Problem 173 - A Complete Resource Book in Chemistry for JEE Main

Question

Solid \(\mathrm{AgNO}_{3}\) is added slowly to a buffer solution of \(\mathrm{pH}\) \(=10\) to precipitate \(\mathrm{AgOH}\). The \(\left[\mathrm{Ag}^{+}\right]\)concentration in the solution is \(\left[\mathrm{K}_{\mathrm{sp}}(\mathrm{AgOH})=10^{-10}\right]\) (a) \(10^{-4} \mathrm{M}\) (b) \(10^{-5} \mathrm{M}\) (c) \(10^{-6} \mathrm{M}\) (d) \(10^{-7} \mathrm{M}\)

Step-by-Step Solution

Answer

The concentration of \(Ag^+\) ions is \(10^{-6} \, \text{M}\) (option c).

Step 1: Understand the Reaction

Silver hydroxide (2;2;\( ext{AgOH}\)) can precipitate in this buffer solution. Its solubility product constant is given as \(K_{sp}(AgOH) = 10^{-10}\). This means that \(Ag^+\) ions react with \(OH^-\) ions to form the equilibrium: \[ Ag^+ + OH^- \rightleftharpoons AgOH(s) \] In this equilibrium, the solubility product is expressed as \(K_{sp} = [Ag^+][OH^-] = 10^{-10}\).

Step 2: Calculate Hydroxide Ion Concentration

We need to calculate the concentration of hydroxide ions \([OH^-]\) in the solution, given that the \(pH = 10\). From the relationship \(pH + pOH = 14\), we can determine \(pOH = 14 - 10 = 4\). Hence, \([OH^-]\) is given by \([OH^-] = 10^{-pOH} = 10^{-4}\,\text{M}\).

Step 3: Solve for Silver Ion Concentration

Substitute the \([OH^-]\) concentration into the \(K_{sp}\) expression to find \([Ag^+]\). With the equation \(K_{sp} = [Ag^+][OH^-]\), and \([OH^-]=10^{-4}\), we have: \[ 10^{-10} = [Ag^+](10^{-4}) \] Solving for \([Ag^+]\), we find: \[ [Ag^+] = \frac{10^{-10}}{10^{-4}} = 10^{-6} \, \text{M} \]

Step 4: Conclusion

The concentration of \(Ag^+\) ions in equilibrium with \(AgOH\) in this buffer solution is \(10^{-6} \, \text{M}\). Therefore, the correct answer is option (c) \(10^{-6} \, \text{M}\).

Key Concepts

Buffer Solution

A buffer solution is a special type of solution that resists changes in its pH when small amounts of an acid or a base are added to it. This property makes buffer solutions very useful in many chemical and biological applications.
A classic buffer solution is made by mixing a weak acid and its conjugate base or a weak base and its conjugate acid. This unique combination helps the solution maintain a relatively stable pH.

• Bicarbonate in blood is a natural buffer that keeps our pH level stable.
• In our exercise, the buffer solution is set at a pH of 10. This means it can absorb excess hydrogen ions (H⁺) or hydroxide ions (OH^-), keeping the pH within the desired range.
• When solid AgNO₃ is added, the buffer works to maintain the pH at 10 by managing the concentrations of ions like OH^-, crucial for calculating the silver ion concentration later.

Understanding buffer solutions is critical in chemistry, as they play vital roles in laboratory settings, industrial processes, and even within living organisms.

pH Calculation

Calculating pH is a fundamental task in chemistry. It involves determining how acidic or basic a solution is. The scale runs from 0 to 14, where 7 is neutral, values below 7 indicate acidity, and values above 7 indicate basicity.
In the context of our buffer solution:

• We are given a pH of 10, indicating a basic solution. From there, the pOH can be calculated using the relationship: \( pH + pOH = 14 \).
• In our example, with a pH of 10, you calculate the pOH as follows: \( pOH = 14 - 10 = 4 \).
• This calculation provides the concentration of hydroxide ions \([OH^-]\) in the solution: \( [OH^-] = 10^{-pOH} = 10^{-4} \, \text{M} \).

Having accurate pH and pOH values is essential in applications ranging from drug formulation to wastewater treatment, ensuring processes are effective and products are safe for use.

Equilibrium Chemistry

Equilibrium chemistry is a fascinating area of study, crucial for understanding how different chemical species interact in solution. It describes a state where the forward and reverse reactions occur at the same rate, maintaining stable concentrations of products and reactants over time.
In our exercise, equilibrium is represented in terms of the precipitation of AgOH:

• The reaction involves silver ions \(Ag^+\) and hydroxide ions \(OH^-\) forming solid silver hydroxide \(AgOH\).
• The solubility product constant \( K_{sp} \) gives the equilibrium expression: \( K_{sp} = [Ag^+][OH^-] \).
• Substituting with the known \([OH^-]\) concentration, the concentration of \([Ag^+]\) is calculated to ensure precipitation occurs when the solution is oversaturated with these ions.

This scenario exemplifies why understanding equilibrium is vital. It allows chemists to predict the directions of reactions, control reactions in industrial processes, and even explain natural phenomena, making it an indispensable tool in the chemist's toolkit.