halving the concentration in this scenario.
• Focus on nitrate ions (\(\text{NO}_3^-\)), which originate from \(\text{AgNO}_3\).

The initial moles of \(\text{NO}_3^-\) matches those from \(\text{AgNO}_3\), giving us a final molarity of \(\frac{0.1V}{2V} = 0.05 \text{ M}\).
This simplified calculation showcases the importance of understanding molarity in solution chemistry.

Stoichiometry

Stoichiometry refers to balancing chemical equations and relates to measuring substance quantities in a chemical reaction. It determines how reactants transform into products using mole-to-mole relationships from the balanced equation.
For the reaction \(\text{AgNO}_3 (aq) + \text{NaCl} (aq) \rightarrow \text{AgCl} (s) + \text{NaNO}_3 (aq)\):

• The stoichiometric coefficients indicate that one mole of \(\text{AgNO}_3\) reacts with one mole of \(\text{NaCl}\) to produce one mole of \(\text{AgCl}\) and one mole of \(\text{NaNO}_3\).
• This 1:1:1:1 ratio controls how much of each substance reacts or forms.

Using stoichiometry, we determine the limiting reactant, product quantities, and ion concentrations, making it a vital concept in predicting chemical behavior.