Problem 165 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the minimum \(\mathrm{pH}\) required to prevent the precipitation of \(\mathrm{ZnS}\) in a solution that is \(0.01 \mathrm{M} \mathrm{ZnCl}_{2}\) and saturated with \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\) ? [Given: \(\mathrm{K}_{\mathrm{sp}}\) of \(\mathrm{ZnS}=10^{-21}, \mathrm{~K}_{\mathrm{a}_{1}} \times \mathrm{K}_{\mathrm{a}_{2}}\) of \(\left.\mathrm{H}_{2} \mathrm{~S}=10^{-20}\right]\) (a) 4 (b) 3 (c) 2 (d) 1

Step-by-Step Solution

Answer

The minimum pH required is 1 (option d).

Step 1: Write the relevant chemical equations

The solubility product expression for the precipitation of ZnS is given by:\[ \text{Zn}^{2+} + \text{S}^{2-} \rightleftharpoons \text{ZnS} \]\[ K_{sp} = [\text{Zn}^{2+}][\text{S}^{2-}] = 10^{-21} \]The dissociation of \( \text{H}_2\text{S} \) in water can be expressed as:\[ \text{H}_2\text{S} \rightleftharpoons \text{HS}^- + \text{H}^+ \rightleftharpoons \text{S}^{2-} + 2\text{H}^+ \]\[ K_{a1} \times K_{a2} = \frac{[\text{S}^{2-}][\text{H}^+]^2}{[\text{H}_2\text{S}]} = 10^{-20} \]

Step 2: Relate given concentrations to the equilibrium expressions

In the solution, the concentration of Zn\(^{2+}\) is \(0.01 \text{ M} \), and we need to find the concentration of \( \text{S}^{2-} \) to use in the \( K_{sp} \) expression.Assuming \( [\text{S}^{2-}] = x \), the \[ K_{sp} = 10^{-21} = (0.01)(x) \]\[ x = \frac{10^{-21}}{0.01} = 10^{-19} \]So, \([\text{S}^{2-}] = 10^{-19} \text{ M} \).

Step 3: Use the equilibrium equation for sulfide ion

We need to use the equation for the dissociation of \( \text{H}_2\text{S} \) to relate \( [\text{S}^{2-}] \) and the \( \text{H}^+ \) concentration:\[ \frac{[\text{S}^{2-}][\text{H}^+]^2}{[\text{H}_2\text{S}]} = 10^{-20} \]Plug the values we have:\[ \frac{10^{-19}([H^+])^2}{0.10} = 10^{-20} \]\[ 10^{-19}[H^+]^2 = 10^{-21} \]\[ [H^+]^2 = 10^{-2} \]\[ [H^+] = 10^{-1} \]

Step 4: Convert [H+] to pH

The pH is related to \([\text{H}^+]\) by the equation:\[ \text{pH} = -\log([\text{H}^+]) \]Given that \([\text{H}^+] = 10^{-1}\), \[ \text{pH} = -\log(10^{-1}) = 1 \]

Step 5: Determine the correct answer

From the calculation, the minimum pH required to prevent precipitation of ZnS is \(1\). Therefore, the correct answer is option (d).

Key Concepts

Solubility Product

In chemistry, solubility product (K_{sp}) is a term that describes the degree to which a compound can dissolve in water to form ions. It is specifically used for sparingly soluble salts.
In our previous problem, we were examining a compound known as Zinc Sulfide (ZnS). This compound splits into Zn^{2+} and S^{2-} ions. The solubility product can be expressed as:\[ K_{sp} = [ ext{Zn}^{2+}][ ext{S}^{2-}] \]For ZnS, the solution has been given a K_{sp} value of 10^{-21}. This means that in equilibrium, the product of the concentrations of the ions produced by ZnS dissolving is 10^{-21}.
Understanding the solubility product is key because it allows us to predict whether a precipitate will form in a reaction. If the ion product of the concentrations exceeds K_{sp}, precipitation occurs. Thus, in the given task, the calculation helps determine conditions to keep ZnS from precipitating by managing S^{2-} ion concentration.

Dissociation Constant

When dealing with acids and bases, the dissociation constant is a critical component. It measures how completely an acid or base ionizes in a solution. The lower the K_a value, the weaker the acid, signifying less dissociation. In the context of our exercise, we looked at the dissociation of H_2S into ions:
\[ ext{H}_2 ext{S} \rightleftharpoons ext{HS}^- + ext{H}^+ \rightleftharpoons ext{S}^{2-} + 2 ext{H}^+ \]
Here, the dissociation constants K_{a1} and K_{a2} are crucial. For this problem, we were given K_{a1} \times K_{a2} = 10^{-20}. This reflects how the sequence of dissociation steps share equilibrium. The given value is vital for calculating the concentration of sulfide ions (S^{2-}), and ultimately the H^+ concentration via the dissociation in the solution.

Equilibrium Expression

Equilibrium expressions link the concentrations of reactants and products when a reaction is at equilibrium. For any chemical equation, this expression helps calculate unknown concentrations based on known values.

In the exercise, two equilibrium expressions played an important role:

• The equilibrium expression for the solubility product: \[ K_{sp} = [ ext{Zn}^{2+}][ ext{S}^{2-}] \]
• The equilibrium for the dissociation of H_2S: \[ rac{[ ext{S}^{2-}][ ext{H}^+]^2}{[ ext{H}_2 ext{S}]} = 10^{-20} \]

By inserting the given concentrations and relating them to the equilibrium expressions, we determined the minimum ext{pH}required to keep ZnS from precipitating.
Understanding equilibrium expressions enables us to balance and predict concentrations, crucial for controlling chemical reactions, as seen in the initial problem.